\(y=\frac{uv}{w}\)
\(\Rightarrow \ln{y}=\ln{\frac{uv}{w}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{u}+\ln{v}-\ln{w}\) ➜ Note \(\because \ln{\frac{xy}{z}}=\ln{x}+\ln{y}-\ln{z}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{u}+\ln{v}-\ln{w})\) ➜ Note \(x\) এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(\ln{u})+\frac{d}{dx}(\ln{v})-\frac{d}{dx}(\ln{w})\)
\(\Rightarrow \frac{dy}{dx}=y\{\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}-\frac{1}{w}\frac{dw}{dx}\}\)
\(\therefore \frac{dy}{dx}=\frac{uv}{w}\left(\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}-\frac{1}{w}\frac{dw}{dx}\right)\) | Note\(\because y=\frac{uv}{w}\)
(proved)

\(y=u^{v}\)
\(\Rightarrow \ln{y}=\ln{u^{v}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=v\ln{u}\) ➜ Note\(\because \ln{x^n}=n\ln{x}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})= \frac{d}{dx}(v\ln{u})\) ➜ Note\(x\) এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}= v\frac{d}{dx}(\ln{u})+\ln{u}\frac{dv}{dx}\) ➜ Note\(\because \frac{d}{dx}(\ln{z})=\frac{1}{z}\frac{dz}{dx}, \ \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{dx}= y\left(v\frac{1}{u}\frac{du}{dx}+\ln{u}\frac{dv}{dx}\right)\)
\(\therefore \frac{dy}{dx}= u^{v}\left(\frac{v}{u}\frac{du}{dx}+\ln{u}\frac{dv}{dx}\right)\) ➜ Note\(\because y=u^{v}\)
(proved)

ধরি,
\(y=(1-x)(1-2x)(1-3x)(1-4x)\)
\(\Rightarrow \ln{y}=\ln{(1-x)(1-2x)(1-3x)(1-4x)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{(1-x)}+\ln{(1-2x)}+\ln{(1-3x)}+\ln{(1-4x)}\) ➜ Note\(\because \ln{xyz}=\ln{x}+\ln{y}+\ln{z}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{\ln{(1-x)}+\ln{(1-2x)}+\ln{(1-3x)}+\ln{(1-4x)}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{1-x}\frac{d}{dx}(1-x)+\frac{1}{1-2x}\frac{d}{dx}(1-2x)+\frac{1}{1-3x}\frac{d}{dx}(1-3x)+\frac{1}{1-4x}\frac{d}{dx}(1-4x)\) ➜ Note \(y, \ (1-x), \ (1-2x), \ (1-3x), \ 1-4x\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{1-x}(0-1)+\frac{1}{1-2x}(0-2)+\frac{1}{1-3x}(0-3)+\frac{1}{1-4x}(0-4)\) ➜ Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=y\left(-\frac{1}{1-x}-2\frac{1}{1-2x}-3\frac{1}{1-3x}-4\frac{1}{1-4x}\right)\)
\(\Rightarrow \frac{dy}{dx}=-y\left(\frac{1}{1-x}+\frac{2}{1-2x}+\frac{3}{1-3x}+\frac{4}{1-4x}\right)\)
\(\therefore \frac{dy}{dx}=-(1-x)(1-2x)(1-3x)(1-4x)\left(\frac{1}{1-x}+\frac{2}{1-2x}+\frac{3}{1-3x}+\frac{4}{1-4x}\right)\) ➜ Note \(\because y=(1-x)(1-2x)(1-3x)(1-4x)\)

ধরি,
\(y=x^{\sin{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{\sin{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\sin{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\sin{x}\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\sin{x}\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(\sin{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\sin{x}\frac{1}{x}+\ln{x}\cos{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \ \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\sin{x}\frac{1}{x}+\ln{x}\cos{x}\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\sin{x}}\left(\frac{\sin{x}}{x}+\ln{x}\cos{x}\right)\) ➜ Note \(\because y=x^{\sin{x}}\)

ধরি,
\(y=x^{\sin{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{\sin{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{\sin{x}}\left(\frac{\sin{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\sin{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\sin{x}}\left(\frac{\sin{x}}{x}.1+\ln{x}\cos{x}\right)\) ➜ Note \(\because \frac{d}{dx}(x)=1, \ \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dy}{dx}=x^{\sin{x}}\left(\frac{\sin{x}}{x}+\ln{x}\cos{x}\right)\)

ধরি,
\(y=(\sin^{-1}{x})^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sin^{-1}{x})^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{\sin^{-1}{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x}\ln{\sin^{-1}{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{d}{dx}(\ln{\sin^{-1}{x}})+\ln{\sin^{-1}{x}}\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{\sin^{-1}{x}}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}+\ln{\sin^{-1}{x}}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{\ln{\sin^{-1}{x}}}{x}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{\ln{\sin^{-1}{x}}}{x}+\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{\sin^{-1}{x}}}{x}+\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}\right)\) ➜ Note \(\because y=(\sin^{-1}{x})^{\ln{x}}\)

ধরি,
\(y=(\sin^{-1}{x})^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}((\sin^{-1}{x})^{\ln{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{\sin^{-1}{x}}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}+\ln{\sin^{-1}{x}}\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \ \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{\sin^{-1}{x}}}{x}+\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}\right)\)

ধরি,
\(y=x^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=(\ln{x})^{2}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x})^{2}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\ln{x}\frac{d}{dx}(\ln{x})\) ➜ Note \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\ln{x}\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=y\frac{2\ln{x}}{x}\)
\(\therefore \frac{dy}{dx}=x^{\ln{x}}\frac{2\ln{x}}{x}\) ➜ Note \(\because y=x^{\ln{x}}\)

ধরি,
\(y=x^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{\ln{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{\ln{x}}\left(\frac{\ln{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\ln{x}}\left(\frac{\ln{x}}{x}.1+\ln{x}\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(x)=x, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=x^{\ln{x}}\left(\frac{\ln{x}}{x}+\frac{\ln{x}}{x}\right)\)
\(\therefore \frac{dy}{dx}=x^{\ln{x}}\frac{2\ln{x}}{x}\)

ধরি,
\(y=x^{\sin^{-1}{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{\sin^{-1}{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\sin^{-1}{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\sin^{-1}{x}\ln{x}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\sin^{-1}{x}\frac{d}{dx}(\ln{x}+\ln{x}\frac{d}{dx}(\sin^{-1}{x}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\sin^{-1}{x}\frac{1}{x}+\ln{x}\times{\frac{1}{\sqrt{1-x^2}}}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}(v), \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{\sin^{-1}{x}}{x}+\frac{\ln{x}}{\sqrt{1-x^2}}\right)\)
\(\therefore \frac{dy}{dx}=x^{\sin^{-1}{x}}\left(\frac{\sin^{-1}{x}}{x}+\frac{\ln{x}}{\sqrt{1-x^2}}\right)\) ➜ Note \(\because y=x^{\sin^{-1}{x}}\)

ধরি,
\(y=x^{\sin^{-1}{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{\sin^{-1}{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{\sin^{-1}{x}}\left(\frac{\sin^{-1}{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\sin^{-1}{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\sin^{-1}{x}}\left(\frac{\sin^{-1}{x}}{x}.1+\ln{x}\times{\frac{1}{\sqrt{1-x^2}}}\right)\) ➜ Note \(\because \frac{d}{dx}(x)=x, \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{dy}{dx}=x^{\sin^{-1}{x}}\left(\frac{\sin^{-1}{x}}{x}+\frac{\ln{x}}{\sqrt{1-x^2}}\right)\)

ধরি,
\(y=x^{x^{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{x^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x^{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{\ln{y}}=\ln{x^{x}\ln{x}}\) ➜ Note আবার, উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{\ln{y}}=\ln{x^{x}}+\ln{\ln{x}}\) ➜ Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow \ln{\ln{y}}=x\ln{x}+\ln{\ln{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{\ln{y}})=\frac{d}{dx}(x\ln{x}+\ln{\ln{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x\ln{x})+\frac{d}{dx}(\ln{\ln{x}})\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)+\frac{d}{dx}(\ln{\ln{x}})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\ln{x}.1+\frac{1}{\ln{x}}\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=1+\ln{x}+\frac{1}{\ln{x}}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=y\ln{y}\left(1+\ln{x}+\frac{1}{x\ln{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x^{x}}x^{x}\ln{x}\left(1+\ln{x}+\frac{1}{x\ln{x}}\right)\) ➜ Note \(\because y=x^{x^{x}}, \ln{y}=x^{x}\ln{x}\)
\(\Rightarrow \frac{dy}{dx}=x^{x^{x}}x^{x}\{\ln{x}+(\ln{x})^2+\frac{1}{x}\}\)
\(\therefore \frac{dy}{dx}=x^{x^{x}}x^{x}\{\frac{1}{x}+\ln{x}(1+\ln{x})\}\)

ধরি,
\(y=x^{x^{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{x^{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{x^{x}}\left(\frac{x^{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(x^{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x^{x}}\{\frac{x^{x}}{x}.1+\ln{x}x^{x}\left(\frac{x}{x}\frac{d}{dx}x+\ln{x}\frac{d}{dx}x\right)\}\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x^{x}}\{\frac{x^{x}}{x}+\ln{x}x^{x}(1.1+\ln{x}.1)\}\) ➜ Note \(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=x^{x^{x}}x^{x}\{\frac{1}{x}+\ln{x}(1+\ln{x})\}\)

ধরি,
\(y=x^{x}\)
\(\Rightarrow \ln{y}=\ln{x^{x}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(x\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\ln{x}.1\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=1+\ln{x}\)
\(\Rightarrow \frac{dy}{dx}=y(1+\ln{x})\)
\(\therefore \frac{dy}{dx}=x^{x}(1+\ln{x})\)

ধরি,
\(y=x^{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{x}\left(\frac{x}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x}(1.1+\ln{x}.1)\) ➜ Note \(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=x^{x}(1+\ln{x})\)

ধরি,
\(y=(\sin{x})^{\cos{x}}+(\cos{x})^{\sin{x}}\)
\(\Rightarrow y=y_{1}+y_{2}\) যেখানে \(y_{1}=(\sin{x})^{\cos{x}}, y_{2}=(\cos{x})^{\sin{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy_{1}}{dx}+\frac{dy_{2}}{dx}y_{2} ....(1)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
এখন,
\(y_{1}=(\sin{x})^{\cos{x}}\)
\(\Rightarrow \ln{y_{1}}=(\sin{x})^{\cos{x}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y_{1}}=\cos{x}\ln{(\sin{x})}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y_{1}})=\frac{d}{dx}\{\cos{x}\ln{(\sin{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y_{1}}\frac{dy_{1}}{dx}=\cos{x}\frac{d}{dx}\{\ln{(\sin{x})}\}+\ln{(\sin{x})}\frac{d}{dx}(\cos{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y_{1}}\frac{dy_{1}}{dx}=\cos{x}\frac{1}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{(\sin{x})}\times{-\sin{x}}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{1}{y_{1}}\frac{dy_{1}}{dx}=\frac{\cos{x}}{\sin{x}}\cos{x}-\sin{x}\ln{(\sin{x})}\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{1}{y_{1}}\frac{dy_{1}}{dx}=\cot{x}\cos{x}-\sin{x}\ln{(\sin{x})}\) ➜ Note \(\because \frac{\cos{x}}{\sin{x}}=\cot{x}\)
\(\Rightarrow \frac{dy_{1}}{dx}=y_{1}(\cot{x}\cos{x}-\sin{x}\ln{(\sin{x})})\)
\(\therefore \frac{dy_{1}}{dx}=(\sin{x})^{\cos{x}}\{\cot{x}\cos{x}-\sin{x}\ln{(\sin{x})}\}\) ➜ Note \(\because y_{1}=(\sin{x})^{\cos{x}}\)
আবার,
\(y_{2}=(\cos{x})^{\sin{x}}\)
\(\Rightarrow \ln{y_{2}}=(\cos{x})^{\sin{x}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y_{2}}=\sin{x}\ln{(\cos{x})}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y_{2}})=\frac{d}{dx}\{\sin{x}\ln{(\cos{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y_{2}}\frac{dy_{2}}{dx}=\sin{x}\frac{d}{dx}\{\ln{(\cos{x})}\}+\ln{(\cos{x})}\frac{d}{dx}(\sin{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y_{2}}\frac{dy_{2}}{dx}=\sin{x}\frac{1}{\cos{x}}\frac{d}{dx}(\cos{x})+\ln{(\cos{x})}\cos{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{1}{y_{2}}\frac{dy_{2}}{dx}=\frac{\sin{x}}{\cos{x}}\times{-\sin{x}}+\cos{x}\ln{(\cos{x})}\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{1}{y_{2}}\frac{dy_{2}}{dx}=-\sin{x}\tan{x}+\cos{x}\ln{(\cos{x})}\) ➜ Note \(\because \frac{\sin{x}}{\cos{x}}=\tan{x}\)
\(\Rightarrow \frac{dy_{2}}{dx}=y_{2}(-\sin{x}\tan{x}+\cos{x}\ln{(\cos{x})}\) \(\therefore \frac{dy_{2}}{dx}=(\cos{x})^{\sin{x}}(-\sin{x}\tan{x}+\cos{x}\ln{(\cos{x})})\) ➜ Note \(\because y_{2}=(\cos{x})^{\sin{x}}\)
\((1)\) সমীকরণে \(\frac{dy_{1}}{dx}\) ও \(\frac{dy_{2}}{dx}\) এর মাণ বসিয়ে।
\(\therefore \frac{dy}{dx}=(\sin{x})^{\cos{x}}\{\cot{x}\cos{x}-\sin{x}\ln{(\sin{x})}\}\)\(+(\cos{x})^{\sin{x}}\{-\sin{x}\tan{x}+\cos{x}\ln{(\cos{x})}\}\)

ধরি,
\(y=(\sin{x})^{\cos{x}}+(\cos{x})^{\sin{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin{x})^{\cos{x}}+(\cos{x})^{\sin{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin{x})^{\cos{x}}\}+\frac{d}{dx}\{(\cos{x})^{\sin{x}}\}\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\frac{\cos{x}}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{\sin{x}}\frac{d}{dx}(\cos{x})\right)\)\(+(\cos{x})^{\sin{x}}\left(\frac{\sin{x}}{\cos{x}}\frac{d}{dx}(\cos{x})+\ln{\cos{x}}\frac{d}{dx}(\sin{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\cot{x}\cos{x}+\ln{\sin{x}}\times{-\sin{x}}\right)\)\(+(\cos{x})^{\sin{x}}\left(-\tan{x}\sin{x}+\cos{x}\ln{\cos{x}}\right)\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)\(\frac{\cos{x}}{\sin{x}}=\cot{x}, \frac{\sin{x}}{\cos{x}}=\tan{x}\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\cot{x}\cos{x}-\sin{x}\ln{\sin{x}}\right)\)\(+(\cos{x})^{\sin{x}}\left(-\tan{x}\sin{x}+\cos{x}\ln{\cos{x}}\right)\)

ধরি,
\(y=\frac{x\sin{x}}{1+\cos{x}}\)
\(\Rightarrow \ln{y}=\ln{(\frac{x\sin{x}}{1+\cos{x}})}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}+\ln{\sin{x}}-\ln{(1+\cos{x})}\) ➜ Note\(\because \ln{\frac{MN}{P}}=\ln{(M)}+\ln{(N)}-\ln{P}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x}+\ln{\sin{x}}-\ln{(1+\cos{x})})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(\ln{x})+\frac{d}{dx}(\ln{\sin{x}})-\frac{d}{dx}(\ln{(1+\cos{x})})\) ➜ Note \(\because \frac{d}{dx}(u+v-w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)-\frac{d}{dx}(w)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{1}{\sin{x}}\frac{d}{dx}(\sin{x})-\frac{1}{1+\cos{x}}\frac{d}{dx}(1+\cos{x})\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{1}{\sin{x}}(\cos{x})-\frac{1}{1+\cos{x}}(0-\sin{x})\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(x)=1, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{\cos{x}}{\sin{x}}+\frac{\sin{x}}{1+\cos{x}}\) \(\Rightarrow \frac{dy}{dx}=y\left(\frac{\sin{x}(1+\cos{x})+x\cos{x}(1+\cos{x})+x\sin{x}\sin{x}}{x\sin{x}(1+\cos{x})}\right)\) \(\Rightarrow \frac{dy}{dx}=\frac{x\sin{x}}{1+\cos{x}}\left(\frac{\sin{x}(1+\cos{x})+x\cos{x}(1+\cos{x})+x\sin^2{x}}{x\sin{x}(1+\cos{x})}\right)\) ➜ Note \(\because y=\frac{x\sin{x}}{1+\cos{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\cos{x}}\left(\frac{\sin{x}+\sin{x}\cos{x}+x\cos{x}+x\cos^2{x}+x\sin^2{x}}{1+\cos{x}}\right)\) \(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\cos{x}}\left(\frac{\sin{x}+\sin{x}\cos{x}+x\cos{x}+x(\sin^2{x}+\cos^2{x})}{1+\cos{x}}\right)\) \(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\cos{x}}\left(\frac{\sin{x}+\sin{x}\cos{x}+x\cos{x}+x.1}{1+\cos{x}}\right)\) ➜ Note \(\because \sin^2{x}+\cos^2{x}=1\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\cos{x}}\left(\frac{\sin{x}(1+\cos{x})+x(1+\cos{x})}{1+\cos{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\cos{x}}\left(\frac{(\sin{x}+x)(1+\cos{x})}{1+\cos{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\cos{x}}(\sin{x}+x)\)
\(\therefore \frac{dy}{dx}=\frac{\sin{x}+x}{1+\cos{x}}\)

ধরি,
\(y=(\cos{x})^{\tan{x}}\)
\(\Rightarrow \ln{y}=\ln{(\cos{x})^{\tan{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\tan{x}\ln{(\cos{x})}\) ➜ Note\(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{\tan{x}\ln{(\cos{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{d}{dx}\{\ln{(\cos{x})}\}+\ln{(\cos{x})}\frac{d}{dx}(\tan{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{1}{\cos{x}}\frac{d}{dx}(\cos{x})+\ln{(\cos{x})}\sec^2{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\tan{x}\frac{\sin{x}}{\cos{x}}+\ln{(\cos{x})}\sec^2{x}\)
\(\Rightarrow \frac{dy}{dx}=y(\ln{(\cos{x})}\sec^2{x}-\tan{x}\tan{x})\)
\(\therefore \frac{dy}{dx}=(\cos{x})^{\tan{x}}(\ln{(\cos{x})}\sec^2{x}-\tan^2{x})\) ➜ Note \(\because y=(\cos{x})^{\tan{x}}\)

ধরি,
\(y=(\cos{x})^{\tan{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\cos{x})^{\tan{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(\frac{\tan{x}}{\cos{x}}\frac{d}{dx}(\cos{x})+\ln{\cos{x}}\frac{d}{dx}(\tan{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(\frac{\tan{x}}{\cos{x}}\times{-\sin{x}}+\ln{\cos{x}}\sec^2{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(-\sin{x}\frac{\tan{x}}{\cos{x}}+\ln{\cos{x}}\sec^2{x}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(-\tan{x}\frac{\sin{x}}{\cos{x}}+\ln{\cos{x}}\sec^2{x}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}(-\tan{x}\tan{x}+\ln{\cos{x}}\sec^2{x})\)
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}(-\tan^2{x}+\ln{\cos{x}}\sec^2{x})\)
\(\therefore \frac{dy}{dx}=(\cos{x})^{\tan{x}}(\ln{\cos{x}}\sec^2{x}-\tan^2{x})\)

ধরি,
\(y=(\sin^{-1}{x})^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sin^{-1}{x})^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{(\sin^{-1}{x})}\) ➜ Note\(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{\ln{x}\ln{(\sin^{-1}{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{d}{dx}\{\ln{(\sin^{-1}{x})}\}+\ln{(\sin^{-1}{x})}\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{(\sin^{-1}{x})}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}\ln{(\sin^{-1}{x})}\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{1}{x}\ln{(\sin^{-1}{x})}\right)\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{1}{x}\ln{(\sin^{-1}{x})}+\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}\right)\) ➜ Note \(\because y=(\sin^{-1}{x})^{\ln{x}}\)

ধরি,
\(y=(\sin^{-1}{x})^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin^{-1}{x})^{\ln{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{\sin^{-1}{x}}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}+\ln{\sin^{-1}{x}}\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{1}{x}\ln{\sin^{-1}{x}}\right)\)

ধরি,
\(y=x^{\cos^{-1}{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{\cos^{-1}{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\cos^{-1}{x}\ln{x}\) ➜ Note\(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\cos^{-1}{x}\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos^{-1}{x}\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(\cos^{-1}{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos^{-1}{x}\frac{1}{x}+\ln{x}\times{-\frac{1}{\sqrt{1-x^2}}}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{\cos^{-1}{x}}{x}-\frac{\ln{x}}{\sqrt{1-x^2}}\right)\)
\(\therefore \frac{dy}{dx}=x^{\cos^{-1}{x}}\left(\frac{\cos^{-1}{x}}{x}-\frac{\ln{x}}{\sqrt{1-x^2}}\right)\) ➜ Note \(\because y=x^{\cos^{-1}{x}}\)

ধরি,
\(y=x^{\cos^{-1}{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{\cos^{-1}{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{\cos^{-1}{x}}\left(\frac{\cos^{-1}{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\cos^{-1}{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\cos^{-1}{x}}\left(\frac{\cos^{-1}{x}}{x}.1+\ln{x}\times{-\frac{1}{\sqrt{1-x^2}}}\right)\) ➜ Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{dy}{dx}=x^{\cos^{-1}{x}}\left(\frac{\cos^{-1}{x}}{x}-\frac{\ln{x}}{\sqrt{1-x^2}}\right)\)

মনে করি,
\(y=\ln{[x-\sqrt{x^2-1}]}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\ln{[x-\sqrt{x^2-1}]})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}\frac{d}{dx}[x-\sqrt{x^2-1}]\) ➜ Note \([x-\sqrt{x^2-1}]\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[\frac{d}{dx}(x)-\frac{d}{dx}(\sqrt{x^2-1})]\) ➜ Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[1-\frac{1}{2\sqrt{x^2-1}}\frac{d}{dx}(x^2-1)]\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\); \((x^2-1)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[1-\frac{1}{2\sqrt{x^2-1}}(2x-0)]\) ➜ Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[1-\frac{1}{2\sqrt{x^2-1}}2x]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[1-\frac{1}{\sqrt{x^2-1}}x]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[1-\frac{x}{\sqrt{x^2-1}}]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[\frac{\sqrt{x^2-1}-x}{\sqrt{x^2-1}}]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2-1}]}[-\frac{x-\sqrt{x^2-1}}{\sqrt{x^2-1}}]\)
\(\therefore \frac{dy}{dx}=-\frac{1}{\sqrt{x^2-1}}\)

মনে করি,
\(y=\ln{[\sqrt{x-a}+\sqrt{x-b}]}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\ln{[\sqrt{x-a}+\sqrt{x-b}]})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[\sqrt{x-a}+\sqrt{x-b}]}\frac{d}{dx}[\sqrt{x-a}+\sqrt{x-b}]\) ➜ Note \([\sqrt{x-a}+\sqrt{x-b}]\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[\sqrt{x-a}+\sqrt{x-b}]}[\frac{d}{dx}(\sqrt{x-a})+\frac{d}{dx}(\sqrt{x-b})]\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[\sqrt{x-a}+\sqrt{x-b}]}[\frac{1}{2\sqrt{x-a}}\frac{d}{dx}(x-a)+\frac{1}{2\sqrt{x-b}}\frac{d}{dx}(x-b)]\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\); \((x-a)\) ও \((x-b)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[\sqrt{x-a}+\sqrt{x-b}]}[\frac{1}{2\sqrt{x-a}}(1-0)+\frac{1}{2\sqrt{x-b}}(1-0)]\) ➜ Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[\sqrt{x-a}+\sqrt{x-b}]}[\frac{1}{2\sqrt{x-a}}+\frac{1}{2\sqrt{x-b}}]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[\sqrt{x-a}+\sqrt{x-b}]}[\frac{\sqrt{x-b}+\sqrt{x-a}}{2\sqrt{(x-a)(x-b)}}]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[\sqrt{x-a}+\sqrt{x-b}]}[\frac{\sqrt{x-a}+\sqrt{x-b}}{2\sqrt{(x-a)(x-b)}}]\)
\(\therefore \frac{dy}{dx}=\frac{1}{2\sqrt{(x-a)(x-b)}}\)

মনে করি,
\(y=\ln{[x+\sqrt{x^2+a^2}]}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\ln{[x+\sqrt{x^2+a^2}]})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2+a^2}]}\frac{d}{dx}[x+\sqrt{x^2+a^2}]\) ➜ Note \([x+\sqrt{x^2+a^2}]\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x-\sqrt{x^2+a^2}]}[\frac{d}{dx}(x)+\frac{d}{dx}(\sqrt{x^2+a^2})]\) ➜ Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2+a^2}]}[1+\frac{1}{2\sqrt{x^2+a^2}}\frac{d}{dx}(x^2+a^2)]\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\); \((x^2+a^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2+a^2}]}[1+\frac{1}{2\sqrt{x^2+a^2}}(2x-0)]\) ➜ Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2+a^2}]}[1+\frac{1}{2\sqrt{x^2+a^2}}2x]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2+a^2}]}[1+\frac{1}{\sqrt{x^2+a^2}}x]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2+a^2}]}[1+\frac{x}{\sqrt{x^2+a^2}}]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2+a^2}]}[\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}]\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{[x+\sqrt{x^2-1}]}[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}]\)
\(\therefore \frac{dy}{dx}=\frac{1}{\sqrt{x^2+a^2}}\)

মনে করি,
\(y=\ln{(ax^2+bx+c)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\ln{(ax^2+bx+c)})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{ax^2+bx+c}\frac{d}{dx}(ax^2+bx+c)\) ➜ Note \(ax^2+bx+c\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{ax^2+bx+c}\{\frac{d}{dx}(ax^2)+\frac{d}{dx}(bx)+\frac{d}{dx}(c)\}\) ➜ Note \(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{ax^2+bx+c}(2ax+b+0)\) ➜ Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{ax^2+bx+c}(2ax+b)\)
\(\therefore \frac{dy}{dx}=\frac{2ax+b}{ax^2+bx+c}\)

মনে করি,
\(y=\ln{\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)}\)
\(\Rightarrow y=\ln{(1+\sqrt{x})}-\ln{(1-\sqrt{x})}\) ➜ Note \(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{\ln{(1+\sqrt{x})}-\ln{(1-\sqrt{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{\ln{(1+\sqrt{x})}\}-\frac{d}{dx}\{\ln{(1-\sqrt{x})}\}\) ➜ Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\sqrt{x}}\frac{d}{dx}(1+\sqrt{x})-\frac{1}{1-\sqrt{x}}\frac{d}{dx}(1-\sqrt{x})\) ➜ Note \(1+\sqrt{x}\), \(1-\sqrt{x}\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\sqrt{x}}\{\frac{d}{dx}(1)+\frac{d}{dx}(\sqrt{x})\}-\frac{1}{1-\sqrt{x}}\{\frac{d}{dx}(1)-\frac{d}{dx}(\sqrt{x})\}\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\), \( \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\sqrt{x}}\{0+\frac{1}{2\sqrt{x}}\}-\frac{1}{1-\sqrt{x}}\{0-\frac{1}{2\sqrt{x}}\}\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{1+\sqrt{x}}\times{\frac{1}{2\sqrt{x}}}+\frac{1}{1-\sqrt{x}}\times{\frac{1}{2\sqrt{x}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}(1+\sqrt{x})}+\frac{1}{2\sqrt{x}(1-\sqrt{x})}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1-\sqrt{x}+1+\sqrt{x}}{2\sqrt{x}(1+\sqrt{x})(1-\sqrt{x})}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{2\sqrt{x}(1-x)}\)
\(\therefore \frac{dy}{dx}=\frac{1}{\sqrt{x}(1-x)}\)

মনে করি,
\(y=\ln{\left(\frac{x^2+x+1}{x^2-x+1}\right)}\)
\(\Rightarrow y=\ln{(x^2+x+1)}-\ln{(x^2-x+1)}\) ➜ Note \(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{\ln{(x^2+x+1)}-\ln{(x^2-x+1)}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{\ln{(x^2+x+1)}\}-\frac{d}{dx}\{\ln{(x^2-x+1)}\}\) ➜ Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x^2+x+1}\frac{d}{dx}(x^2+x+1)-\frac{1}{x^2-x+1}\frac{d}{dx}(x^2+x+1)\) ➜ Note \(x^2+x+1\), \(x^2-x+1\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x^2+x+1}\{\frac{d}{dx}(x^2)+\frac{d}{dx}(x)+\frac{d}{dx}(1)\}-\frac{1}{x^2-x+1}\{\frac{d}{dx}(x^2)-\frac{d}{dx}(x)+\frac{d}{dx}(1)\}\) ➜ Note \(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\), \( \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x^2+x+1}(2x+1+0)-\frac{1}{x^2-x+1}\{2x-1+0\}\) ➜ Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x^2+x+1}(2x+1)-\frac{1}{x^2-x+1}(2x-1)\)
\(\Rightarrow \frac{dy}{dx}=\frac{2x+1}{x^2+x+1}-\frac{2x-1}{x^2-x+1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{(2x+1)(x^2-x+1)-(2x-1)(x^2+x+1)}{(x^2+x+1)(x^2-x+1)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2x^3-2x^2+2x+x^2-x+1-2x^3-2x^2-2x+x^2+x+1}{(x^2+1+x)(x^2+1-x)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2-2x^2}{(x^2+1)^2-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2(1-x^2)}{x^4+2x^2+1-x^2}\)
\(\therefore \frac{dy}{dx}=\frac{2(1-x^2)}{x^4+x^2+1}\)

মনে করি,
\(y=\ln{\left(\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}\right)}\)
\(\Rightarrow y=\ln{\left(\frac{1-\cos{x}}{1+\cos{x}}\right)^{\frac{1}{2}}}\)
\(\Rightarrow y=\frac{1}{2}\ln{\left(\frac{1-\cos{x}}{1+\cos{x}}\right)}\) ➜ Note \(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow y=\frac{1}{2}\{\ln{(1-\cos{x})}-\ln{(1+\cos{x})}\}\) ➜ Note \(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\{\ln{(1-\cos{x})}-\ln{(1+\cos{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\{\frac{d}{dx}\ln{(1-\cos{x})}-\frac{d}{dx}\ln{(1+\cos{x})}\}\) ➜ Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\{\frac{1}{1-\cos{x}}\frac{d}{dx}(1-\cos{x})-\frac{1}{1+\cos{x}}\frac{d}{dx}(1+\cos{x})\}\) ➜ Note \(1-\cos{x}\), \(1+\cos{x}\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{1-\cos{x}}\{\frac{d}{dx}(1)-\frac{d}{dx}(\cos{x})\}-\frac{1}{1+\cos{x}}\{\frac{d}{dx}(1)+\frac{d}{dx}(\cos{x})\}\right]\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\), \( \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\{\frac{1}{1-\cos{x}}(0+\sin{x})-\frac{1}{1+\cos{x}}(0-\sin{x})\}\) ➜ Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\{\frac{1}{1-\cos{x}}\sin{x}+\frac{1}{1+\cos{x}}\sin{x}\}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\sin{x}\frac{1+\cos{x}+1-\cos{x}}{(1-\cos{x})(1+\cos{x})}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\sin{x}\frac{2}{1-\cos^2{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}\sin{x}\frac{2}{\sin^2{x}}\) ➜ Note \(\because 1-\cos^2{A}=\sin^2{A}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{\sin{x}}\)
\(\therefore \frac{dy}{dx}= cosec{x}\)

মনে করি,
\(y=\ln{\{e^x\left(\frac{x-1}{x+1}\right)^{\frac{3}{2}}\}}\)
\(\Rightarrow y=\ln{e^x}+\ln{\left(\frac{x-1}{x+1}\right)^{\frac{3}{2}}}\) ➜ Note \(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow y=\ln{e^x}+\frac{3}{2}\ln{\left(\frac{x-1}{x+1}\right)}\) ➜ Note \(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow y=\ln{e^x}+\frac{3}{2}\{\ln{(x-1)}-\ln{(x+1)}\}\) ➜ Note \(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left[\ln{e^x}+\frac{3}{2}\{\ln{(x-1)}-\ln{(x+1)}\}\right]\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\ln{e^x})+\frac{3}{2}\frac{d}{dx}\{\ln{(x-1)}-\ln{(x+1)}\}\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\ln{e^x})+\frac{3}{2}\{\frac{d}{dx}\ln{(x-1)}-\frac{d}{dx}\ln{(x+1)}\}\) ➜ Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{e^x}\frac{d}{dx}(e^x)+\frac{3}{2}\{\frac{1}{x-1}\frac{d}{dx}(x-1)-\frac{1}{x+1}\frac{d}{dx}(x+1)\}\) ➜ Note \(e^x\), \(x-1\), \(x+1\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{e^x}e^x+\frac{3}{2}\{\frac{1}{x-1}.1-\frac{1}{x+1}.1\}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=1+\frac{3}{2}\{\frac{1}{x-1}-\frac{1}{x+1}\}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=1+\frac{3}{2}\frac{x+1-x+1}{(x-1)(x+1)}\)
\(\Rightarrow \frac{dy}{dx}=1+\frac{3}{2}\frac{2}{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=1+\frac{3}{x^2-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{x^2-1+3}{x^2-1}\)
\(\therefore \frac{dy}{dx}=\frac{x^2+2}{x^2-1}\)

মনে করি,
\(y=\ln{\left(\frac{1+x}{1-x}\right)^{\frac{1}{4}}}-\frac{1}{2}\tan^{-1}{x}\)
\(\Rightarrow y=\frac{1}{4}\ln{\left(\frac{1+x}{1-x}\right)}-\frac{1}{2}\tan^{-1}{x}\) ➜ Note \(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow y=\frac{1}{4}\{\ln{(1+x)}-\ln{(1-x)}\}-\frac{1}{2}\tan^{-1}{x}\) ➜ Note \(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{4}\frac{d}{dx}\{\ln{(1+x)}-\ln{(1-x)}\}-\frac{1}{2}\frac{d}{dx}(\tan^{-1}{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে। \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{4}\{\frac{d}{dx}\ln{(1+x)}-\frac{d}{dx}\ln{(1-x)}\}-\frac{1}{2}\frac{d}{dx}(\tan^{-1}{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে। \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{4}\{\frac{1}{1+x}\frac{d}{dx}(1+x)-\frac{1}{1-x}\frac{d}{dx}(1-x)\}-\frac{1}{2}\frac{1}{1+x^2}\) ➜ Note \(1+x\), \(1-x\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{4}\{\frac{1}{1+x}.1-\frac{1}{1-x}(-1)\}-\frac{1}{2(1+x^2)}\) ➜ Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{4}\{\frac{1}{1+x}+\frac{1}{1-x}\}-\frac{1}{2(1+x^2)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{4}.\frac{1-x+1+x}{(1+x)(1-x)}-\frac{1}{2(1+x^2)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{4(1-x^2)}-\frac{1}{2(1+x^2)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2(1-x^2)}-\frac{1}{2(1+x^2)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1+x^2-1+x^2}{2(1-x^2)(1+x^2)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2x^2}{2(1-x^4)}\)
\(\therefore \frac{dy}{dx}=\frac{x^2}{1-x^4}\)

মনে করি,
\(y=\log_{x}a\)
\(\Rightarrow y=\frac{\ln{a}}{\ln{x}}\) ➜ Note \(\because \log_{p}q=\frac{\ln{q}}{\ln{p}}\)
\(\Rightarrow y=\ln{a}\frac{1}{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\ln{a}\frac{1}{\ln{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\ln{a}\frac{d}{dx}\left(\frac{1}{\ln{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=\ln{a}\left(-\frac{1}{(\ln{x})^2}\right)\frac{d}{dx}(\ln{x})\) ➜ Note \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\frac{1}{x})=-\frac{1}{x^2}\)
\(\Rightarrow \frac{dy}{dx}=\ln{a}\left(-\frac{1}{(\ln{x})^2}\right)\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\therefore \frac{dy}{dx}=-\frac{\ln{a}}{x(\ln{x})^2}\)

মনে করি,
\(y=\log_{x}2x\)
\(\Rightarrow y=\frac{\ln{2x}}{\ln{x}}\) ➜ Note \(\because \log_{p}q=\frac{\ln{q}}{\ln{p}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\ln{2x}}{\ln{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{x}\frac{d}{dx}(\ln{2x})-\ln{2x}\frac{d}{dx}(\ln{x})}{(\ln{x})^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{x}\frac{1}{2x}\frac{d}{dx}(2x)-\ln{2x}\frac{1}{x}}{(\ln{x})^2}\) ➜ Note \(2x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{x}\frac{1}{2x}.2-\ln{2x}\frac{1}{x}}{(\ln{x})^2}\) ➜ Note \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{x}\frac{1}{x}-\ln{2x}\frac{1}{x}}{(\ln{x})^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x}\times{\frac{\ln{x}-\ln{2x}}{(\ln{x})^2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{\frac{x}{2x}}}{x(\ln{x})^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{\frac{1}{2}}}{x(\ln{x})^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{2^{-1}}}{x(\ln{x})^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{-\ln{2}}{x(\ln{x})^2}\) ➜ Note \(\because \ln{M^n}=n\ln{(M)}\)
\(\therefore \frac{dy}{dx}=-\frac{\ln{2}}{x(\ln{x})^2}\)

মনে করি,
\(y=\log_{\cos{x}}\tan{x}\)
\(\Rightarrow y=\frac{\ln{\tan{x}}}{\ln{\cos{x}}}\) ➜ Note \(\because \log_{p}q=\frac{\ln{q}}{\ln{p}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\ln{\tan{x}}}{\ln{\cos{x}}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{\cos{x}}\frac{d}{dx}(\ln{\tan{x}})-\ln{\tan{x}}\frac{d}{dx}(\ln{\cos{x}})}{(\ln{\cos{x}})^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{\cos{x}}\frac{1}{\tan{x}}\frac{d}{dx}(\tan{x})-\ln{(\tan{x})}\frac{1}{\cos{x}}\frac{d}{dx}(\cos{x})}{(\ln{\cos{x}})^2}\) ➜ Note \(\tan{x}\), \(\cos{x}\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{(\cos{x})}\frac{1}{\tan{x}}\sec^2{x}-\ln{(\tan{x})}\frac{1}{\cos{x}}(-\sin{x})}{(\ln{\cos{x}})^2}\) ➜ Note \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{(\cos{x})}\cot{x}\sec^2{x}+\ln{(\tan{x})}\frac{\sin{x}}{\cos{x}}}{(\ln{\cos{x}})^2}\) ➜ Note \(\because \frac{1}{\tan{x}}=\cot{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{(\cos{x})}\frac{\cos{x}}{\sin{x}}\frac{1}{\cos^2{x}}+\ln{(\tan{x})}\tan{x}}{(\ln{\cos{x}})^2}\) ➜ Note \(\because \frac{\sin{x}}{\cos{x}}=\tan{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{(\cos{x})}\frac{1}{\sin{x}}\frac{1}{\cos{x}}+\ln{(\tan{x})}\tan{x}}{(\ln{\cos{x}})^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\ln{(\cos{x})} cosec{x}\sec{x}+\tan{x}\ln{(\tan{x})}}{(\ln{\cos{x}})^2}\) ➜ Note \(\because \frac{1}{\sin{x}}=cosec{x}, \frac{1}{\cos{x}}=\sec{x}\)
\(\therefore \frac{dy}{dx}=\frac{ cosec{x}\sec{x}\ln{(\cos{x})}+\tan{x}\ln{(\tan{x})}}{(\ln{\cos{x}})^2}\)

মনে করি,
\(y=\log_{a}x+\log_{x}a\)
\(\Rightarrow y=\frac{\ln{x}}{\ln{a}}+\frac{\ln{a}}{\ln{x}}\) ➜ Note \(\because \log_{p}q=\frac{\ln{q}}{\ln{p}}\)
\(\Rightarrow y=\frac{1}{\ln{a}}\ln{x}+\ln{a}\frac{1}{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{\ln{a}}\ln{x}+\ln{a}\frac{1}{\ln{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{\ln{a}}\frac{d}{dx}(\ln{x})+\ln{a}\frac{d}{dx}\left(\frac{1}{\ln{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{\ln{a}}\frac{1}{x}+\ln{a}\left(-\frac{1}{(\ln{x})^2}\right)\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x\ln{a}}-\frac{\ln{a}}{(\ln{x})^2}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{x\ln{a}}-\frac{\ln{a}}{x(\ln{x})^2}\)
\(\therefore \frac{dy}{dx}=\frac{(\ln{x})^2-(\ln{a})^2}{x(\ln{x})^2\ln{a}}\)

ধরি,
\(y=\frac{e^{-x}(3x+5)}{7x-1}\)
\(\Rightarrow \ln{y}=\ln{\left(\frac{e^{-x}(3x+5)}{7x-1}\right)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{e^{-x}}+\ln{(3x+5)}-\ln{(7x-1)}\) ➜ Note\(\because \ln{\frac{MN}{P}}=\ln{(M)}+\ln{(N)}-\ln{P}\)
\(\Rightarrow \ln{y}=-x+\ln{(3x+5)}-\ln{(7x-1)}\) ➜ Note\(\because \ln{e^{x}}=x\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{-x+\ln{(3x+5)}-\ln{(7x-1)}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(-x)+\frac{d}{dx}\ln{(3x+5)}-\frac{d}{dx}\ln{(7x-1)}\) ➜ Note \(\because \frac{d}{dx}(u+v-w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)-\frac{d}{dx}(w)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-1+\frac{1}{3x+5}\frac{d}{dx}(3x+5)-\frac{1}{7x-1}\frac{d}{dx}(7x-1)\) ➜ Note \(3x+5\), \(7x-1\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-1+\frac{1}{3x+5}\{\frac{d}{dx}(3x)+\frac{d}{dx}(5)\}-\frac{1}{7x-1}\{\frac{d}{dx}(7x)-\frac{d}{dx}(1)\}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-1+\frac{1}{3x+5}\{3\frac{d}{dx}(x)+\frac{d}{dx}(5)\}-\frac{1}{7x-1}\{7\frac{d}{dx}(x)-\frac{d}{dx}(1)\}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-1+\frac{1}{3x+5}\{3.1+0\}-\frac{1}{7x-1}\{7.1-0\}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-1+\frac{1}{3x+5}.3-\frac{1}{7x-1}.7\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{3}{3x+5}-\frac{7}{7x-1}-1\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{e^{-x}(3x+5)}{7x-1}\left(\frac{21x-3-21x-35-(3x+5)(7x-1)}{(3x+5)(7x-1)}\right)\) ➜ Note \(\because y=\frac{e^{-x}(3x+5)}{7x-1}\)
\(\Rightarrow \frac{dy}{dx}=\frac{e^{-x}(3x+5)}{7x-1}\left(\frac{-38-21x^2+3x-35x+5}{(3x+5)(7x-1)}\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{e^{-x}}{7x-1}\left(\frac{-21x^2-32x-33}{7x-1}\right)\)
\(\therefore \frac{dy}{dx}=-e^{-x}\left(\frac{21x^2+32x+33}{(7x-1)^2}\right)\)

ধরি,
\(y=\frac{(x+1)^2\sqrt{x-1}}{(x+4)^3e^x}\)
\(\Rightarrow \ln{y}=\ln{\left(\frac{(x+1)^2\sqrt{x-1}}{(x+4)^3e^x}\right)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{(x+1)^2}+\ln{(\sqrt{x-1})}-\ln{(x+4)^3}-\ln{e^x}\) ➜ Note\(\because \ln{\frac{MN}{PQ}}=\ln{(M)}+\ln{(N)}-\ln{P}-\ln{Q}\)
\(\Rightarrow \ln{y}=\ln{(x+1)^2}+\ln{(\sqrt{x-1})}-\ln{(x+4)^3}-x\) ➜ Note\(\because \ln{e^{x}}=x\)
\(\Rightarrow \ln{y}=\ln{(x+1)^2}+\ln{(x-1)^{\frac{1}{2}}}-\ln{(x+4)^3}-x\)
\(\Rightarrow \ln{y}=2\ln{(x+1)}+\frac{1}{2}\ln{(x-1)}-3\ln{(x+4)}-x\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{2\ln{(x+1)}+\frac{1}{2}\ln{(x-1)}-3\ln{(x+4)}-x\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\{2\ln{(x+1)}\}+\frac{d}{dx}\{\frac{1}{2}\ln{(x-1)}\}-\frac{d}{dx}\{3\ln{(x+4)}\}-\frac{d}{dx}(x)\) ➜ Note \(\because \frac{d}{dx}(u+v-w-x)=\frac{d}{dx}{u}+\frac{d}{dx}{v}-\frac{d}{dx}{w}-\frac{d}{dx}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\frac{d}{dx}\ln{(x+1)}+\frac{1}{2}\frac{d}{dx}\ln{(x-1)}-3\frac{d}{dx}\ln{(x+4)}-\frac{d}{dx}(x)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\frac{1}{(x+1)}\frac{d}{dx}(x+1)+\frac{1}{2}\frac{1}{x-1}\frac{d}{dx}(x-1)-\frac{1}{x+4}\frac{d}{dx}(x+4)-1\) ➜ Note \(x+1\), \(x-1\), \(x+4\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\frac{1}{(x+1)}(1+0)+\frac{1}{2}\frac{1}{x-1}(1-0)-\frac{1}{x+4}(1+0)-1\) ➜ Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{1}{x+4}-1\right)\)
\(\therefore \frac{dy}{dx}=\frac{(x+1)^2\sqrt{x-1}}{(x+4)^3e^x}\left(\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{1}{x+4}-1\right)\) ➜ Note \(\because y=\frac{(x+1)^2\sqrt{x-1}}{(x+4)^3e^x}\)

ধরি,
\(y=x^3\sqrt{\frac{x^2+4}{x^2+3}}\)
\(\Rightarrow \ln{y}=\ln{\left(x^3\sqrt{\frac{x^2+4}{x^2+3}}\right)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x^3}+\ln{\left(\sqrt{\frac{x^2+4}{x^2+3}}\right)}\) ➜ Note\(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow \ln{y}=\ln{x^3}+\ln{\left(\frac{x^2+4}{x^2+3}\right)^{\frac{1}{2}}}\)
\(\Rightarrow \ln{y}=3\ln{x}+\frac{1}{2}\ln{\left(\frac{x^2+4}{x^2+3}\right)}\) ➜ Note\(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=3\ln{x}+\frac{1}{2}\{\ln{(x^2+4)}-\ln{(x^2+3)}\}\) ➜ Note\(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left[3\ln{x}+\frac{1}{2}\{\ln{(x^2+4)}-\ln{(x^2+3)}\}\right]\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=3\frac{d}{dx}(\ln{x})+\frac{1}{2}\frac{d}{dx}\{\ln{(x^2+4)}-\ln{(x^2+3)}\}\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=3\frac{d}{dx}(\ln{x})+\frac{1}{2}\left(\frac{d}{dx}\ln{(x^2+4)}-\frac{d}{dx}\ln{(x^2+3)}\right)\) ➜ Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=3\frac{1}{x}+\frac{1}{2}\left(\frac{1}{x^2+4}\frac{d}{dx}(x^2+4)-\frac{1}{x^2+3}\frac{d}{dx}(x^2+3)\right)\) ➜ Note \(x^2+4\), \(x^2+3\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=3\frac{1}{x}+\frac{1}{2}\left(\frac{1}{x^2+4}(2x+0)-\frac{1}{x^2+3}(2x+0)\right)\) ➜ Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{3}{x}+\frac{1}{2}.2\left(\frac{x}{x^2+4}-\frac{x}{x^2+3}\right)\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{3}{x}+\frac{x}{x^2+4}-\frac{x}{x^2+3}\right)\)
\(\therefore \frac{dy}{dx}=x^3\sqrt{\frac{x^2+4}{x^2+3}}\left(\frac{3}{x}+\frac{x}{x^2+4}-\frac{x}{x^2+3}\right)\) ➜ Note \(\because y=x^3\sqrt{\frac{x^2+4}{x^2+3}}\)

ধরি,
\(y=\frac{x\cos^{-1}{x}}{\sqrt{1-x^2}}\)
\(\Rightarrow \ln{y}=\ln{\left(\frac{x\cos^{-1}{x}}{\sqrt{1-x^2}}\right)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}+\ln{\cos^{-1}{x}}-\ln{\sqrt{1-x^2}}\) ➜ Note\(\because \ln{\frac{MN}{p}}=\ln{(M)}+\ln{(N)}-\ln{P}\)
\(\Rightarrow \ln{y}=\ln{x}+\ln{\cos^{-1}{x}}-\ln{(1-x^2)^{\frac{1}{2}}}\)
\(\Rightarrow \ln{y}=\ln{x}+\ln{\cos^{-1}{x}}-\frac{1}{2}\ln{(1-x^2)}\) ➜ Note\(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left(\ln{x}+\ln{\cos^{-1}{x}}-\frac{1}{2}\ln{(1-x^2)}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x})+\frac{d}{dx}(\ln{\cos^{-1}{x}})-\frac{1}{2}\frac{d}{dx}\{\ln{(1-x^2)}\}\) ➜ Note \(\because \frac{d}{dx}(u+v-w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)-\frac{d}{dx}(w)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{1}{\cos^{-1}{x}}\frac{d}{dx}(\cos^{-1}{x})-\frac{1}{2}\frac{1}{1-x^2}\frac{d}{dx}(1-x^2)\) ➜ Note \(\cos^{-1}{x}\), \(1-x^2\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{1}{\cos^{-1}{x}}\left(-\frac{1}{\sqrt{1-x^2}}\right)-\frac{1}{2}\frac{1}{1-x^2}(0-2x)\) ➜ Note \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{1}{x}-\frac{1}{\sqrt{1-x^2}\cos^{-1}{x}}+\frac{x}{1-x^2}\right)\)
\(\therefore \frac{dy}{dx}=\frac{x\cos^{-1}{x}}{\sqrt{1-x^2}}\left(\frac{1}{x}-\frac{1}{\sqrt{1-x^2}\cos^{-1}{x}}+\frac{x}{1-x^2}\right)\) ➜ Note \(\because y=\frac{x\cos^{-1}{x}}{\sqrt{1-x^2}}\)

ধরি,
\(y=\frac{e^{x^2}\tan^{-1}{x}}{\sqrt{1+x^2}}\)
\(\Rightarrow \ln{y}=\ln{\left(\frac{e^{x^2}\tan^{-1}{x}}{\sqrt{1+x^2}}\right)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{e^{x^2}}+\ln{\tan^{-1}{x}}-\ln{\sqrt{1+x^2}}\) ➜ Note\(\because \ln{\frac{MN}{p}}=\ln{(M)}+\ln{(N)}-\ln{P}\)
\(\Rightarrow \ln{y}=x^2+\ln{\tan^{-1}{x}}-\ln{(1+x^2)^{\frac{1}{2}}}\) ➜ Note\(\because \ln{e^x}=x\)
\(\Rightarrow \ln{y}=x^2+\ln{\tan^{-1}{x}}-\frac{1}{2}\ln{(1+x^2)}\) ➜ Note\(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left(x^2+\ln{\tan^{-1}{x}}-\frac{1}{2}\ln{(1+x^2)}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(x^2)+\frac{d}{dx}(\ln{\tan^{-1}{x}})-\frac{1}{2}\frac{d}{dx}\{\ln{(1+x^2)}\}\) ➜ Note \(\because \frac{d}{dx}(u+v-w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)-\frac{d}{dx}(w)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2x+\frac{1}{\tan^{-1}{x}}\frac{d}{dx}(\tan^{-1}{x})-\frac{1}{2}\frac{1}{1+x^2}\frac{d}{dx}(1+x^2)\) ➜ Note \(\tan^{-1}{x}\), \(1+x^2\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2x+\frac{1}{\tan^{-1}{x}}\left(\frac{1}{1+x^2}\right)-\frac{1}{2}\frac{1}{1+x^2}(0+2x)\) ➜ Note \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=y\left(2x+\frac{1}{(1+x^2)\tan^{-1}{x}}-\frac{x}{1+x^2}\right)\)
\(\therefore \frac{dy}{dx}=\frac{e^{x^2}\tan^{-1}{x}}{\sqrt{1+x^2}}\left(2x+\frac{1}{(1+x^2)\tan^{-1}{x}}-\frac{x}{1+x^2}\right)\) ➜ Note \(\because y=\frac{e^{x^2}\tan^{-1}{x}}{\sqrt{1+x^2}}\)

ধরি,
\(y=\frac{(x^2+1)^2}{\sqrt[3]{x^2}}\)
\(\Rightarrow \ln{y}=\ln{\left(\frac{(x^2+1)^2}{\sqrt[3]{x^2}}\right)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{(x^2+1)^2}-\ln{\sqrt[3]{x^2}}\) ➜ Note\(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \ln{y}=\ln{(x^2+1)^2}-\ln{(x^2)^{\frac{1}{3}}}\)
\(\Rightarrow \ln{y}=\ln{(x^2+1)^2}-\ln{(x)^{\frac{2}{3}}}\)
\(\Rightarrow \ln{y}=2\ln{(x^2+1)}-\frac{2}{3}\ln{(x)}\) ➜ Note\(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left(2\ln{(x^2+1)}-\frac{2}{3}\ln{(x)}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\frac{d}{dx}\ln{(x^2+1)}-\frac{2}{3}\frac{d}{dx}\ln{(x)}\) ➜ Note\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\frac{1}{x^2+1}\frac{d}{dx}(x^2+1)-\frac{2}{3}\frac{1}{x}\) ➜ Note\(x^2+1\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\frac{1}{x^2+1}(2x+0)-\frac{2}{3}\frac{1}{x}\) ➜ Note\(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{12x^2-2x^2-2}{3x(x^2+1)}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{10x^2-2}{3x(x^2+1)}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\frac{5x^2-1}{3x(x^2+1)}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{2}{3}\times{\frac{5x^2-1}{x(x^2+1)}}\)
\(\Rightarrow \frac{dy}{dx}=y\times{\frac{2}{3}}\times{\frac{5x^2-1}{x(x^2+1)}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{(x^2+1)^2}{\sqrt[3]{x^2}}\times{\frac{2}{3}}\times{\frac{5x^2-1}{x(x^2+1)}}\) ➜ Note \(\because y=\frac{(x^2+1)^2}{\sqrt[3]{x^2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}\frac{(x^2+1)(5x^2-1)}{x.x^{\frac{2}{3}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}\frac{5x^4+5x^2-x^2-1}{x^{\frac{2}{3}+1}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}\frac{5x^4+4x^2-1}{x^{\frac{2+3}{3}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}\frac{5x^4+4x^2-1}{x^{\frac{5}{3}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}\left(\frac{5x^4}{x^{\frac{5}{3}}}+\frac{4x^2}{x^{\frac{5}{3}}}-\frac{1}{x^{\frac{5}{3}}}\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}\left(5x^{4-\frac{5}{3}}+4x^{2-\frac{5}{3}}-x^{-\frac{5}{3}}\right)\) ➜ Note \(\because \frac{x^m}{x^n}=x^{m-n}, \frac{1}{x^m}=x^{-m}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{3}\left(5x^{\frac{12-5}{3}}+4x^{\frac{6-5}{3}}-x^{-\frac{5}{3}}\right)\)
\(\therefore \frac{dy}{dx}=\frac{2}{3}\left(5x^{\frac{7}{3}}+4x^{\frac{1}{3}}-x^{-\frac{5}{3}}\right)\)

ধরি,
\(y=\left(\frac{x}{1+\sqrt{1-x^2}}\right)^{n}\)
\(\Rightarrow \ln{y}=\ln{\left(\frac{x}{1+\sqrt{1-x^2}}\right)^{n}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=n\ln{\left(\frac{x}{1+\sqrt{1-x^2}}\right)}\) ➜ Note\(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=n\{\ln{x}-\ln{(1+\sqrt{1-x^2})}\}\) ➜ Note\(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left[n\{\ln{x}-\ln{(1+\sqrt{1-x^2})}\}\right]\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\{\frac{d}{dx}\ln{x}-\frac{d}{dx}\ln{(1+\sqrt{1-x^2})}\}\) ➜ Note\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\{\frac{1}{x}-\frac{1}{1+\sqrt{1-x^2}}\frac{d}{dx}(1+\sqrt{1-x^2})\}\) ➜ Note\((1+\sqrt{1-x^2})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\left[\frac{1}{x}-\frac{1}{1+\sqrt{1-x^2}}\{0+\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}(1-x^2)\}\right]\) ➜ Note\((1-x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\{\frac{1}{x}-\frac{1}{1+\sqrt{1-x^2}}\times{\frac{1}{2\sqrt{1-x^2}}}(0-2x)\}\) ➜ Note\(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\{\frac{1}{x}+\frac{x}{(1+\sqrt{1-x^2})\sqrt{1-x^2}}\}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\frac{(1+\sqrt{1-x^2})\sqrt{1-x^2}+x^2}{x(1+\sqrt{1-x^2})(1-x^2)}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\frac{\sqrt{1-x^2}+1-x^2+x^2}{x(1+\sqrt{1-x^2})\sqrt{1-x^2}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\frac{1+\sqrt{1-x^2}}{x(1+\sqrt{1-x^2})\sqrt{1-x^2}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=n\frac{1}{x\sqrt{1-x^2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{n}{x\sqrt{1-x^2}}\times{y}\)
\(\therefore \frac{dy}{dx}=\frac{n}{x\sqrt{1-x^2}}\left(\frac{x}{1+\sqrt{1-x^2}}\right)^{n}\) ➜ Note \(\because y=\left(\frac{x}{1+\sqrt{1-x^2}}\right)^{n}\)

ধরি,
\(y=\frac{x\log{x}}{\sqrt{1+x^2}}\)
\(\Rightarrow \ln{y}=\ln{\left(\frac{x\log{x}}{\sqrt{1+x^2}}\right)}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}+\ln{\log{x}}-\ln{\sqrt{1+x^2}}\) ➜ Note\(\because \ln{\frac{MN}{p}}=\ln{(M)}+\ln{(N)}-\ln{P}\)
\(\Rightarrow \ln{y}=\ln{x}+\ln{\log{x}}-\ln{(1+x^2)^{\frac{1}{2}}}\)
\(\Rightarrow \ln{y}=\ln{x}+\ln{\log{x}}-\frac{1}{2}\ln{(1+x^2)}\) ➜ Note\(\because \ln{M^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left(\ln{x}+\ln{\log{x}}-\frac{1}{2}\ln{(1+x^2)}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x})+\frac{d}{dx}(\ln{\log{x}})-\frac{1}{2}\frac{d}{dx}(\ln{(1+x^2)})\) ➜ Note \(\because \frac{d}{dx}(u+v-w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)-\frac{d}{dx}(w)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{1}{\log{x}}\frac{d}{dx}(\log{x})-\frac{1}{2}\frac{1}{1+x^2}\frac{d}{dx}(1+x^2)\) ➜ Note \(\log{x}\), \(1+x^2\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \( \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{1}{\log{x}}\left(\frac{\log_{10}e}{x}\right)-\frac{1}{2}\frac{1}{1+x^2}(0+2x)\) ➜ Note \(\because \frac{d}{dx}(\log{x})=\frac{\log_{10}e}{x}, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{1}{x}+\frac{\log_{10}e}{x\log{x}}-\frac{x}{1+x^2}\right)\)
\(\therefore \frac{dy}{dx}=\frac{x\log{x}}{\sqrt{1+x^2}}\left(\frac{1}{x}+\frac{\log_{10}e}{x\log{x}}-\frac{x}{1+x^2}\right)\) ➜ Note \(\because y=\frac{x\log{x}}{\sqrt{1+x^2}}\)

ধরি,
\(y=(\sin{x})^{\tan{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sin{x})^{\tan{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\tan{x}\ln{\sin{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\tan{x}\ln{\sin{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{d}{dx}(\ln{\sin{x}})+\ln{\sin{x}}\frac{d}{dx}(\tan{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{1}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{\sin{x}}\sec^2{x}\) ➜ Note \(\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \ \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{1}{\sin{x}}\cos{x}+\ln{\sin{x}}\sec^2{x}\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{\cos{x}}{\sin{x}}+\sec^2{x}\ln{\sin{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\cot{x}+\sec^2{x}\ln{\sin{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{1}{\tan{x}}+\sec^2{x}\ln{\sin{x}}\)
\(\Rightarrow \frac{dy}{dx}=y(1+\sec^2{x}\ln{\sin{x}})\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\tan{x}}(1+\sec^2{x}\ln{\sin{x}})\) ➜ Note \(\because y=(\sin{x})^{\tan{x}}\)

ধরি,
\(y=(\sin{x})^{\tan{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin{x})^{\tan{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\tan{x}}\left(\frac{\tan{x}}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{(\sin{x})}\frac{d}{dx}(\tan{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\tan{x}}\left(\frac{\tan{x}}{\sin{x}}.\cos{x}+\ln{(\sin{x})}\sec^2{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \ \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\tan{x}}\left(\tan{x}\frac{\cos{x}}{\sin{x}}+\sec^2{x}\ln{\sin{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\tan{x}}\left(\tan{x}\cot{x}+\sec^2{x}\ln{\sin{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\tan{x}}\left(\tan{x}\frac{1}{\tan{x}}+\sec^2{x}\ln{\sin{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\tan{x}}\left(1+\sec^2{x}\ln{\sin{x}}\right)\)

ধরি,
\(y=(\sin^{-1}{x})^{x}\)
\(\Rightarrow \ln{y}=\ln{(\sin^{-1}{x})^{x}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x\ln{\sin^{-1}{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(x\ln{\sin^{-1}{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{\sin^{-1}{x}})+\ln{\sin^{-1}{x}}\frac{d}{dx}(x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{\sin^{-1}{x}}.1\) ➜ Note \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \ \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}+\ln{\sin^{-1}{x}}\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{x}{\sqrt{1-x^2}\sin{x}}+\ln{\sin^{-1}{x}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{x}{\sqrt{1-x^2}\sin{x}}+\ln{\sin^{-1}{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{x})^{x}\left(\frac{x}{\sqrt{1-x^2}\sin{x}}+\ln{\sin^{-1}{x}}\right)\) ➜ Note \(\because y=(\sin^{-1}{x})^{x}\)

ধরি,
\(y=(\sin^{-1}{x})^{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin^{-1}{x})^{x}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{x}\left(\frac{x}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{(\sin^{-1}{x})}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{x}\left(\frac{x}{\sin^{-1}{x}}.\frac{1}{\sqrt{1-x^2}}+\ln{(\sin^{-1}{x})}.1\right)\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \ \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{x})^{x}\left(\frac{x}{\sqrt{1-x^2}\sin^{-1}{x}}+\ln{\sin^{-1}{x}}\right)\)

ধরি,
\(y=(\sin{x})^{\cos{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sin{x})^{\cos{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\cos{x}\ln{\sin{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\cos{x}\ln{\sin{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos{x}\frac{d}{dx}(\ln{\sin{x}})+\ln{\sin{x}}\frac{d}{dx}(\cos{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos{x}\frac{1}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{\sin{x}}.(-\sin{x})\) ➜ Note \(\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \ \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos{x}\frac{1}{\sin{x}}\cos{x}-\sin{x}\ln{\sin{x}}\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos^2{x} \ cosec{x}-\sin{x}\ln{\sin{x}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\cos^2{x} \ cosec{x}-\sin{x}\ln{\sin{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\cos^2{x} \ cosec{x}-\sin{x}\ln{\sin{x}}\right)\) ➜ Note \(\because y=(\sin{x})^{\cos{x}}\)

ধরি,
\(y=(\sin{x})^{\cos{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin{x})^{\cos{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\frac{\cos{x}}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{(\sin{x})}\frac{d}{dx}(\cos{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\frac{\cos{x}}{\sin{x}}.\cos{x}+\ln{(\sin{x})}.(-\sin{x})\right)\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \ \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\cos^2{x} \ cosec{x}-\sin{x}\ln{\sin{x}}\right)\)

ধরি,
\(y=(\cos^{-1}{x})^{x}\)
\(\Rightarrow \ln{y}=\ln{(\cos^{-1}{x})^{x}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x\ln{\cos^{-1}{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(x\ln{\cos^{-1}{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{\cos^{-1}{x}})+\ln{\cos^{-1}{x}}\frac{d}{dx}(x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{\cos^{-1}{x}}\frac{d}{dx}(\cos^{-1}{x})+\ln{\cos^{-1}{x}}.1\) ➜ Note \(\cos^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \ \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{\cos^{-1}{x}}\frac{-1}{\sqrt{1-x^2}}+\ln{\cos^{-1}{x}}\) ➜ Note \(\because \frac{d}{dx}(\cos^{-1}{x})=\frac{-1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\frac{x}{\sqrt{1-x^2}\cos{x}}+\ln{\cos^{-1}{x}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(-\frac{x}{\sqrt{1-x^2}\cos{x}}+\ln{\cos^{-1}{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\cos^{-1}{x})^{x}\left(\ln{\cos^{-1}{x}}-\frac{x}{\sqrt{1-x^2}\cos{x}}\right)\) ➜ Note \(\because y=(\cos^{-1}{x})^{x}\)

ধরি,
\(y=(\cos^{-1}{x})^{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\cos^{-1}{x})^{x}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\cos^{-1}{x})^{x}\left(\frac{x}{\cos^{-1}{x}}\frac{d}{dx}(\cos^{-1}{x})+\ln{(\cos^{-1}{x})}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\cos^{-1}{x})^{x}\left(\frac{x}{\cos^{-1}{x}}.\frac{-1}{\sqrt{1-x^2}}+\ln{(\cos^{-1}{x})}.1\right)\) ➜ Note \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}, \ \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=(\cos^{-1}{x})^{x}\left(\ln{\cos^{-1}{x}}-\frac{x}{\sqrt{1-x^2}\cos^{-1}{x}}\right)\)

ধরি,
\(y=(\ln{x})^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{(\ln{x})^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{\ln{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x}\ln{\ln{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{d}{dx}(\ln{\ln{x}})+\ln{\ln{x}}\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\ln{x}}\frac{d}{dx}(\ln{x})+\ln{\ln{x}}.\frac{1}{x}\) ➜ Note \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\ln{x}}\frac{1}{x}+\ln{\ln{x}}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\ln{\ln{x}}\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{1}{x}+\ln{\ln{x}}\frac{1}{x}\right)\)
\(\therefore \frac{dy}{dx}=(\ln{x})^{\ln{x}}\frac{1}{x}\{1+\ln{\ln{x}}\}\) ➜ Note \(\because y=(\ln{x})^{\ln{x}}\)

ধরি,
\(y=(\ln{x})^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\ln{x})^{\ln{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\ln{x})^{\ln{x}}\left(\frac{\ln{x}}{\ln{x}}\frac{d}{dx}(\ln{x})+\ln{(\ln{x})}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\ln{x})^{\ln{x}}\left(\frac{\ln{x}}{\ln{x}}.\frac{1}{x}+\ln{(\ln{x})}.\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=(\ln{x})^{\ln{x}}\left(\frac{1}{x}+\ln{(\ln{x})}.\frac{1}{x}\right)\)
\(\therefore \frac{dy}{dx}=(\ln{x})^{\ln{x}}\frac{1}{x}\{1+\ln{(\ln{x})}\}\)

ধরি,
\(y=(\sin{x})^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sin{x})^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{\sin{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x}\ln{\sin{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{d}{dx}(\ln{\sin{x}})+\ln{\sin{x}}\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{\sin{x}}.\frac{1}{x}\) ➜ Note \(\sin{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin{x}}\cos{x}+\ln{\sin{x}}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{\cos{x}}{\sin{x}}+\frac{1}{x}\ln{\sin{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\cot{x}+\frac{1}{x}\ln{\sin{x}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\cot{x}\ln{x}+\frac{1}{x}\ln{\sin{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(\cot{x}\ln{x}+\frac{1}{x}\ln{(\sin{x})}\right)\) ➜ Note \(\because y=(\sin{x})^{\ln{x}}\)

ধরি,
\(y=(\sin{x})^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin{x})^{\ln{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{(\sin{x})}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin{x}}.\cos{x}+\ln{(\sin{x})}.\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(\ln{x}\frac{\cos{x}}{\sin{x}}+\frac{1}{x}\ln{(\sin{x})}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(\ln{x}\cot{x}+\frac{1}{x}\ln{(\sin{x})}\right)\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(\cot{x}\ln{x}+\frac{1}{x}\ln{(\sin{x})}\right)\)

ধরি,
\(y=(\cot{x})^{\tan{x}}\)
\(\Rightarrow \ln{y}=\ln{(\cot{x})^{\tan{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\tan{x}\ln{\cot{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\tan{x}\ln{\cot{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{d}{dx}(\ln{\cot{x}})+\ln{\cot{x}}\frac{d}{dx}(\tan{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{1}{\cot{x}}\frac{d}{dx}(\cot{x})+\ln{\cot{x}}.\sec^2{x}\) ➜ Note \(\cot{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\tan{x}(-cosec^2{x})+\sec^2{x}\ln{\cot{x}}\) ➜ Note \(\because \frac{d}{dx}(\cot{x})=-cosec^2{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\tan^2{x} cosec^2{x}+\sec^2{x}\ln{\cot{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\frac{\sin^2{x}}{\cos^2{x}}\times{\frac{1}{sin^2{x}}}+\sec^2{x}\ln{\cot{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\frac{\sin^2{x}}{\cos^2{x}}\times{\frac{1}{sin^2{x}}}+\sec^2{x}\ln{\cot{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=y\{-\sec^2{x} +\sec^2{x}\ln{(\cot{x})}\}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=(\cot{x})^{\tan{x}}\{\sec^2{x}\ln{(\cot{x})}-\sec^2{x}\}\) ➜ Note \(\because y=(\cot{x})^{\tan{x}}\)
\(\therefore \frac{1}{y}\frac{dy}{dx}=(\cot{x})^{\tan{x}}\sec^2{x}\{\ln{(\cot{x})}-1\}\)

ধরি,
\(y=(\cot{x})^{\tan{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\cot{x})^{\tan{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\cot{x})^{\tan{x}}\left(\frac{\tan{x}}{\cot{x}}\frac{d}{dx}(\cot{x})+\ln{(\cot{x})}\frac{d}{dx}(\tan{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\cot{x})^{\tan{x}}\left(\frac{\tan{x}}{\cot{x}}(-cosec^2{x})+\ln{(\cot{x})}.\sec^2{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\cot{x})=-cosec^2{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{dy}{dx}=(\cot{x})^{\tan{x}}\left(-\tan{x}\tan{x} cosec^2{x}+\sec^2{x}\ln{(\cot{x})}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(-\tan^2{x} cosec^2{x}+\sec^2{x}\ln{(\cot{x})}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\left(-\frac{\sin^2{x}}{\cos^2{x}}\times{\frac{1}{sin^2{x}}}+\sec^2{x}\ln{(\cot{x})}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\{-\sec^2{x}+\sec^2{x}\ln{(\cot{x})}\}\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\ln{x}}\{\sec^2{x}\ln{(\cot{x})}-\sec^2{x}\}\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\ln{x}}\sec^2{x}\{\ln{(\cot{x})}-\sec^2{x}\}\)

ধরি,
\(y=x^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=(\ln{x})^2\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{(\ln{x})^2\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\ln{x}\frac{d}{dx}(\ln{x})\) ➜ Note \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{d}{dx}(\ln{y})=2\ln{x}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=y\frac{2\ln{x}}{x}\)
\(\therefore \frac{d}{dx}(\ln{y})=x^{\ln{x}}\frac{2\ln{x}}{x}\) ➜ Note \(\because y=x^{\ln{x}}\)

ধরি,
\(y=x^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^{\ln{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{\ln{x}}\left(\frac{\ln{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\ln{x}}\left(\frac{\ln{x}}{x}.1+\ln{x}.\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=x^{\ln{x}}\left(\frac{\ln{x}}{x}+\frac{\ln{x}}{x}\right)\)
\(\therefore \frac{dy}{dx}=x^{\ln{x}}\frac{2\ln{x}}{x}\)

ধরি,
\(y=(\ln{x})^{x}\)
\(\Rightarrow \ln{y}=\ln{(\ln{x})^{x}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x\ln{(\ln{x})}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{x\ln{(\ln{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}\{\ln{(\ln{x})}\}+\ln{(\ln{x})}\frac{d}{dx}(x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{\ln{x}}\frac{d}{dx}(\ln{x})+\ln{(\ln{x})}.1\) ➜ Note \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{\ln{x}}\frac{1}{x}+\ln{(\ln{x})}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{\ln{x}}+\ln{(\ln{x})}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{1}{\ln{x}}+\ln{(\ln{x})}\right)\)
\(\therefore \frac{dy}{dx}=(\ln{x})^{x}\left(\frac{1}{\ln{x}}+\ln{(\ln{x})}\right)\) ➜ Note \(\because y=(\ln{x})^{x}\)

ধরি,
\(y=(\ln{x})^{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\ln{x})^{x}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\ln{x})^{x}\left(\frac{x}{\ln{x}}\frac{d}{dx}(\ln{x})+\ln{\ln{x}}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\ln{x})^{x}\left(\frac{x}{\ln{x}}.\frac{1}{x}+\ln{\ln{x}}.1\right)\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=(\ln{x})^{x}\left(\frac{1}{\ln{x}}+\ln{\ln{x}}\right)\)

ধরি,
\(y=(\tan{x})^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{(\tan{x})^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{(\tan{x})}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{\ln{x}\ln{(\tan{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{d}{dx}\{\ln{(\tan{x})}\}+\ln{(\tan{x})}\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\tan{x}}\frac{d}{dx}(\tan{x})+\ln{(\tan{x})}\frac{1}{x}\) ➜ Note \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\tan{x}}\sec^2{x}+\frac{1}{x}\ln{(\tan{x})}\) ➜ Note\(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\cot{x}\sec^2{x}+\frac{1}{x}\ln{(\tan{x})}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\times{\frac{\cos{x}}{\sin{x}}}\times{\frac{1}{\cos^2{x}}}+\frac{1}{x}\ln{(\tan{x})}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\times{\frac{1}{\sin{x}\cos{x}}}+\frac{1}{x}\ln{(\tan{x})}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=2\ln{x}\times{\frac{1}{2\sin{x}\cos{x}}}+\frac{1}{x}\ln{(\tan{x})}\)
\(\Rightarrow \frac{dy}{dx}=y\left(2\ln{x}\times{\frac{1}{\sin{2x}}}+\frac{1}{x}\ln{(\tan{x})}\right)\)
\(\therefore \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(2\ln{x} \ cosec{2x}+\frac{1}{x}\ln{(\tan{x})}\right)\) ➜ Note \(\because y=(\tan{x})^{\ln{x}}\)

ধরি,
\(y=(\tan{x})^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\tan{x})^{\ln{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(\frac{\ln{x}}{\tan{x}}\frac{d}{dx}(\tan{x})+\ln{\tan{x}}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(\ln{x}.\cot{x}.\sec^2{x}+\ln{\tan{x}}.\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(\ln{x}.\frac{\cos{x}}{\sin{x}}.\frac{1}{\cos^2{x}}+\frac{1}{x}\ln{\tan{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(\ln{x}.\frac{1}{\sin{x}\cos{x}}+\frac{1}{x}\ln{\tan{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(2\ln{x}.\frac{1}{2\sin{x}\cos{x}}+\frac{1}{x}\ln{\tan{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(2\ln{x}.\frac{1}{\sin{2x}}+\frac{1}{x}\ln{\tan{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\tan{x})^{\ln{x}}\left(2\ln{x} \ cosec{2x}+\frac{1}{x}\ln{\tan{x}}\right)\)

ধরি,
\(y=(\cos{x})^{\tan{x}}\)
\(\Rightarrow \ln{y}=\ln{(\cos{x})^{\tan{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\tan{x}\ln{(\cos{x})}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\{\tan{x}\ln{(\cos{x})}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{d}{dx}\{\ln{(\cos{x})}\}+\ln{(\cos{x})}\frac{d}{dx}(\tan{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\tan{x}\frac{1}{\cos{x}}\frac{d}{dx}(\cos{x})+\ln{(\cos{x})}\sec^2{x}\) ➜ Note \(\cos{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\tan{x}\frac{\sin{x}}{\cos{x}}+\sec^2{x}\ln{(\cos{x})}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=-\tan{x}\tan{x}+\sec^2{x}\ln{(\cos{x})}\)
\(\Rightarrow \frac{dy}{dx}=y\{-\tan^2{x}+\sec^2{x}\ln{(\cos{x})}\}\)
\(\therefore \frac{dy}{dx}=(\cos{x})^{\tan{x}}\{\sec^2{x}\ln{(\cos{x})}-\tan^2{x}\}\) ➜ Note \(\because y=(\cos{x})^{\tan{x}}\)

ধরি,
\(y=(\cos{x})^{\tan{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\cos{x})^{\tan{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(\frac{\tan{x}}{\cos{x}}\frac{d}{dx}(\cos{x})+\ln{(\cos{x})}\frac{d}{dx}(\tan{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(-\tan{x}\frac{\sin{x}}{\cos{x}}+\ln{(\cos{x})}\sec^2{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(-\tan^2{x}+\sec^2{x}\ln{(\cos{x})}\right)\)
\(\therefore \frac{dy}{dx}=(\cos{x})^{\tan{x}}\left(\sec^2{x}\ln{(\cos{x})-\tan^2{x}}\right)\)

ধরি,
\(y=(\sin^{-1}{x})^{\ln{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sin^{-1}{x})^{\ln{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\ln{x}\ln{\sin^{-1}{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{x}\ln{\sin^{-1}{x}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{d}{dx}(\ln{\sin^{-1}{x}})+\ln{\sin^{-1}{x}}\frac{d}{dx}(\ln{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{\sin^{-1}{x}}.\frac{1}{x}\) ➜ Note \(\sin^{-1}{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{x}\frac{1}{\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}+\frac{1}{x}\ln{\sin^{-1}{x}}\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{1}{x}\ln{\sin^{-1}{x}}\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{1}{x}\ln{\sin^{-1}{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{\ln{x}})^{x}\left(\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{\ln{\sin^{-1}{x}}}{x}\right)\) ➜ Note \(\because y=(\sin^{-1}{x})^{\ln{x}}\)

ধরি,
\(y=(\sin^{-1}{x})^{\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{(\sin^{-1}{x})^{\ln{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{(\sin^{-1}{x})}\frac{d}{dx}(\ln{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sin^{-1}{x}}.\frac{1}{\sqrt{1-x^2}}+\ln{(\sin^{-1}{x})}.\frac{1}{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \ \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{1}{x}\ln{\sin^{-1}{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin^{-1}{x})^{\ln{x}}\left(\frac{\ln{x}}{\sqrt{1-x^2}\sin^{-1}{x}}+\frac{\ln{\sin^{-1}{x}}}{x}\right)\)

ধরি,
\(y=x^{x}\)
\(\Rightarrow \ln{y}=\ln{x^{x}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(x\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\ln{x}.1\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=1+\ln{x}\)
\(\Rightarrow \frac{dy}{dx}=y(1+\ln{x})\)
\(\therefore \frac{dy}{dx}=x^{x}(1+\ln{x})\) ➜ Note \(\because y=x^{x}\)

ধরি,
\(y=x^{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{x}\left(\frac{x}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x}(1.1+\ln{x}.1)\) ➜ Note \(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=x^{x}(1+\ln{x})\)

ধরি,
\(y=e^{2\ln{(\tan{5x})}}\)
\(\Rightarrow y=e^{\ln{(\tan^2{5x})}}\) ➜ Note \(\because n\ln{(M)}=\ln{M^{n}}\)
\(\Rightarrow y=\tan^2{5x}\) ➜ Note \(\because e^{\ln{x}}=x\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(\tan^2{5x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=2\tan{5x}\frac{d}{dx}(\tan{5x})\) ➜ Note \(\tan{5x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{dy}{dx}=2\tan{5x}\sec^2{5x}\frac{d}{dx}(5x)\) ➜ Note \(5x\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{dy}{dx}=2\tan{5x}\sec^2{5x}.5\)
\(\therefore \frac{dy}{dx}=10\tan{5x}\sec^2{5x}\)

ধরি,
\(y=(\sec{x})^{x^{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sec{x})^{x^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x^{x}\ln{(\sec{x})}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{\ln{y}}=\ln{x^{x}\ln{(\sec{x})}}\) ➜ Note আবার, উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{\ln{y}}=\ln{x^{x}}+\ln{\ln{(\sec{x})}}\) ➜ Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow \ln{\ln{y}}=x\ln{x}+\ln{\ln{(\sec{x})}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{\ln{y}})=\frac{d}{dx}\{x\ln{x}+\ln{\ln{(\sec{x})}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{\ln{y}})=\frac{d}{dx}(x\ln{x})+\frac{d}{dx}\{\ln{\ln{(\sec{x})}}\}\) ➜ Note \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)+\frac{1}{\ln{(\sec{x})}}\frac{d}{dx}\{\ln{(\sec{x})}\}\) ➜ Note \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\) এবং \(\ln{\sec{x}}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\ln{x}.1+\frac{1}{\ln{(\sec{x})}}\frac{1}{\sec{x}}\frac{d}{dx}(\sec{x})\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\) এবং \(\sec{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{1}{y\ln{y}}\frac{dy}{dx}=1+\ln{x}+\frac{1}{\sec{x}\ln{(\sec{x})}}\sec{x}\tan{x}\)
\(\Rightarrow \frac{1}{y\ln{y}}\frac{dy}{dx}=1+\ln{x}+\frac{\tan{x}}{\ln{(\sec{x})}}\)
\(\Rightarrow \frac{dy}{dx}=y\ln{y}\{1+\ln{x}+\frac{\tan{x}}{\ln{(\sec{x})}}\}\)
\(\Rightarrow \frac{dy}{dx}=(\sec{x})^{x^{x}}x^{x}\ln{(\sec{x})}\{1+\ln{x}+\frac{\tan{x}}{\ln{(\sec{x})}}\}\) ➜ Note \(\because y=(\sec{x})^{x^{x}}, \ln{y}=x^{x}\ln{(\sec{x})}\)
\(\Rightarrow \frac{dy}{dx}=(\sec{x})^{x^{x}}x^{x}\{\ln{(\sec{x})}(1+\ln{x})+\tan{x}\}\)
\(\therefore \frac{dy}{dx}=(\sec{x})^{x^{x}}x^{x}\{\tan{x}+(1+\ln{x})\ln{(\sec{x})}\}\)

মনে করি,
\(y=x^{x}\ln{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(x^{x}\ln{x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{x}\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}\left(x^{x}\right)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^{x}\frac{1}{x}+\ln{x}\left[x^{x}\{\frac{x}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(x)\}\right]\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(=x^{x}.x^{-1}+\ln{x}\left[x^{x}\{1.1+\ln{x}.1\}\right]\) ➜ Note \(\because \frac{d}{dx}(x)=1\)
\(=x^{x-1}+x^{x}\ln{x}(1+\ln{x})\)
\(=x^{x-1}+x^{x-1}.x\ln{x}(1+\ln{x})\)
\(\therefore \frac{dy}{dx}=x^{x-1}\{1+x\ln{x}(1+\ln{x})\}\)

মনে করি,
\(y=e^{x^2}+x^{x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(e^{x^2}+x^{x^2}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(e^{x^2}\right)+\frac{d}{dx}\left(x^{x^2}\right)\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(=e^{x^2}\frac{d}{dx}(x^2)+x^{x^2}\left(\frac{x^2}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(x^2)\right)\) ➜ Note \(x^2\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(=e^{x^2}.2x+x^{x^2}(x.1+\ln{x}.2x)\) ➜ Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(x)=1\)
\(=2xe^{x^2}+x^{x^2}(x+2x\ln{x})\)
\(=2xe^{x^2}+x^{x^2}x(1+2\ln{x})\)
\(\therefore \frac{dy}{dx}=2xe^{x^2}+x^{x^2+1}(1+\ln{x^2})\) ➜ Note \(\because n\ln{(M)}=\ln{M^n}\)

ধরি,
\(y=x^{x^{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{x^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x^{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{\ln{y}}=\ln{x^{x}\ln{x}}\) ➜ Note আবার, উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{\ln{y}}=\ln{x^{x}}+\ln{\ln{x}}\) ➜ Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow \ln{\ln{y}}=x\ln{x}+\ln{\ln{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{\ln{y}})=\frac{d}{dx}\{x\ln{x}+\ln{\ln{x}}\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{\ln{y}})=\frac{d}{dx}(x\ln{x})+\frac{d}{dx}\{\ln{\ln{x}}\}\) ➜ Note \(\frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)+\frac{1}{\ln{x}}\frac{d}{dx}\{\ln{x}\}\) ➜ Note \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\) এবং \(\ln{x}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\ln{x}.1+\frac{1}{\ln{x}}\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y\ln{y}}\frac{dy}{dx}=1+\ln{x}+\frac{1}{x\ln{x}}\)
\(\Rightarrow \frac{1}{y\ln{y}}\frac{dy}{dx}=1+\ln{x}+\frac{1}{x\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=y\ln{y}\{1+\ln{x}+\frac{1}{x\ln{x}}\}\)
\(\Rightarrow \frac{dy}{dx}=x^{x^{x}}x^{x}\ln{x}\left(1+\ln{x}+\frac{1}{x\ln{x}}\right)\) ➜ Note \(\because y=x^{x^{x}}, \ln{y}=x^{x}\ln{x}\)
\(\therefore \frac{dy}{dx}=x^{x^{x}}x^{x}\left(\ln{x}(1+\ln{x})+\frac{1}{x}\right)\)

ধরি,
\(y=e^{x^{x}}\)
\(\Rightarrow \ln{y}=\ln{e^{x^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x^{x}\ln{e}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=x^{x}.1\) ➜ Note \(\because \ln{e}=1\)
\(\Rightarrow \ln{y}=x^{x}\)
\(\Rightarrow \ln{\ln{y}}=\ln{x^{x}}\) ➜ Note আবার, উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{\ln{y}}=x\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{\ln{y}})=\frac{d}{dx}(x\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)\) ➜ Note \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{\ln{y}}\frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\ln{x}.1\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y\ln{y}}\frac{dy}{dx}=1+\ln{x}\)
\(\Rightarrow \frac{dy}{dx}=y\ln{y}(1+\ln{x})\)
\(\therefore \frac{dy}{dx}=e^{x^{x}}x^{x}(1+\ln{x})\) ➜ Note \(\because y=e^{x^{x}}, \ln{y}=x^{x}\)

ধরি,
\(y=x^{e^{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{e^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=e^{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(e^{x}\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=e^{x}\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}((e^{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=e^{x}\frac{1}{x}+\ln{x}e^{x}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=e^{x}\left(\frac{1}{x}+\ln{x}\right)\)
\(\Rightarrow \frac{dy}{dx}=ye^x\left(\frac{1}{x}+\ln{x}\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{e^{x}}e^x\left(\frac{1}{x}+\ln{x}\right)\) ➜ Note \(\because y=x^{e^{x}}\)

ধরি,
\(y=x^{e^{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^{e^{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=x^{e^{x}}\left(\frac{e^x}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(e^x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{e^{x}}\left(\frac{e^x}{x}.1+\ln{x}e^x\right)\) ➜ Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(e^x)=e^x\)
\(\therefore \frac{dy}{dx}=x^{e^{x}}e^x\left(\frac{1}{x}+\ln{x}\right)\)

ধরি,
\(y=x^{e^{x}}\)
\(\Rightarrow \ln{y}=\ln{e^{e^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=e^{x}\ln{e}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=e^{x}.1\) ➜ Note \(\because \ln{e}=1\)
\(\Rightarrow \ln{y}=e^{x}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(e^{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=e^{x}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x\)
\(\Rightarrow \frac{dy}{dx}=ye^{x}\)
\(\therefore \frac{dy}{dx}=e^{e^{x}}e^{x}\) ➜ Note \(\because y=e^{e^{x}}\)

ধরি,
\(y=a^{a^{x}}\)
\(\Rightarrow \ln{y}=\ln{a^{a^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=a^{x}\ln{a}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=\ln{a}a^{x}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{a}a^{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=\ln{a}\frac{d}{dx}(a^{x})\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{a}a^{x}\ln{a}\) ➜ Note \(\because \frac{d}{dx}(a^x)=a^x\ln{a}\)
\(\Rightarrow \frac{dy}{dx}=y(\ln{a})^2a^{x}\)
\(\therefore \frac{dy}{dx}=a^{a^{x}}a^{x}(\ln{a})^2\) ➜ Note \(\because y=a^{a^{x}}\)

ধরি,
\(y=x^{\frac{1}{x}}\)
\(\Rightarrow \ln{y}=\ln{x^{\frac{1}{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\frac{1}{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=\frac{\ln{x}}{x}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left(\frac{\ln{x}}{x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{x\frac{d}{dx}(\ln{x})-\ln{x}\frac{d}{dx}(x)}{x^2}\) ➜ Note \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{x\frac{1}{x}-\ln{x}.1}{x^2}\) ➜ Note\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1-\ln{x}}{x^2}\) ➜ Note\(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=y\left(\frac{1-\ln{x}}{x^2}\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{\frac{1}{x}}x^{-2}(1-\ln{x})\)
\(\therefore \frac{dy}{dx}=x^{\frac{1}{x}-2}(1-\ln{x})\)

ধরি,
\(y=(\sqrt{x})^{\sqrt{x}}\)
\(\Rightarrow \ln{y}=\ln{(\sqrt{x})^{\sqrt{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=\sqrt{x}\ln{\sqrt{x}}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=\sqrt{x}\ln{x^{\frac{1}{2}}}\)
\(\Rightarrow \ln{y}=\frac{1}{2}\sqrt{x}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}\left(\frac{1}{2}\sqrt{x}\ln{x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(\sqrt{x}\ln{x})\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left(\sqrt{x}\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(\sqrt{x})\right)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left(\sqrt{x}\frac{1}{x}+\ln{x}\frac{1}{2\sqrt{x}}\right)\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left(\frac{\sqrt{x}}{x}+\frac{\ln{x}}{2\sqrt{x}}\right)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{x}}\left(\frac{x}{x}+\frac{\ln{x}}{2}\right)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{x}}\left(1+\frac{1}{2}\ln{x}\right)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{x}}\left(1+\ln{x^{\frac{1}{2}}}\right)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{x}}\left(1+\ln{\sqrt{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}y\frac{1}{\sqrt{x}}\left(1+\ln{\sqrt{x}}\right)\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2}(\sqrt{x})^{\sqrt{x}}\frac{1}{\sqrt{x}}\left(1+\ln{\sqrt{x}}\right)\) |Note\(\because y=(\sqrt{x})^{\sqrt{x}}\)
\(\therefore \frac{dy}{dx}=(\sqrt{x})^{\sqrt{x}}\frac{1}{2\sqrt{x}}\left(1+\ln{\sqrt{x}}\right)\)

ধরি,
\(y=a^{x^{x}}\)
\(\Rightarrow \ln{y}=\ln{a^{x^{x}}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x^{x}\ln{a}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=\ln{a}x^{x}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(\ln{a}x^{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{y})=\ln{a}\frac{d}{dx}(x^{x})\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=\ln{a}x^{x}\left(\frac{x}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=y\ln{a}x^{x}(1.1+\ln{x}.1)\) ➜ Note \(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=a^{x^{x}}x^{x}\ln{a}(1+\ln{x})\) ➜ Note \(\because y=a^{x^{x}}\)

ধরি,
\(y=(x^{x})^{x}\)
\(\Rightarrow y=x^{x.x}\)
\(\Rightarrow y=x^{x^2}\)
\(\Rightarrow \ln{y}=\ln{x^{x^2}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=x^{2}\ln{x}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(x^{2}\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x^2\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x^{2})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x^{2}\frac{1}{x}+\ln{x}.2x\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=x+2x\ln{x}\)
\(\Rightarrow \frac{dy}{dx}=y\{x(1+2\ln{x})\}\)
\(\therefore \frac{dy}{dx}=(x^{x})^{x}\{x(1+\ln{x^2})\}\) ➜ Note \(\because y=(x^{x})^{x}\)

ধরি,
\(y=(ax)^{bx}\)
\(\Rightarrow \ln{y}=\ln{(ax)^{bx}}\) ➜ Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{y}=bx\ln{ax}\) ➜ Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{y}=bx(\ln{a}+\ln{x})\) ➜ Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow \ln{y}=b\ln{a}x+bx\ln{x}\)
\(\Rightarrow \frac{d}{dx}(\ln{y})=\frac{d}{dx}(b\ln{a}x+bx\ln{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=b\ln{a}\frac{d}{dx}(x)+b\frac{d}{dx}(x\ln{x})\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=b\ln{a}.1+b\{x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=b\ln{a}+b\{x\frac{1}{x}+\ln{x}.1\}\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=b\ln{a}+b\{1+\ln{x}\}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=b(\ln{a}+1+\ln{x})\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}=b(1+\ln{a}+\ln{x})\)
\(\Rightarrow \frac{dy}{dx}=y\{b(1+\ln{ax})\}\)
\(\therefore \frac{dy}{dx}=(ax)^{bx}\{b(1+\ln{ax})\}\) ➜ Note \(\because y=(ax)^{bx}\)

ধরি,
\(y=x^{x}+x^{\frac{1}{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(x^{x}+x^{\frac{1}{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x^{x})+\frac{d}{dx}(x^{\frac{1}{x}})\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=x^{x}\left(\frac{x}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(x)\right)\)\(+x^{\frac{1}{x}}\left(\frac{\frac{1}{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\frac{1}{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x}(1.1+\ln{x}.1)\)\(+x^{\frac{1}{x}}\left(\frac{1}{x.x}.1+\ln{x}(-\frac{1}{x^2})\right)\) ➜ Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}\left(\frac{1}{x}\right)=\left(-\frac{1}{x^2}\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x}(1+\ln{x})+x^{\frac{1}{x}}\left(\frac{1}{x^2}-\frac{\ln{x}}{x^2}\right)\)
\(\Rightarrow \frac{dy}{dx}=x^{x}(1+\ln{x})+x^{\frac{1}{x}}\frac{1}{x^2}(1-\ln{x})\)
\(\Rightarrow \frac{dy}{dx}=x^{x}(1+\ln{x})+x^{\frac{1}{x}}x^{-2}(1-\ln{x})\)
\(\therefore \frac{dy}{dx}=x^{x}(1+\ln{x})+x^{\frac{1}{x}-2}(1-\ln{x})\)

ধরি,
\(y=(\tan{x})^{x}+x^{\tan{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left((\tan{x})^{x}+x^{\tan{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left((\tan{x})^{x}\right)+\frac{d}{dx}(x^{\tan{x}})\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{x}\left(\frac{x}{\tan{x}}\frac{d}{dx}(\tan{x})+\ln{\tan{x}}\frac{d}{dx}(x)\right)\)\(+(x)^{\tan{x}}\left(\frac{\tan{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\tan{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\tan{x})^{x}(\frac{x}{\tan{x}}.\sec^2{x}+\ln{\tan{x}}.1)\)\(+x^{\tan{x}}\left(\frac{\tan{x}}{x}.1+\ln{x}.\sec^2{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}, \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=(\tan{x})^{x}(x\cot{x}\sec^2{x}+\ln{\tan{x}})\)\(+x^{\tan{x}}\left(\frac{\tan{x}}{x}+\sec^2{x}\ln{x}\right)\)

ধরি,
\(y=a^{x}+b^{x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(a^{x}+b^{x^2}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(a^{x})+\frac{d}{dx}(b^{x^2})\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=a^{x}\ln{a}+b^{x^2}\ln{b}\frac{d}{dx}(x^2)\) ➜ Note \(\because \frac{d}{dx}(a^{x})=a^{x}\ln{a}\) এবং \(x^2\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{dy}{dx}=a^{x}\ln{a}+b^{x^2}\ln{b}.2x\) ➜ Note \(\because \frac{d}{dx}(x^{2})=2x\)
\(\therefore \frac{dy}{dx}=a^{x}\ln{a}+2xb^{x^2}\ln{b}\)

ধরি,
\(y=(\ln{x})^{x}+x^{\sin{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left((\ln{x})^{x}+x^{\sin{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left((\ln{x})^{x}\right)+\frac{d}{dx}(x^{\sin{x}})\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=(\ln{x})^{x}\left(\frac{x}{\ln{x}}\frac{d}{dx}(\ln{x})+\ln{\ln{x}}\frac{d}{dx}(x)\right)\)\(+(x)^{\sin{x}}\left(\frac{\sin{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\sin{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\ln{x})^{x}\left(\frac{x}{\ln{x}}.\frac{1}{x}+\ln{\ln{x}}.1\right)\)\(+x^{\sin{x}}\left(\frac{\sin{x}}{x}.1+\ln{x}.\cos{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dy}{dx}=(\ln{x})^{x}\left(\frac{1}{\ln{x}}+\ln{\ln{x}}\right)\)\(+x^{\sin{x}}\left(\frac{\sin{x}}{x}+\cos{x}\ln{x}\right)\)

ধরি,
\(y=x^{\cos^{-1}{x}}+(\sin^{-1}{x})^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left(x^{\cos^{-1}{x}}+(\sin^{-1}{x})^{x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^{\cos^{-1}{x}}\right)+\frac{d}{dx}\left((\sin^{-1}{x})^{x}\right)\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=(x)^{\cos^{-1}{x}}\left(\frac{\cos^{-1}{x}}{x}\frac{d}{dx}(x)+\ln{x}\frac{d}{dx}(\cos^{-1}{x})\right)\)\(+(\sin^{-1}{x})^{x}\left(\frac{x}{\sin^{-1}{x}}\frac{d}{dx}(\sin^{-1}{x})+\ln{\sin^{-1}{x}}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(x)^{\cos^{-1}{x}}\left(\frac{\cos^{-1}{x}}{x}.1+\ln{x}.(-\frac{1}{\sqrt{1-x^2}})\right)\)\(+(\sin^{-1}{x})^{x}\left(\frac{x}{\sin^{-1}{x}}.\frac{1}{\sqrt{1-x^2}}+\ln{\sin^{-1}{x}}.1\right)\) ➜ Note \(\because \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=(x)^{\cos^{-1}{x}}\left(\frac{\cos^{-1}{x}}{x}-\frac{\ln{x}}{\sqrt{1-x^2}}\right)\)
\(+(\sin^{-1}{x})^{x}\left(\frac{x}{\sqrt{1-x^2}\sin^{-1}{x}}+\ln{\sin^{-1}{x}}\right)\)

ধরি,
\(y=(\sin{x})^{\cos{x}}+(\cos{x})^{\sin{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left((\sin{x})^{\cos{x}}+(\cos{x})^{\sin{x}}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left((\sin{x})^{\cos{x}}\right)+\frac{d}{dx}\left((\cos{x})^{\sin{x}}\right)\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\frac{\cos{x}}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{\sin{x}}\frac{d}{dx}(\cos{x})\right)\)\(+(\cos{x})^{\sin{x}}\left(\frac{\sin{x}}{\cos{x}}\frac{d}{dx}(\cos{x})+\ln{\cos{x}}\frac{d}{dx}(\sin{x})\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\cot{x}\cos{x}+\ln{\sin{x}}(-\sin{x})\right)\)\(+(\cos{x})^{\sin{x}}\left(\tan{x}(-\sin{x})+\ln{\cos{x}}\cos{x}\right)\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\cot{x}\cos{x}-\sin{x}\ln{\sin{x}}\right)\)
\(+(\cos{x})^{\sin{x}}\left(-\tan{x}\sin{x}+\cos{x}\ln{\cos{x}}\right)\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{\cos{x}}\left(\cot{x}\cos{x}-\sin{x}\ln{\sin{x}}\right)\)
\(+(\cos{x})^{\sin{x}}\left(\cos{x}\ln{\cos{x}}-\tan{x}\sin{x}\right)\)

ধরি,
\(y=(1+x^2)^{x^2}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left((1+x^2)^{x^2}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(1+x^2)^{x^2}\left(\frac{x^2}{1+x^2}\frac{d}{dx}(1+x^2)+\ln{(1+x^2)}\frac{d}{dx}(x^2)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(1+x^2)^{x^2}\left(\frac{x^2}{1+x^2}(0+2x)+\ln{(1+x^2)}.2x\right)\) ➜ Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=(1+x^2)^{x^2}\left(\frac{2x^3}{1+x^2}+2x\ln{(1+x^2)}\right)\)
\(\therefore \frac{dy}{dx}=2x(1+x^2)^{x^2}\left(\frac{x^2}{1+x^2}+\ln{(1+x^2)}\right)\)

ধরি,
\(y=(1+x^2)^{2x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left((1+x^2)^{2x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(1+x^2)^{2x}\left(\frac{2x}{1+x^2}\frac{d}{dx}(1+x^2)+\ln{(1+x^2)}\frac{d}{dx}(2x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(1+x^2)^{2x}\left(\frac{2x}{1+x^2}(0+2x)+\ln{(1+x^2)}.2\right)\) ➜ Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(2x)=2\)
\(\therefore \frac{dy}{dx}=(1+x^2)^{2x}\left(\frac{4x^2}{1+x^2}+2\ln{(1+x^2)}\right)\)

ধরি,
\(y=(\sin{x})^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left((\sin{x})^{x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{x}\left(\frac{x}{\sin{x}}\frac{d}{dx}(\sin{x})+\ln{(\sin{x})}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{x}\left(\frac{x}{\sin{x}}\cos{x}+\ln{(\sin{x})}.1\right)\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=(\sin{x})^{x}\left(\frac{x\cos{x}}{\sin{x}}+\ln{(\sin{x})}\right)\)
\(\therefore \frac{dy}{dx}=(\sin{x})^{x}\{x\cot{x}+\ln{(\sin{x})}\}\)

ধরি,
\(y=(1+x)^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\left((1+x)^{x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=(1+x)^{x}\left(\frac{x}{1+x}\frac{d}{dx}(1+x)+\ln{(1+x)}\frac{d}{dx}(x)\right)\) ➜ Note \(\because \frac{d}{dx}(u^v)=u^v\left(\frac{v}{u}\frac{d}{dx}(u)+\ln{u}\frac{d}{dx}(v)\right)\)
\(\Rightarrow \frac{dy}{dx}=(1+x)^{x}\left(\frac{x}{1+x}(0+1)+\ln{(1+x)}.1\right)\) ➜ Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dx}=(1+x)^{x}\left(\frac{x}{1+x}+\ln{(1+x)}\right)\)

দেওয়া আছে,
\(y=\sqrt{\sin{x}+\sqrt{\sin{x}+\sqrt{\sin{x}+.......\infty}}}\)
\(\Rightarrow y=\sqrt{\sin{x}+y}\) ➜ Note \(\because y=\sqrt{\sin{x}+\sqrt{\sin{x}+\sqrt{\sin{x}+.......\infty}}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{\sin{x}+y})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{\sin{x}+y}}\frac{d}{dx}(\sin{x}+y)\) ➜ Note \((\sin{x}+y)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2y}\{\frac{d}{dx}(\sin{x})+\frac{d}{dx}(y)\}\) ➜ Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), y=\sqrt{\sin{x}+y}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2y}\{\cos{x}+\frac{dy}{dx}\}\) ➜ Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\cos{x}}{2y}+\frac{1}{2y}\frac{dy}{dx}\)
\(\Rightarrow \frac{dy}{dx}-\frac{1}{2y}\frac{dy}{dx}=\frac{\cos{x}}{2y}\)
\(\Rightarrow \frac{dy}{dx}\left(1-\frac{1}{2y}\right)=\frac{\cos{x}}{2y}\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{2y-1}{2y}\right)=\frac{\cos{x}}{2y}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\cos{x}}{2y}\times{\frac{2y}{2y-1}}\)
\(\therefore \frac{dy}{dx}=\frac{\cos{x}}{2y-1}\)
(Showed)

দেওয়া আছে,
\(y=\sqrt{\cos{x}\sqrt{\cos{x}\sqrt{\cos{x}.......\infty}}}\)
\(\Rightarrow y=\sqrt{\cos{x}\times{y}}\) ➜ Note \(\because y=\sqrt{\cos{x}\sqrt{\cos{x}\sqrt{\cos{x}.......\infty}}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{\cos{x}\times{y}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{\cos{x}y}}\frac{d}{dx}(\cos{x}\times{y})\) ➜ Note \((\cos{x}\times{y})\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2y}\{\cos{x}\frac{d}{dx}(y)+y\frac{d}{dx}(\cos{x})\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), y=\sqrt{\cos{x}\times{y}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2y}\{\cos{x}\frac{dy}{dx}+y(-\sin{x})\}\) ➜ Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{2y}\{\cos{x}\frac{dy}{dx}-y\sin{x}\}\) ➜ Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\cos{x}}{2y}\frac{dy}{dx}-\frac{1}{2}\sin{x}\)
\(\Rightarrow \frac{dy}{dx}-\frac{\cos{x}}{2y}\frac{dy}{dx}=-\frac{1}{2}\sin{x}\)
\(\Rightarrow \frac{dy}{dx}\left(1-\frac{\cos{x}}{2y}\right)=-\frac{1}{2}\sin{x}\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{2y-\cos{x}}{2y}\right)=-\frac{1}{2}\sin{x}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{2}\sin{x}\times{\frac{2y}{2y-\cos{x}}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y\sin{x}}{2y-\cos{x}}\)
\(\therefore \frac{dy}{dx}=\frac{y\sin{x}}{\cos{x}-2y}\)
(Showed)

দেওয়া আছে,
\(y=\sqrt{1-x^2}\sin^{-1}{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{1-x^2}\sin^{-1}{x})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\sqrt{1-x^2}\frac{d}{dx}(\sin^{-1}{x})+\sin^{-1}{x}\frac{d}{dx}(\sqrt{1-x^2})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{dx}=\sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}}+\sin^{-1}{x}\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}(1-x^2)\) ➜ Note \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}, \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\) এবং \((1-x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে ।
\(\Rightarrow \frac{dy}{dx}=1+\sin^{-1}{x}\frac{1}{2\sqrt{1-x^2}}(0-2x)\) ➜ Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=1-\frac{x\sin^{-1}{x}}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{dy}{dx}=1-\frac{x\sqrt{1-x^2}\sin^{-1}{x}}{\sqrt{1-x^2}\times{\sqrt{1-x^2}}}\) ➜ Note লব ও হরের সহিত \(\sqrt{1-x^2}\) গুণ করে।
\(\Rightarrow \frac{dy}{dx}=1-\frac{xy}{1-x^2}\) ➜ Note \(\because y=\sqrt{1-x^2}\sin^{-1}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1-x^2-xy}{1-x^2}\)
\(\Rightarrow (1-x^2)\frac{dy}{dx}=1-x^2-xy\)
\(\therefore (1-x^2)\frac{dy}{dx}=1-xy-x^2\)
(Showed)