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

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

দেওয়া আছে,
\(x^y=e^{x-y}\)
\(\Rightarrow \ln{x^y}=\ln{e^{x-y}}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow y\ln{x}=(x-y)\ln{e}\) ➜Note\(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow y\ln{x}=x-y\) ➜Note\(\because \ln{e}=1\)
\(\Rightarrow y+y\ln{x}=x\)
\(\Rightarrow y(1+\ln{x})=x\)
\(\Rightarrow y=\frac{x}{1+\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dx}\left(\frac{x}{1+\ln{x}}\right)\) |Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{(1+\ln{x})\frac{dy}{dx}(x)-x\frac{d}{dx}(1+\ln{x})}{(1+\ln{x})^2}\) |Note \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{(1+\ln{x}).1-x(0+\frac{1}{x})}{(1+\ln{x})^2}\) |Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1+\ln{x}-1}{(1+\ln{x})^2}\)
\(\therefore \frac{dy}{dx}=\frac{\ln{x}}{(1+\ln{x})^2}\)

দেওয়া আছে,
\(\sqrt{x}+\sin{y}=x^2\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x}+\sin{y})=\frac{d}{dx}(x^2)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\sqrt{x})+\frac{d}{dx}(\sin{y})=2x\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{1}{2\sqrt{x}}+\cos{y}\frac{d}{dx}(y)=2x\) ➜Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}+\cos{y}\frac{dy}{dx}=2x\)
\(\Rightarrow \cos{y}\frac{dy}{dx}=2x-\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \cos{y}\frac{dy}{dx}=\frac{4x\sqrt{x}-1}{2\sqrt{x}}\)
\(\therefore \frac{dy}{dx}=\frac{4x\sqrt{x}-1}{2\sqrt{x}\cos{y}}\)

দেওয়া আছে,
\(4x^4-x^2y+2y^3=0\)
\(\Rightarrow \frac{d}{dx}(4x^4-x^2y+2y^3)=\frac{d}{dx}(0)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(4x^4)-\frac{d}{dx}(x^2y)+\frac{d}{dx}(2y^3)=0\) ➜Note \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow 16x^3-\left(x^2\frac{d}{dx}(y)+y\frac{d}{dx}(x^2)\right)+6y^2\frac{d}{dx}(y)=0\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow 16x^3-\left(x^2\frac{dy}{dx}+y.2x\right)+6y^2\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 16x^3-x^2\frac{dy}{dx}-2xy+6y^2\frac{dy}{dx}=0\)
\(\Rightarrow -x^2\frac{dy}{dx}+6y^2\frac{dy}{dx}=2xy-16x^3\)
\(\Rightarrow \frac{dy}{dx}(6y^2-x^2)=2xy-16x^3\)
\(\Rightarrow \frac{dy}{dx}=\frac{2xy-16x^3}{6y^2-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{-2x(8x^2-y)}{-(x^2-6y^2)}\)
\(\therefore \frac{dy}{dx}=\frac{2x(8x^2-y)}{x^2-6y^2}\)

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

দেওয়া আছে,
\(x=a(\theta-\sin{\theta})\) এবং \(y=a(1-\cos{\theta})\)
\(\Rightarrow x=a(\theta-\sin{\theta})\)
\(\Rightarrow \frac{dx}{d\theta}=\frac{d}{d\theta}a(\theta-\sin{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a\left(\frac{d}{d\theta}(\theta)-\frac{d}{d\theta}(\sin{\theta})\right)\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\theta}=a(1-\cos{\theta})\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dx}{d\theta}=a(1-\cos{\theta}) ....(1)\)
আবার,
\(y=a(1-\cos{\theta})\)
\(\Rightarrow y=a(1-\cos{\theta})\)
\(\Rightarrow \frac{dy}{d\theta}=\frac{d}{d\theta}a(1-\cos{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\left(\frac{d}{d\theta}(1)-\frac{d}{d\theta}(\cos{\theta})\right)\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=a(0+\sin{\theta})\) ➜Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\therefore \frac{dy}{d\theta}=a\sin{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=a\sin{\theta}\times{\frac{1}{a(1-\cos{\theta})}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\theta}}{1-\cos{\theta}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{2\sin^2{\frac{\theta}{2}}}\) ➜Note \(\because \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}, 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}\)
\(\therefore \frac{dy}{dx}=\cot{\frac{\theta}{2}}\)

দেওয়া আছে,
\(x=at^2\) এবং \(y=2at\)
\(\Rightarrow x=at^2\)
\(\Rightarrow \frac{dx}{dt}=a\frac{d}{dt}(t^2)\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=2at\) ➜Note \(\because \frac{d}{dx}(x^2)=2x\)
\(\therefore \frac{dx}{dt}=2at ....(1)\)
আবার,
\(y=2at\)
\(\Rightarrow y=2at\)
\(\Rightarrow \frac{dy}{dt}=2a\frac{d}{dt}(t)\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=2a\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dt}=2a ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=2a\times{\frac{1}{2at}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\therefore \frac{dy}{dx}=\frac{1}{t}\)

দেওয়া আছে,
\(\tan{y}=\frac{2t}{1-t^2}\) এবং \(\sin{x}=\frac{2t}{1+t^2}\)
\(\Rightarrow \tan{y}=\frac{2t}{1-t^2}\)
\(\Rightarrow \tan{y}=\frac{2\tan{\theta}}{1-\tan^2{\theta}}\)
ধরি,
\(\tan{\theta}=t\Rightarrow \theta=\tan^{-1}(t)\)
\(\Rightarrow \tan{y}=\tan{2\theta}\)
\(\Rightarrow y=2\theta\)
\(\Rightarrow y=2\tan^{-1}{t}\)
\(\Rightarrow \frac{dy}{dt}=2\frac{d}{dt}(\tan^{-1}{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=2\frac{1}{1+t^2}\) ➜Note \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\therefore \frac{dy}{dt}=\frac{2}{1+t^2} ....(1)\)
আবার,
\(\sin{x}=\frac{2t}{1+t^2}\)
\(\Rightarrow \sin{x}=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\)
ধরি, \(\tan{\theta}=t\Rightarrow \theta=\tan^{-1}(t)\)
\(\Rightarrow \sin{x}=\sin{2\theta}\)
\(\Rightarrow x=2\theta\)
\(\Rightarrow x=2\tan^{-1}{t}\)
\(\Rightarrow \frac{dx}{dt}=2\frac{d}{dt}(\tan^{-1}{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=2\frac{1}{1+t^2}\) ➜Note \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\therefore \frac{dx}{dt}=\frac{2}{1+t^2} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{1+t^2}\times{\frac{1}{\frac{2}{1+t^2}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\therefore \frac{dy}{dx}=1\)

দেওয়া আছে,
\(x=a\cos^3{\theta}\) এবং \(y=a\sin^3{\theta}\)
\(\Rightarrow x=a\cos^3{\theta}\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\cos^3{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=3\cos^2{\theta}\frac{d}{d\theta}(\cos{\theta})\) ➜Note\(\cos{\theta}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dx}{d\theta}=-3\cos^2{\theta}\sin{\theta}\) ➜Note\(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\therefore \frac{dx}{d\theta}=-3\cos^2{\theta}\sin{\theta} ....(1)\)
আবার,
\(\Rightarrow y=a\sin^3{\theta}\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\sin^3{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=3\sin^2{\theta}\frac{d}{d\theta}(\sin{\theta})\) ➜Note\(\sin{\theta}\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dy}{d\theta}=3\sin^2{\theta}\cos{\theta}\) ➜Note\(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dy}{d\theta}=3\sin^2{\theta}\cos{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=3\sin^2{\theta}\cos{\theta}\times{\frac{1}{-3\cos^2{\theta}\sin{\theta}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\sin{\theta}\times{\frac{1}{-\cos{\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{\sin{\theta}}{\cos{\theta}}\)
\(\therefore \frac{dy}{dx}=-\tan{\theta}\)

দেওয়া আছে,
\(x=e^t\cos{t}\) এবং \(y=e^t\sin{t}\)
\(\Rightarrow x=e^t\cos{t}\)
\(\Rightarrow \frac{dx}{dt}=\frac{d}{dt}(e^t\cos{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=e^t\frac{d}{dt}(\cos{t})+\cos{t}\frac{d}{dt}(e^t)\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dx}{dt}=-e^t\sin{t}+\cos{t}e^t\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\therefore \frac{dx}{dt}=e^t(\cos{t}-\sin{t}) ....(1)\)
আবার,
\(\Rightarrow y=e^t\sin{t}\)
\(\Rightarrow \frac{dy}{dt}=\frac{d}{dt}(e^t\sin{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=e^t\frac{d}{dt}(\sin{t})+\sin{t}\frac{d}{dt}(e^t)\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{dt}=e^t\cos{t}+\sin{t}e^t\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\therefore \frac{dy}{dt}=e^t(\cos{t}+\sin{t}) ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=e^t(\cos{t}+\sin{t})\times{\frac{1}{e^t(\cos{t}-\sin{t})}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=(\cos{t}+\sin{t})\times{\frac{1}{(\cos{t}-\sin{t})}}\)
\(\therefore \frac{dy}{dx}=\frac{\sin{t}+\cos{t}}{\cos{t}-\sin{t}}\)

দেওয়া আছে,
\(x^yy^x=1\)
\(\Rightarrow \ln{x^yy^x}=\ln{1}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{x^y}+\ln{y^x}=0\) ➜Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}, \ln{1}=0\)
\(\Rightarrow y\ln{x}+x\ln{y}=0\) ➜Note \(\because \ln{(M)^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(y\ln{x}+x\ln{y})=\frac{d}{dx}(0)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y\ln{x})+\frac{d}{dx}(x\ln{y})=0\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(y)+x\frac{d}{dx}(\ln{y})+\ln{y}\frac{d}{dx}(x)=0\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y\frac{1}{x}+\ln{x}\frac{dy}{dx}+x\frac{1}{y}\frac{dy}{dx}+\ln{y}.1=0\) ➜Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{y}{x}+\ln{x}\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\ln{y}=0\)
\(\Rightarrow \ln{x}\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}=-\ln{y}-\frac{y}{x}\)
\(\Rightarrow \frac{dy}{dx}\left(\ln{x}+\frac{x}{y}\right)=-\left(\ln{y}+\frac{y}{x}\right)\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{y\ln{x}+x}{y}\right)=-\frac{x\ln{y}+y}{x}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{x\ln{y}+y}{x}\times{\frac{y}{y\ln{x}+x}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{-y\ln{x}+y}{x}\times{\frac{y}{-x\ln{y}+x}}\) ➜Note \(\because y\ln{x}+x\ln{y}=0\Rightarrow x\ln{y}=-y\ln{x}\Rightarrow y\ln{x}=-x\ln{y}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y-y\ln{x}}{x}\times{\frac{y}{x-x\ln{y}}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(1-\ln{x})}{x}\times{\frac{y}{x(1-\ln{y})}}\)
\(\therefore \frac{dy}{dx}=-\frac{y^2}{x^2}.\frac{1-ln{x}}{1-\ln{y}}\)

দেওয়া আছে,
\(x^4+x^2y^2+y^4=1\)
\(\Rightarrow \frac{d}{dx}(x^4+x^2y^2+y^4)=\frac{d}{dx}(1)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^4)+\frac{d}{dx}(x^2y^2)+\frac{d}{dx}(y^4)=0\) ➜Note \(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w), \frac{d}{dx}(c)=0\)
\(\Rightarrow 4x^3+x^2\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x^2)+4y^3\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x^4)=4x^3, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(y^4)=4y^3\frac{dy}{dx}\)
\(\Rightarrow 4x^3+x^2.2y\frac{dy}{dx}+y^2.2x+4y^3\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(y^2)=2y\frac{dy}{dx}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 4x^3+2x^2y\frac{dy}{dx}+2xy^2+4y^3\frac{dy}{dx}=0\)
\(\Rightarrow 2x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}=-4x^3-2xy^2\)
\(\Rightarrow \frac{dy}{dx}(2x^2y+4y^3)=-2x(2x^2+y^2)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{2x(2x^2+y^2)}{(2x^2y+4y^3)}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{2x(2x^2+y^2)}{2y(x^2+2y^2)}\)
\(\therefore \frac{dy}{dx}=-\frac{x(2x^2+y^2)}{y(x^2+2y^2)}\)

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

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

দেওয়া আছে,
\(e^{xy}-4xy=2\)
\(\Rightarrow \frac{d}{dx}(e^{xy}-4xy)=\frac{d}{dx}(2)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(e^{xy})-4\frac{d}{dx}(xy)=0\) ➜Note\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow e^{xy}\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)-4\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)=0\) ➜Note\((xy)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow e^{xy}\left(x\frac{dy}{dx}+y.1\right)-4\left(x\frac{dy}{dx}+y.1\right)=0\) ➜Note\(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow xe^{xy}\frac{dy}{dx}-4x\frac{dy}{dx}=4y-ye^{xy}\)
\(\Rightarrow \frac{dy}{dx}(xe^{xy}-4x)=4y-ye^{xy}\)
\(\Rightarrow \frac{dy}{dx}=\frac{4y-ye^{xy}}{(xe^{xy}-4x)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{y(4-e^{xy})}{-x(4-e^{xy})}\)
\(\therefore \frac{dy}{dx}=-\frac{y}{x}\)

দেওয়া আছে,
\(x^2-xy+y^2=3\)
\(\Rightarrow \frac{d}{dx}(x^2-xy+y^2)=\frac{d}{dx}(3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^2)-\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=0\) ➜Note\(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w), \frac{d}{dx}(c)=0\)
\(\Rightarrow 2x-\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)+2y\frac{dy}{dx}=0\) ➜Note\(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\)
\(\Rightarrow 2x-x\frac{dy}{dx}-y+2y\frac{dy}{dx}=0\)
\(\Rightarrow -x\frac{dy}{dx}+2y\frac{dy}{dx}=y-2x\)
\(\Rightarrow \frac{dy}{dx}(2y-x)=y-2x\)
\(\Rightarrow \frac{dy}{dx}=\frac{-(2x-y)}{-(x-2y)}\)
\(\therefore \frac{dy}{dx}=\frac{2x-y}{x-2y}\)

দেওয়া আছে,
\(x^3y+xy^3=2\)
\(\Rightarrow \frac{d}{dx}(x^3y+xy^3)=\frac{d}{dx}(2)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^3y)+\frac{d}{dx}(xy^3)=0\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow x^3\frac{d}{dx}(y)+y\frac{d}{dx}(x^3)+x\frac{d}{dx}(y^3)+y^3\frac{d}{dx}(x)=0\) ➜Note\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow x^3\frac{dy}{dx}+y.3x^2+x.3y^2\frac{dy}{dx}+y^3.1=0\) ➜Note\(\because \frac{d}{dx}(x^3)=3x^2, \frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}, \frac{d}{dx}(x)=1\)
\(\Rightarrow x^3\frac{dy}{dx}+3x^2y+3xy^2\frac{dy}{dx}+y^3=0\)
\(\Rightarrow x^3\frac{dy}{dx}+3xy^2\frac{dy}{dx}=-y^3-3x^2y\)
\(\Rightarrow \frac{dy}{dx}(x^3+3xy^2)=-y(3x^2+y^2)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(3x^2+y^2)}{(x^3+3xy^2)}\)
\(\therefore \frac{dy}{dx}=-\frac{y(3x^2+y^2)}{x(x^2+3y^2)}\)

দেওয়া আছে,
\(x^3+y^3=3axy\)
\(\Rightarrow \frac{d}{dx}(x^3+y^3)=\frac{d}{dx}(3axy)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^3)+\frac{d}{dx}(y^3)=3a\frac{d}{dx}(xy)\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow 3x^2+3y^2\frac{dy}{dx}=3a\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)\) ➜Note\(\because \frac{d}{dx}(x^3)=3x^2, \frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow 3x^2+3y^2\frac{dy}{dx}=3a\left(x\frac{dy}{dx}+y.1\right)\) ➜Note\(\because \frac{d}{dx}(y)=\frac{dy}{dx}, \frac{d}{dx}(x)=1\)
\(\Rightarrow 3x^2+3y^2\frac{dy}{dx}=3ax\frac{dy}{dx}+3ay\)
\(\Rightarrow 3y^2\frac{dy}{dx}-3ax\frac{dy}{dx}=3ay-3x^2\)
\(\Rightarrow \frac{dy}{dx}(3y^2-3ax)=3ay-3x^2\)
\(\Rightarrow \frac{dy}{dx}=\frac{3ay-3x^2}{(3y^2-3ax)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{3(ay-x^2)}{3(y^2-ax)}\)
\(\therefore \frac{dy}{dx}=\frac{ay-x^2}{y^2-ax}\)

দেওয়া আছে,
\(\sqrt{x}+\sqrt{y}=\sqrt{a}\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x}+\sqrt{y})=\frac{d}{dx}(\sqrt{a})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\sqrt{x})+\frac{d}{dx}(\sqrt{y})=0\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0\) ➜Note\(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(\sqrt{y})=\frac{1}{2\sqrt{y}}\frac{dy}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{y}}\frac{dy}{dx}=-\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{2\sqrt{x}}\times{\frac{2\sqrt{y}}{1}}\)
\(\therefore \frac{dy}{dx}=--\frac{\sqrt{y}}{\sqrt{x}}\)

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

দেওয়া আছে,
\(x+y=xy^2\)
\(\Rightarrow \frac{d}{dx}(x+y)=\frac{d}{dx}(xy^2)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x)+\frac{d}{dx}(y)=x\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x)\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\), \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow 1+\frac{dy}{dx}=x.2y\frac{dy}{dx}+y^2.1\) ➜Note\(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(y^2)=2y\frac{dy}{dx}\)
\(\Rightarrow 1+\frac{dy}{dx}=2xy\frac{dy}{dx}+y^2\)
\(\Rightarrow \frac{dy}{dx}-2xy\frac{dy}{dx}=y^2-1\)
\(\Rightarrow \frac{dy}{dx}(1-2xy)=y^2-1\)
\(\therefore \frac{dy}{dx}=\frac{y^2-1}{1-2xy}\)

দেওয়া আছে,
\(x^ay^b=(x-y)^{a+b}\)
\(\Rightarrow \ln{x^ay^b}=\ln{(x-y)^{a+b}}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{x^a}+\ln{y^b}=\ln{(x-y)^{a+b}}\) ➜Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow a\ln{x}+b\ln{y}=(a+b)\ln{(x-y)}\) ➜Note \(\because \ln{(M)^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(a\ln{x}+b\ln{y})=(a+b)\frac{d}{dx}\{\ln{(x-y)}\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow a\frac{d}{dx}(\ln{x})+b\frac{d}{dx}(\ln{y})=(a+b)\frac{d}{dx}\{\ln{(x-y)}\}\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow a\frac{1}{x}+b\frac{1}{y}\frac{dy}{dx}=(a+b)\frac{1}{x-y}\frac{d}{dx}(x-y)\) ➜Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\) এবং \((x-y)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{a}{x}+\frac{b}{y}\frac{dy}{dx}=\frac{a+b}{x-y}\left(1-\frac{dy}{dx}\right)\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{a}{x}+\frac{b}{y}\frac{dy}{dx}=\frac{a+b}{x-y}-\frac{a+b}{x-y}\frac{dy}{dx}\)
\(\Rightarrow \frac{b}{y}\frac{dy}{dx}+\frac{a+b}{x-y}\frac{dy}{dx}=\frac{a+b}{x-y}-\frac{a}{x}\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{b}{y}+\frac{a+b}{x-y}\right)=\frac{a+b}{x-y}-\frac{a}{x}\)
\(\Rightarrow \frac{dy}{dx}\frac{bx-by+ay+by}{y(x-y)}=\frac{ax+bx-ax+ay}{x(x-y)}\)
\(\Rightarrow \frac{dy}{dx}\frac{bx+ay}{y(x-y)}=\frac{bx+ay}{x(x-y)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{bx+ay}{x(x-y)}\times{\frac{y(x-y)}{bx+ay}}\)
\(\therefore \frac{dy}{dx}=\frac{y}{x}\)

দেওয়া আছে,
\(x^yy^x=1\)
\(\Rightarrow \ln{x^yy^x}=\ln{1}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{x^y}+\ln{y^x}=0\) ➜Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}, \ln{1}=0\)
\(\Rightarrow y\ln{x}+x\ln{y}=0\) ➜Note \(\because \ln{(M)^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(y\ln{x}+x\ln{y})=\frac{d}{dx}(0)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y\ln{x})+\frac{d}{dx}(x\ln{y})=0\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow y\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(y)+x\frac{d}{dx}(\ln{y})+\ln{y}\frac{d}{dx}(x)=0\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y\frac{1}{x}+\ln{x}\frac{dy}{dx}+x\frac{1}{y}\frac{dy}{dx}+\ln{y}.1=0\) ➜Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{y}{x}+\ln{x}\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\ln{y}=0\)
\(\Rightarrow \ln{x}\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}=-\ln{y}-\frac{y}{x}\)
\(\Rightarrow \frac{dy}{dx}\left(\ln{x}+\frac{x}{y}\right)=-\left(\ln{y}+\frac{y}{x}\right)\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{y\ln{x}+x}{y}\right)=-\frac{x\ln{y}+y}{x}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{x\ln{y}+y}{x}\times{\frac{y}{y\ln{x}+x}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{-y\ln{x}+y}{x}\times{\frac{y}{-x\ln{y}+x}}\) ➜Note \(\because y\ln{x}+x\ln{y}=0\Rightarrow x\ln{y}=-y\ln{x}\Rightarrow y\ln{x}=-x\ln{y}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y-y\ln{x}}{x}\times{\frac{y}{x-x\ln{y}}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(1-\ln{x})}{x}\times{\frac{y}{x(1-\ln{y})}}\)
\(\therefore \frac{dy}{dx}=-\frac{y^2}{x^2}.\frac{1-ln{x}}{1-\ln{y}}\)

দেওয়া আছে,
\(xy+x^2y^2=c\)
\(\Rightarrow \frac{d}{dx}(xy+x^2y^2)=\frac{d}{dx}(c)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(xy)+\frac{d}{dx}(x^2y^2)=0\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow x\frac{d}{dx}(y)+y\frac{d}{dx}(x)+x^2\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x^2)=0\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow x\frac{dy}{dx}+y.1+x^2.2y\frac{dy}{dx}+y^2.2x=0\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(y^2)=2y\frac{dy}{dx}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow x\frac{dy}{dx}+y+2x^2y\frac{dy}{dx}+2xy^2=0\)
\(\Rightarrow x\frac{dy}{dx}+2x^2y\frac{dy}{dx}=-2xy^2-y\)
\(\Rightarrow \frac{dy}{dx}(x+2x^2y)=-y(2xy+1)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(2xy+1)}{(x+2x^2y)}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(2xy+1)}{x(2xy+1)}\)
\(\therefore \frac{dy}{dx}=-\frac{y}{x}\)

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

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

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

দেওয়া আছে,
\(x^4+x^2y^2+y^4=0\)
\(\Rightarrow \frac{d}{dx}(x^4+x^2y^2+y^4)=\frac{d}{dx}(0)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^4)+\frac{d}{dx}(x^2y^2)+\frac{d}{dx}(y^4)=0\) ➜Note \(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\), \(\frac{d}{dx}(0)=0\)
\(\Rightarrow 4x^3+\left(x^2\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x^2)\right)+4y^3\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x^4)=4x^3, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^4)=4y^3\frac{dy}{dx}\)
\(\Rightarrow 4x^3+\left(x^2.2y\frac{dy}{dx}+y^2.2x\right)+4y^3\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(y^2)=2y\frac{dy}{dx}\)
\(\Rightarrow 4x^3+2x^2y\frac{dy}{dx}+2xy^2+4y^3\frac{dy}{dx}=0\)
\(\Rightarrow \frac{dy}{dx}(2x^2y+4y^3)=-2xy^2-4x^3\)
\(\Rightarrow \frac{dy}{dx}=\frac{-2xy^2-4x^3}{2x^2y+4y^3}\)
\(\Rightarrow \frac{dy}{dx}=\frac{-2x(2x^2+y^2)}{2y(x^2+2y^2)}\)
\(\therefore \frac{dy}{dx}=-\frac{x(2x^2+y^2)}{y(x^2+2y^2)}\)

দেওয়া আছে,
\(x^4+y^4=3axy\)
\(\Rightarrow \frac{d}{dx}(x^4+y^4)=\frac{d}{dx}(3axy)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^4)+\frac{d}{dx}(y^4)=3a\frac{d}{dx}(xy)\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow 4x^3+4y^3\frac{dy}{dx}=3a\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)\) ➜Note \(\because \frac{d}{dx}(x^4)=4x^3, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^4)=4y^3\frac{dy}{dx}\)
\(\Rightarrow 4x^3+4y^3\frac{dy}{dx}=3a\left(x\frac{dy}{dx}+y.1\right)\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow 4x^3+4y^3\frac{dy}{dx}=3ax\frac{dy}{dx}+3ay\)
\(\Rightarrow 4y^3\frac{dy}{dx}-3ax\frac{dy}{dx}=3ay-4x^3\)
\(\Rightarrow \frac{dy}{dx}(4y^3-3ax)=3ay-4x^3\)
\(\Rightarrow \frac{dy}{dx}=\frac{3ay-4x^3}{4y^3-3ax}\)
\(\Rightarrow \frac{dy}{dx}=\frac{xy(3ay-4x^3)}{xy(4y^3-3ax})\) ➜Note ডান পার্শে লব ও হর কে \(xy\) দ্বারা গুণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{y(3axy-4x^4)}{x(4y^4-3axy})\)
\(\Rightarrow \frac{dy}{dx}=\frac{y(x^4+y^4-4x^4)}{x(4y^4-x^4-y^4})\) ➜Note \(\because x^4+y^4=3axy\)
\(\therefore \frac{dy}{dx}=\frac{y(y^4-3x^4)}{x(3y^4-x^4})\)

দেওয়া আছে,
\(x^my^n=(x-y)^{m+n}\)
\(\Rightarrow \ln{x^my^n}=\ln{(x-y)^{m+n}}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{x^m}+\ln{y^n}=\ln{(x-y)^{m+n}}\) ➜Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow m\ln{x}+n\ln{y}=(m+n)\ln{(x-y)}\) ➜Note \(\because \ln{(M)^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(m\ln{x}+n\ln{y})=(m+n)\frac{d}{dx}\{\ln{(x-y)}\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow m\frac{d}{dx}(\ln{x})+n\frac{d}{dx}(\ln{y})=(m+n)\frac{d}{dx}\{\ln{(x-y)}\}\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow m\frac{1}{x}+n\frac{1}{y}\frac{dy}{dx}=(m+n)\frac{1}{x-y}\frac{d}{dx}(x-y)\) ➜Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\) এবং \((x-y)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x-y}\left(1-\frac{dy}{dx}\right)\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x-y}-\frac{m+n}{x-y}\frac{dy}{dx}\)
\(\Rightarrow \frac{n}{y}\frac{dy}{dx}+\frac{m+n}{x-y}\frac{dy}{dx}=\frac{m+n}{x-y}-\frac{m}{x}\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{n}{y}+\frac{m+n}{x-y}\right)=\frac{m+n}{x-y}-\frac{m}{x}\)
\(\Rightarrow \frac{dy}{dx}\frac{nx-ny+my+ny}{y(x-y)}=\frac{mx+nx-mx+my}{x(x-y)}\)
\(\Rightarrow \frac{dy}{dx}\frac{nx+my}{y(x-y)}=\frac{nx+my}{x(x-y)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{nx+my}{x(x-y)}\times{\frac{y(x-y)}{nx+my}}\)
\(\therefore \frac{dy}{dx}=\frac{y}{x}\)

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

দেওয়া আছে,
\(1+xy^2+x^2y=0\)
\(\Rightarrow \frac{d}{dx}(1+xy^2+x^2y)=\frac{d}{dx}(0)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(1)+\frac{d}{dx}(xy^2)+\frac{d}{dx}(x^2y)=0\) ➜Note \(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\)
\(\Rightarrow 0+x\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x)+x^2\frac{d}{dx}(y)+y\frac{d}{dx}(x^2)=0\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(\Rightarrow x.2y\frac{dy}{dx}+y^2.1+x^2\frac{dy}{dx}+y.2x=0\) ➜Note \(\because \frac{d}{dx}(y^2)=2y\frac{dy}{dx}, \frac{d}{dx}(x)=1, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 2xy\frac{dy}{dx}+y^2+x^2\frac{dy}{dx}+2xy=0\)
\(\Rightarrow 2xy\frac{dy}{dx}+x^2\frac{dy}{dx}=-2xy-y^2\)
\(\Rightarrow \frac{dy}{dx}(2xy+x^2)=-y(y+2x)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(y+2x)}{(2xy+x^2)}\)
\(\therefore \frac{dy}{dx}=-\frac{y(y+2x)}{x(x+2y)}\)

দেওয়া আছে,
\(ax^2+2hxy+by^2+2gx+2fy+c=0\)
\(\Rightarrow \frac{d}{dx}(ax^2+2hxy+by^2+2gx+2fy+c)=\frac{d}{dx}(0)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow a\frac{d}{dx}(x^2)+2h\frac{d}{dx}(xy)+b\frac{d}{dx}(y^2)+2g\frac{d}{dx}(x)+2f\frac{d}{dx}(y)+2f\frac{d}{dx}(c)=0\) ➜Note \(\because \frac{d}{dx}(M+N+P+Q+R+S)=\frac{d}{dx}(M)+\frac{d}{dx}(N)\)\(+\frac{d}{dx}(P)+\frac{d}{dx}(Q)+\frac{d}{dx}(R)+\frac{d}{dx}(S)\)
\(\Rightarrow a.2x+2h\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)+b.2y\frac{dy}{dx}+2g.1+2f\frac{dy}{dx}+0=0\) ➜Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow 2ax+2h\left(x\frac{dy}{dx}+y.1\right)+2by\frac{dy}{dx}+2g+2f\frac{dy}{dx}=0\)
\(\Rightarrow 2ax+2hx\frac{dy}{dx}+2hy+2by\frac{dy}{dx}+2g+2f\frac{dy}{dx}=0\)
\(\Rightarrow 2hx\frac{dy}{dx}+2by\frac{dy}{dx}+2f\frac{dy}{dx}=-2ax-2hy-2g\)
\(\Rightarrow \frac{dy}{dx}(2hx+2by+2f)=-2ax-2hy-2g\)
\(\Rightarrow \frac{dy}{dx}=-\frac{2(ax+hy+g)}{2hx+2by+2f}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{2(ax+hy+g)}{2(hx+by+f)}\)
\(\therefore \frac{dy}{dx}=-\frac{ax+hy+g}{hx+by+f}\)

দেওয়া আছে,
\(xy+x^2y^2=3\)
\(\Rightarrow \frac{d}{dx}(xy+x^2y^2)=\frac{d}{dx}(3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(xy)+\frac{d}{dx}(x^2y^2)=0\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow x\frac{d}{dx}(y)+y\frac{d}{dx}(x)+x^2\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x^2)=0\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow x\frac{dy}{dx}+y.1+x^2.2y\frac{dy}{dx}+y^2.2x=0\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(y^2)=2y\frac{dy}{dx}, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow x\frac{dy}{dx}+y+2x^2y\frac{dy}{dx}+2xy^2=0\)
\(\Rightarrow x\frac{dy}{dx}+2x^2y\frac{dy}{dx}=-2xy^2-y\)
\(\Rightarrow \frac{dy}{dx}(x+2x^2y)=-y(2xy+1)\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(2xy+1)}{(x+2x^2y)}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y(2xy+1)}{x(2xy+1)}\)
\(\therefore \frac{dy}{dx}=-\frac{y}{x}\)

দেওয়া আছে,
\(3x^4-x^2y+2y^3=1\)
\(\Rightarrow \frac{d}{dx}(3x^4-x^2y+2y^3)=\frac{d}{dx}(1)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 3\frac{d}{dx}(x^4)-\frac{d}{dx}(x^2y)+2\frac{d}{dx}(y^3)=0\) ➜Note \(\because \frac{d}{dx}(u-v+w)=\frac{d}{dx}(u)-\frac{d}{dx}(v)+\frac{d}{dx}(w)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow 3.4x^3-\left(x^2\frac{d}{dx}(y)+y\frac{d}{dx}(x^2)\right)+2.3y^2\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x^4)=4x^3, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}\)
\(\Rightarrow 12x^3-\left(x^2\frac{dy}{dx}+y.2x\right)+6y^2\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 12x^3-x^2\frac{dy}{dx}-2xy+6y^2\frac{dy}{dx}=0\)
\(\Rightarrow -x^2\frac{dy}{dx}+6y^2\frac{dy}{dx}=2xy-12x^3\)
\(\Rightarrow \frac{dy}{dx}(-x^2+6y^2)=2xy-12x^3\)
\(\Rightarrow \frac{dy}{dx}=\frac{2xy-12x^3}{-x^2+6y^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{-2x(6x^2-y)}{-(x^2-6y^2)}\)
\(\therefore \frac{dy}{dx}=\frac{2x(6x^2-y)}{x^2-6y^2}\)

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

দেওয়া আছে,
\(x^py^q=(x-y)^{p+q}\)
\(\Rightarrow \ln{x^py^q}=\ln{(x-y)^{p+q}}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{x^p}+\ln{y^q}=\ln{(x-y)^{p+q}}\) ➜Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(N)}\)
\(\Rightarrow p\ln{x}+q\ln{y}=(p+q)\ln{(x-y)}\) ➜Note \(\because \ln{(M)^n}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(p\ln{x}+q\ln{y})=(p+q)\frac{d}{dx}\{\ln{(x-y)}\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow p\frac{d}{dx}(\ln{x})+q\frac{d}{dx}(\ln{y})=(p+q)\frac{d}{dx}\{\ln{(x-y)}\}\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow p\frac{1}{x}+q\frac{1}{y}\frac{dy}{dx}=(p+q)\frac{1}{x-y}\frac{d}{dx}(x-y)\) ➜Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}\) এবং \((x-y)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে।
\(\Rightarrow \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{p+q}{x-y}\left(1-\frac{dy}{dx}\right)\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{p+q}{x-y}-\frac{p+q}{x-y}\frac{dy}{dx}\)
\(\Rightarrow \frac{q}{y}\frac{dy}{dx}+\frac{p+q}{x-y}\frac{dy}{dx}=\frac{p+q}{x-y}-\frac{p}{x}\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{q}{y}+\frac{p+q}{x-y}\right)=\frac{p+q}{x-y}-\frac{p}{x}\)
\(\Rightarrow \frac{dy}{dx}\frac{qx-qy+py+qy}{y(x-y)}=\frac{px+qx-px+py}{x(x-y)}\)
\(\Rightarrow \frac{dy}{dx}\frac{qx+py}{y(x-y)}=\frac{qx+py}{x(x-y)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{qx+py}{x(x-y)}\times{\frac{y(x-y)}{qx+py}}\)
\(\therefore \frac{dy}{dx}=\frac{y}{x}\)

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

দেওয়া আছে,
\(y=\sqrt{x\sqrt{x\sqrt{x ......\infty}}}\)
\(\Rightarrow y=\sqrt{xy}\) ➜Note \(\because y=\sqrt{x\sqrt{x\sqrt{x ......\infty}}}\)
\(\Rightarrow y^2=xy\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow y^2-xy=0\)
\(\Rightarrow y(y-x)=0\)
\(\Rightarrow y-x=0, y\ne{0}\)
\(\Rightarrow y=x\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\therefore \frac{dy}{dx}=1\) ➜Note \(\because \frac{d}{dx}(x)=1\)

দেওয়া আছে,
\(x^{y^n}=y^{x^n}\)
\(\Rightarrow \ln{x^{y^n}}=\ln{y^{x^n}}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow y^n\ln{x}=x^n\ln{y}\) ➜Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(y^n\ln{x})=\frac{d}{dx}(x^n\ln{y})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow y^n\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(y^n)=x^n\frac{d}{dx}(\ln{y})+\ln{y}\frac{d}{dx}(x^n)\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow y^n\frac{1}{x}+\ln{x}.ny^{n-1}\frac{dy}{dx}=x^n\frac{1}{y}\frac{dy}{dx}+\ln{y}.nx^{n-1}\) ➜Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{y^n}{x}+ny^{n-1}\ln{x}\frac{dy}{dx}=\frac{x^n}{y}\frac{dy}{dx}+nx^{n-1}\ln{y}\)
\(\Rightarrow ny^{n-1}\ln{x}\frac{dy}{dx}-\frac{x^n}{y}\frac{dy}{dx}=nx^{n-1}\ln{y}-\frac{y^n}{x}\)
\(\Rightarrow \frac{dy}{dx}\left(ny^{n-1}\ln{x}-\frac{x^n}{y}\right)=nx^{n-1}\ln{y}-\frac{y^n}{x}\)
\(\Rightarrow \frac{dy}{dx}\frac{ny^{n}\ln{x}-x^n}{y}=\frac{nx^{n}\ln{y}-y^n}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{nx^{n}\ln{y}-y^n}{x}\times{\frac{y}{ny^{n}\ln{x}-x^n}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{nyy^{n}\ln{x}-y^{n+1}}{nxx^{n}\ln{y}-x^{n+1}}\) ➜Note \(\because y^n\ln{x}=x^n\ln{y}\)
\(\Rightarrow \frac{dy}{dx}=\frac{ny^{n+1}\ln{x}-y^{n+1}}{nx^{n+1}\ln{y}-x^{n+1}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{y^{n+1}(n\ln{x}-1)}{x^{n+1}(n\ln{y}-1)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{y^{n+1}}{x^{n+1}}.\frac{n\ln{x}-1}{n\ln{y}-1}\)
\(\therefore \frac{dy}{dx}=\left(\frac{y}{x}\right)^{n+1}.\frac{n\ln{x}-1}{n\ln{y}-1}\)
(Showed)

দেওয়া আছে,
\(x^3y+x^2y^2+xy^3=3\)
\(\Rightarrow \frac{d}{dx}(x^3y+x^2y^2+xy^3)=\frac{d}{dx}(3)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^3y)+\frac{d}{dx}(x^2y^2)+\frac{d}{dx}(xy^3)=0\) ➜Note \(\because \frac{d}{dx}(u+v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)\), \(\frac{d}{dx}(c)=0\)
\(\Rightarrow x^3\frac{d}{dx}(y)+y\frac{d}{dx}(x^3)+x^2\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x^2)+x\frac{d}{dx}(y^3)+y^3\frac{d}{dx}(x)=0\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow x^3\frac{dy}{dx}+y.3x^2+x^2.2y\frac{dy}{dx}+y^2.2x+x.3y^2\frac{dy}{dx}+y^3.1=0\) ➜Note \(\because \frac{d}{dx}(x^3)=3x^2, \frac{d}{dx}(y^2)=2y\frac{dy}{dx}\), \(\frac{d}{dx}(2^2)=2x, \frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}, \frac{dy}{dx}(x)=1\)
\(\Rightarrow x^3\frac{dy}{dx}+3x^2y+2x^2y\frac{dy}{dx}+2xy^2+3xy^2\frac{dy}{dx}+y^3=0\)
\(\Rightarrow x^3\frac{dy}{dx}+2x^2y\frac{dy}{dx}+3xy^2\frac{dy}{dx}=-y^3-3x^2y-2xy^2\)
\(\Rightarrow \frac{dy}{dx}(x^3+2x^2y+3xy^2)=-y^3-3x^2y-2xy^2\)
\(\Rightarrow \frac{dy}{dx}=-\frac{y^3+3x^2y+2xy^2}{x^3+2x^2y+3xy^2}\)
এখন,
\((1, 1)\) বিন্দুতে
\(\frac{dy}{dx}=-\frac{(1)^3+3.(1)^2.1+2.1.(1)^2}{(1)^3+2.(1)^2.1+3.1.(1)^2}\)
\(=-\frac{1+3.1.1+2.1.1}{1+2.1.1+3.1.1}\)
\(=-\frac{1+3+2}{1+2+3}\)
\(=-\frac{6}{6}\)
\(=-1\)
(proved)

দেওয়া আছে,
\(f(x)=\left(\frac{a+x}{b+x}\right)^{a+b+2x}\)
\(\Rightarrow \ln{f(x)}=\ln{\left(\frac{a+x}{b+x}\right)^{a+b+2x}}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow \ln{f(x)}=(a+b+2x)\ln{\left(\frac{a+x}{b+x}\right)}\) ➜Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow \ln{f(x)}=(a+b+2x)\{\ln{(a+x)}-\ln{(b+x)}\}\) ➜Note \(\because \ln{\frac{M}{N}}=\ln{(M)}-\ln{(N)}\)
\(\Rightarrow \ln{f(x)}=(a+b+2x)\ln{(a+x)}-(a+b+2x)\ln{(b+x)}\)
\(\Rightarrow \frac{d}{dx}(\ln{f(x)})=\frac{d}{dx}\{(a+b+2x)\ln{(a+x)}-(a+b+2x)\ln{(b+x)}\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=\frac{d}{dx}\{(a+b+2x)\ln{(a+x)}\}-\frac{d}{dx}\{(a+b+2x)\ln{(b+x)}\}\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\), \(\frac{d}{dx}(\ln{f(x)})=\frac{1}{f(x)}f^{'}(x)\)
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=(a+b+2x)\frac{d}{dx}\{\ln{(a+x)}\}+\ln{(a+x)}\frac{d}{dx}(a+b+2x)-\)\(\left[(a+b+2x)\frac{d}{dx}\{\ln{(b+x)}\}+\ln{(b+x)}\frac{d}{dx}(a+b+2x)\right]\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=(a+b+2x)\frac{1}{a+x}\frac{d}{dx}(a+x)+\ln{(a+x)}(0+0+2.1)-\)\(\left[(a+b+2x)\frac{1}{b+x}\frac{d}{dx}(b+x)+\ln{(b+x)}(0+0+2.1)\right]\) ➜Note\((a+x)\), \((b+x)\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(c)=0\), \(\frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=(a+b+2x)\frac{1}{a+x}(0+1)+2\ln{(a+x)}-\)\(\left[(a+b+2x)\frac{1}{b+x}(0+1)+2\ln{(b+x)}\right]\)
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=\frac{(a+b+2x)}{a+x}+2\ln{(a+x)}-\frac{(a+b+2x)}{b+x}-2\ln{(b+x)}\)
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=\frac{(a+b+2x)}{a+x}-\frac{(a+b+2x)}{b+x}+2\{\ln{(a+x)}-\ln{(b+x)}\}\)
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=(a+b+2x)\left(\frac{1}{a+x}-\frac{1}{b+x}\right)+2\{\ln{(a+x)}-\ln{(b+x)}\}\)
\(\Rightarrow \frac{1}{f(x)}f^{'}(x)=(a+b+2x)\frac{b+x-a-x}{(a+x)(b+x)}+2\ln{\frac{a+x}{b+x}}\)
\(\Rightarrow f^{'}(x)=\left[(a+b+2x)\frac{b-a}{(a+x)(b+x)}+2\ln{\frac{a+x}{b+x}}\right]\times{f(x)}\)
\(\Rightarrow f^{'}(x)=\left[(a+b+2x)\frac{b-a}{(a+x)(b+x)}+2\ln{\frac{a+x}{b+x}}\right]\)\(\times{\left(\frac{a+x}{b+x}\right)^{a+b+2x}}\) ➜Note \(\because f(x)=\left(\frac{a+x}{b+x}\right)^{a+b+2x}\)
এখন,
\(x=0\) বসিয়ে
\(\Rightarrow f^{'}(0)=\left[(a+b+2.0)\frac{b-a}{(a+0)(b+0)}+2\ln{\frac{a+0}{b+0}}\right]\)\(\times{\left(\frac{a+0}{b+0}\right)^{a+b+2.0}}\)
\(\Rightarrow f^{'}(0)=\left[(a+b)\frac{b-a}{ab}+2\ln{\frac{a}{b}}\right]\left(\frac{a}{b}\right)^{a+b}\)
\(\Rightarrow f^{'}(0)=\left(\frac{(b+a)(b-a)}{ab}+2\ln{\frac{a}{b}}\right)\left(\frac{a}{b}\right)^{a+b}\)
\(\Rightarrow f^{'}(0)=\left(\frac{b^2-a^2}{ab}+2\ln{\frac{a}{b}}\right)\left(\frac{a}{b}\right)^{a+b}\)
\(\therefore f^{'}(0)=\left(2\ln{\frac{a}{b}}+\frac{b^2-a^2}{ab}\right)\left(\frac{a}{b}\right)^{a+b}\)
(proved)

দেওয়া আছে,
\(x\sqrt{(1+y)}+y\sqrt{(1+x)}=0\)
\(\Rightarrow x\sqrt{(1+y)}=-y\sqrt{(1+x)}\)
\(\Rightarrow x^2(1+y)=y^2(1+x)\)
\(\Rightarrow x^2+x^2y=y^2+xy^2\)
\(\Rightarrow x^2+x^2y-y^2-xy^2=0\)
\(\Rightarrow x^2-y^2+x^2y-xy^2=0\)
\(\Rightarrow (x+y)(x-y)+xy(x-y)=0\)
\(\Rightarrow (x-y)(x+y+xy)=0\)
\(\Rightarrow (x-y)\ne{0}, (x+y+xy)=0\)
\(\Rightarrow x+y+xy=0\)
\(\Rightarrow y(1+x)=-x\)
\(\Rightarrow y=-\frac{x}{1+x}\)
\(\Rightarrow \frac{d}{dx}(y)=-\frac{d}{dx}(\frac{x}{1+x})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=-\frac{(1+x)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x)}{(1+x)^2}\) ➜Note \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{(1+x).1-x(0+1)}{(1+x)^2}\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1+x-x}{(1+x)^2}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{(1+x)^2}\)
(Showed)

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

দেওয়া আছে,
\(e^{x-y}=x^y\)
\(\Rightarrow \ln{(e^{x-y})}=\ln{x^y}\) ➜Note উভয় পার্শে \(\ln\) সংযোজন করে।
\(\Rightarrow (x-y)\ln{(e)}=y\ln{x}\) ➜Note \(\because \ln{M^{n}}=n\ln{(M)}\)
\(\Rightarrow x-y=y\ln{x}\) ➜Note \(\because \ln{(e)}=1\)
\(\Rightarrow y\ln{x}=x-y\)
\(\Rightarrow y\ln{x}+y=x\)
\(\Rightarrow y(\ln{x}+1)=x\)
\(\Rightarrow y=\frac{x}{1+\ln{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{1+\ln{x}}\right)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{(1+\ln{x})\frac{d}{dx}(x)-x\frac{d}{dx}(1+\ln{x})}{(1+\ln{x})^2}\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{dx}=\frac{(1+\ln{x}).1-x(0+\frac{1}{x})}{(1+\ln{x})^2}\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1+\ln{x}-1}{(1+\ln{x})^2}\)
\(\therefore \frac{dy}{dx}=\frac{\ln{x}}{(1+\ln{x})^2}\)

দেওয়া আছে,
\(\tan{(x+y)}=a\)
\(\Rightarrow x+y=\tan^{-1}{a}\)
\(\Rightarrow \frac{d}{dx}(x+y)=\frac{d}{dx}(\tan^{-1}{a})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x)+\frac{d}{dx}(y)=0\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow 1+\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\therefore 1+\frac{dy}{dx}=-1\)

দেওয়া আছে,
\(\sin^{-1}{(x+y)}=a\)
\(\Rightarrow x+y=\sin{a}\)
\(\Rightarrow \frac{d}{dx}(x+y)=\frac{d}{dx}(\sin{a})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x)+\frac{d}{dx}(y)=0\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow 1+\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\therefore 1+\frac{dy}{dx}=-1\)

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

দেওয়া আছে,
\(e^x+e^y=e^a\)
\(\Rightarrow \frac{d}{dx}(e^x+e^y)=\frac{d}{dx}(e^a)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(e^x)+\frac{d}{dx}(e^y)=0\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow e^x+e^y\frac{dy}{dx}=0\) ➜Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(e^y)=e^y\frac{dy}{dx}\)
\(\Rightarrow e^y\frac{dy}{dx}=-e^x\)
\(\Rightarrow \frac{dy}{dx}=-\frac{e^x}{e^y}\)
\(\Rightarrow \frac{dy}{dx}=-e^x.e^{-y}\)
\(\therefore \frac{dy}{dx}=-e^{x-y}\)

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

দেওয়া আছে,
\(\ln{(x^ny^n)}=x^n+y^n\)
\(\Rightarrow \ln{(x^n)}+\ln{(y^n)}=x^n+y^n\) ➜Note \(\because \ln{(MN)}=\ln{(M)}+ln{(N)}\)
\(\Rightarrow \frac{d}{dx}\{\ln{(x^n)}+\ln{(y^n)}\}=\frac{d}{dx}(x^n+y^n)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}\{\ln{(x^n)}\}+\frac{d}{dx}\{\ln{(y^n)}=\frac{d}{dx}(x^n)+\frac{d}{dx}(y^n)\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{1}{x^n}\frac{d}{dx}(x^n)+\frac{1}{y^n}\frac{d}{dx}(y^n)=nx^{n-1}+ny^{n-1}\frac{dy}{dx}\) ➜Note \(x^n\), \(y^n\) কে পর্যায়ক্রমে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{1}{x^n}.nx^{n-1}+\frac{1}{y^n}.ny^{n-1}\frac{dy}{dx}=nx^{n-1}+ny^{n-1}\frac{dy}{dx}\) ➜Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{nx^{n-1}}{x^{n-1}.x}+\frac{ny^{n-1}}{y^{n-1}.y}\frac{dy}{dx}=nx^{n-1}+ny^{n-1}\frac{dy}{dx}\)
\(\Rightarrow \frac{n}{x}+\frac{n}{y}\frac{dy}{dx}=nx^{n-1}+ny^{n-1}\frac{dy}{dx}\)
\(\Rightarrow \frac{n}{y}\frac{dy}{dx}-ny^{n-1}\frac{dy}{dx}=nx^{n-1}-\frac{n}{x}\)
\(\Rightarrow \frac{dy}{dx}.\frac{n-ny^n}{y}=\frac{nx^n-n}{x}\)
\(\Rightarrow \frac{dy}{dx}.\frac{n(1-y^n)}{y}=\frac{n(x^n-1)}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{n(x^n-1)}{x}\times{\frac{y}{n(1-y^n)}}\)
\(\therefore \frac{dy}{dx}=\frac{y(x^n-1)}{x(1-y^n)}\)

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

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

দেওয়া আছে,
\(y=\tan{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{\tan{(x+y)}\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\sec^2{(x+y)}\frac{d}{dx}(x+y)\) ➜Note \((x+y)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{dy}{dx}=\sec^2{(x+y)}\left(1+\frac{dy}{dx}\right)\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=\sec^2{(x+y)}+\sec^2{(x+y)}\frac{dy}{dx}\)
\(\Rightarrow \frac{dy}{dx}-\sec^2{(x+y)}\frac{dy}{dx}=\sec^2{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}(1-\sec^2{(x+y)})=\sec^2{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\sec^2{(x+y)}}{1-\sec^2{(x+y)}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1+\tan^2{(x+y)}}{1-1-\tan^2{(x+y)}}\) ➜Note \(\because \sec^2(x)=1+\tan^2{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1+y^2}{-y^2}\) ➜Note \(\because y=\tan{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1+y^2}{y^2}\)

দেওয়া আছে,
\(\ln{xy}=x+y\)
\(\Rightarrow \ln{x}+\ln{y}=x+y\) ➜Note \(\because \ln{(MN)}=\ln{(M)}+\ln{(M)}\)
\(\Rightarrow \frac{d}{dx}(\ln{x}+\ln{y})=\frac{d}{dx}(x+y)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(\ln{x})+\frac{d}{dx}(\ln{y})=1+\frac{dy}{dx}\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}\) ➜Note \(\because \frac{d}{dx}(\ln{x})=\frac{1}{x}, \frac{d}{dx}(\ln{y})=\frac{1}{y}\frac{dy}{dx}\)
\(\Rightarrow \frac{1}{y}\frac{dy}{dx}-\frac{dy}{dx}=1-\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}\left(\frac{1}{y}-1\right)=1-\frac{1}{x}\)
\(\Rightarrow \frac{dy}{dx}\frac{1-y}{y}=\frac{x-1}{x}\)
\(\Rightarrow \frac{dy}{dx}=\frac{x-1}{x}\times{\frac{y}{1-y}}\)
\(\therefore \frac{dy}{dx}=\frac{y(x-1)}{x(1-y)}\)

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

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

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

দেওয়া আছে,
\(e^{xy}-4xy=c\)
\(\Rightarrow \frac{d}{dx}(e^{xy}-4xy)=\frac{d}{dx}(c)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(e^{xy})-4\frac{d}{dx}(xy)=0\) ➜Note\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v), \frac{d}{dx}(c)=0\)
\(\Rightarrow e^{xy}\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)-4\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)=0\) ➜Note\((xy)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow e^{xy}\left(x\frac{dy}{dx}+y.1\right)-4\left(x\frac{dy}{dx}+y.1\right)=0\) ➜Note\(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow xe^{xy}\frac{dy}{dx}-4x\frac{dy}{dx}=4y-ye^{xy}\)
\(\Rightarrow \frac{dy}{dx}(xe^{xy}-4x)=4y-ye^{xy}\)
\(\Rightarrow \frac{dy}{dx}=\frac{4y-ye^{xy}}{(xe^{xy}-4x)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{y(4-e^{xy})}{-x(4-e^{xy})}\)
\(\therefore \frac{dy}{dx}=-\frac{y}{x}\)

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

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

দেওয়া আছে,
\(y=\cot{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\{\cot{(x+y)}\}\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=-cosec^2{(x+y)}\frac{d}{dx}(x+y)\) ➜Note \((x+y)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cot{x})=-cosec^2{x}\)
\(\Rightarrow \frac{dy}{dx}=-cosec^2{(x+y)}\left(1+\frac{dy}{dx}\right)\) ➜Note \(\because \frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{dx}=-cosec^2{(x+y)}-cosec^2{(x+y)}\frac{dy}{dx}\)
\(\Rightarrow \frac{dy}{dx}+cosec^2{(x+y)}\frac{dy}{dx}=-cosec^2{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}(1+cosec^2{(x+y)})=-cosec^2{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}(1+1+\cot^2{(x+y)})=-1-\cot^2{(x+y)}\) ➜Note \(\because cosec^2{x}=1+\cot^2{x}\)
\(\Rightarrow \frac{dy}{dx}(2+y^2)=-1-y^2\) ➜Note \(\because y=\cot{(x+y)}\)
\(\Rightarrow \frac{dy}{dx}(2+y^2)=-(1+y^2)\)
\(\therefore \frac{dy}{dx}=-\frac{1+y^2}{2+y^2}\)

দেওয়া আছে,
\(x^2=5y^2+\sin{y}\)
\(\Rightarrow 5y^2+\sin{y}=x^2\)
\(\Rightarrow \frac{d}{dx}(5y^2+\sin{y})=\frac{d}{dx}(x^2)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow 5\frac{d}{dx}(y^2)+\frac{d}{dx}(\sin{y})=2x\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v), \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow 5.2y\frac{dy}{dx}+\cos{y}\frac{dy}{dx}=2x\) ➜Note \(\because \frac{d}{dx}(y^2)=2y\frac{dy}{dx}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{dy}{dx}(10y+\cos{y})=2x\)
\(\therefore \frac{dy}{dx}=\frac{2x}{10y+\cos{y}}\)

দেওয়া আছে,
\(\tan{y}=\sin{x}\)
\(\Rightarrow \tan{y}=\tan{\frac{x}{\sqrt{1-x^2}}}\) ➜Note \(\because \sin{x}=\tan{\frac{x}{\sqrt{1-x^2}}}\)
\(\Rightarrow y=\frac{x}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{x}{\sqrt{1-x^2}}\right)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-x^2}\frac{d}{dx}(x)-x\frac{d}{dx}(\sqrt{1-x^2})}{(\sqrt{1-x^2})^2}\) ➜Note \(\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-x^2}.1-x\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}(1-x^2)}{1-x^2}\) ➜Note \((1-x^2)\) কে \(x\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-x^2}-x\frac{1}{2\sqrt{1-x^2}}(0-2x)}{1-x^2}\) ➜Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\frac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\frac{1}{\sqrt{1-x^2}}}{1-x^2}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{(1-x^2)\sqrt{1-x^2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{(1-x^2)(1-x^2)^{\frac{1}{2}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{(1-x^2)^{1+\frac{1}{2}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{1}{(1-x^2)^{\frac{2+1}{2}}}\)
\(\therefore \frac{dy}{dx}=\frac{1}{(1-x^2)^{\frac{3}{2}}}\)
(Showed)

দেওয়া আছে,
\(y=\sqrt{\cos{x}+\sqrt{\cos{x}+\sqrt{\cos{x} ......\infty}}}\)
\(\Rightarrow y=\sqrt{\cos{x}+y}\) ➜Note \(\because y=\sqrt{\cos{x}+\sqrt{\cos{x}+\sqrt{\cos{x} ......\infty}}}\)
\(\Rightarrow y^2=\cos{x}+y\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow y^2-y=\cos{x}\)
\(\Rightarrow \frac{d}{dx}(y^2-y)=\frac{d}{dx}(\cos{x})\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y^2)-\frac{d}{dx}(y)=-\sin{x}\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\), \(\frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow 2y\frac{dy}{dx}-\frac{dy}{dx}=-\sin{x}\) ➜Note \(\because \frac{d}{dx}(y^2)=2y\frac{dy}{dx}\)
\(\therefore (2y-1)\frac{dy}{dx}+\sin{x}=0\)
(Showed)

দেওয়া আছে,
\(x=a(\theta+\sin{\theta})\) এবং \(y=a(1+\cos{\theta})\)
\(\Rightarrow x=a(\theta+\sin{\theta})\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\theta+\sin{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\theta)+a\frac{d}{d\theta}(\sin{\theta})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\theta}=a.1+a\cos{\theta}\) ➜Note\(\because \frac{d}{d\theta}(\theta)=1, \frac{d}{d\theta}(\sin{\theta})=\cos{\theta}\)
\(\therefore \frac{dx}{d\theta}=a(1+\cos{\theta}) ....(1)\)
আবার,
\(\Rightarrow y=a(1+\cos{\theta})\)
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(1+\cos{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(1)+a\frac{d}{d\theta}(\cos{\theta})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=0-a\sin{\theta}\) ➜Note\(\because \frac{d}{d\theta}(c)=0, \frac{d}{d\theta}(\cos{\theta})=-\sin{\theta}\)
\(\therefore \frac{dy}{d\theta}=-a\sin{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=-a\sin{\theta}\times{\frac{1}{a(1+\cos{\theta})}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=-\frac{\sin{\theta}}{1+\cos{\theta}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{2\cos^2{\frac{\theta}{2}}}\) ➜Note\(\because \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}, 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}\)
\(\therefore \frac{dy}{dx}=-\tan{\frac{\theta}{2}}\)

দেওয়া আছে,
\(x=a(\cos{\theta}+\theta\sin{\theta}), y=a(\sin{\theta}-\theta\cos{\theta})\) যখন \(\theta=\frac{3\pi}{4}\)
\(\Rightarrow x=a(\cos{\theta}+\theta\sin{\theta})\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\cos{\theta}+\theta\sin{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\cos{\theta})+a\frac{d}{d\theta}(\theta\sin{\theta})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\theta}=-a\sin{\theta}+a\{\theta\frac{d}{d\theta}(\sin{\theta})+\sin{\theta}\frac{d}{d\theta}(\theta)\}\) ➜Note\(\because \frac{d}{d\theta}(\cos{\theta})=-\sin{\theta}, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dx}{d\theta}=-a\sin{\theta}+a\{\theta\cos{\theta}+\sin{\theta}.1\}\) ➜Note\(\because \frac{d}{d\theta}(\sin{\theta})=\cos{\theta}, \frac{d}{d\theta}(\theta)=1\)
\(\Rightarrow \frac{dx}{d\theta}=a\{-\sin{\theta}+\theta\cos{\theta}+\sin{\theta}\}\)
\(\therefore \frac{dx}{d\theta}=a\theta\cos{\theta} ....(1)\)
আবার,
\(\Rightarrow y=a(\sin{\theta}-\theta\cos{\theta})\)
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(\sin{\theta}-\theta\cos{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(\sin{\theta})-a\frac{d}{d\theta}(\theta\cos{\theta})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=a\cos{\theta}-a\{\theta\frac{d}{d\theta}(\cos{\theta})+\cos{\theta}\frac{d}{d\theta}(\theta)\}\) ➜Note\(\because \frac{d}{d\theta}(\sin{\theta})=\cos{\theta}, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{d\theta}=a\cos{\theta}-a\{-\theta\sin{\theta}+\cos{\theta}.1\}\) ➜Note\(\because \frac{d}{d\theta}(\cos{\theta})=-\sin{\theta}, \frac{d}{d\theta}(\theta)=1\)
\(\Rightarrow \frac{dy}{d\theta}=a\{\cos{\theta}+\theta\sin{\theta}-\cos{\theta}\}\)
\(\therefore \frac{dy}{d\theta}=a\theta\sin{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=a\theta\sin{\theta}\times{\frac{1}{a\theta\cos{\theta}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \frac{dy}{dx}=\tan{\theta}\)
\(\Rightarrow \frac{dy}{dx}=\tan{\frac{3\pi}{4}}\) ➜Noteযখন \(\theta=\frac{3\pi}{4}\)
\(\Rightarrow \frac{dy}{dx}=\tan{\pi-\frac{\pi}{4}}\)
\(\Rightarrow \frac{dy}{dx}=-\tan{\frac{\pi}{4}}\) ➜Note দ্বিতীয় চৌকোণে \(\tan{\theta}\) ঋণাত্মক।
\(\therefore \frac{dy}{dx}=-1\) ➜Note \(\because \tan{\frac{\pi}{4}}=1\)

দেওয়া আছে,
\(x=\tan^{-1}\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\), \(y=\tan^{-1}\frac{\cos{\theta}}{1+\sin{\theta}}\)
\(\Rightarrow x=\tan^{-1}\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}\)
\(\Rightarrow x=\tan^{-1}\sqrt{\frac{2\sin^2{\frac{\theta}{2}}}{2\cos^2{\frac{\theta}{2}}}}\) ➜Note\(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, 1+\cos{A}=2\cos^2{\frac{A}{2}}\)
\(\Rightarrow x=\tan^{-1}\sqrt{\frac{\sin^2{\frac{\theta}{2}}}{\cos^2{\frac{\theta}{2}}}}\)
\(\Rightarrow x=\tan^{-1}\left(\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}\right)\)
\(\Rightarrow x=\tan^{-1}\left(\tan{\frac{\theta}{2}}\right)\)
\(\Rightarrow x=\frac{\theta}{2}\)
\(\Rightarrow x=\frac{1}{2}\theta\)
\(\Rightarrow \frac{dx}{d\theta}=\frac{1}{2}\frac{d}{d\theta}(\theta)\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=\frac{1}{2}.1\) ➜Note \(\because \frac{d}{d\theta}(\theta)=1\)
\(\therefore \frac{dx}{d\theta}=\frac{1}{2} ....(1)\)
আবার,
\(\Rightarrow y=\tan^{-1}\frac{\cos{\theta}}{1+\sin{\theta}}\)
\(\Rightarrow y=\tan^{-1}\frac{\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}}{\sin^2{\frac{\theta}{2}}+\cos^2{\frac{\theta}{2}}+2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}\) ➜Note\(\because \cos{A}=\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}\), \(\sin^2{\frac{A}{2}}+\cos^2{\frac{A}{2}}=1, \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(\Rightarrow y=\tan^{-1}\frac{(\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}})(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}})}{(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}})^2}\) ➜Note\(\because a^2+b^2+2ab=(a+b)^2, a^2-b^2=(a+b)(a-b)\)
\(\Rightarrow y=\tan^{-1}\frac{\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}}\)
\(\Rightarrow y=\tan^{-1}\frac{1-\tan{\frac{\theta}{2}}}{\tan{\frac{\theta}{2}}+1}\) ➜Note লব ও হরের সহিত \(\cos{\frac{\theta}{2}}\) ভাগ করে।
\(\Rightarrow y=\tan^{-1}\frac{\tan{\frac{\pi}{4}}-\tan{\frac{\theta}{2}}}{1+\tan{\frac{\pi}{4}}\tan{\frac{\theta}{2}}}\) ➜Note \(\because \tan{\frac{\pi}{4}}=1\)
\(\Rightarrow y=\tan^{-1}\{\tan{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\}\) ➜Note \(\because \tan{(A-B)}=\frac{\tan{A}-\tab{B}}{1+\tan{A}\tan{B}}\)
\(\Rightarrow y=\frac{\pi}{4}-\frac{\theta}{2}\)
\(\Rightarrow y=\frac{\pi}{4}-\frac{1}{2}\theta\)
\(\Rightarrow \frac{dy}{d\theta}=\frac{d}{d\theta}\left(\frac{\pi}{4}-\frac{1}{2}\theta\right)\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=\frac{d}{d\theta}\left(\frac{\pi}{4}\right)-\frac{1}{2}\frac{d}{d\theta}\left(\theta\right)\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=0-\frac{1}{2}.1\) ➜Note \(\because \frac{d}{dx}(c)=0,\frac{d}{dx}(x)=1\)
\(\Rightarrow \frac{dy}{d\theta}=-\frac{1}{2}....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{1}{2}\times{\frac{1}{\frac{1}{2}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\therefore \frac{dy}{dx}=-1\)

দেওয়া আছে,
\(x=a\cos{\theta}+b\sin{\theta}\), \(y=a\sin{\theta}-b\cos{\theta}\)
\(\Rightarrow x=a\cos{\theta}+b\sin{\theta}\)
\(\Rightarrow \frac{dx}{d\theta}=\frac{d}{d\theta}(a\cos{\theta}+b\sin{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\cos{\theta})+b\frac{d}{d\theta}(\sin{\theta})\) ➜Note \(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\theta}=-a\sin{\theta}+b\cos{\theta}\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{dx}{d\theta}=b\cos{\theta}-a\sin{\theta} ....(1)\)
আবার,
\(\Rightarrow y=a\sin{\theta}-b\cos{\theta}\)
\(\Rightarrow \frac{dy}{d\theta}=\frac{d}{d\theta}(a\sin{\theta}-b\cos{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(\sin{\theta})-b\frac{d}{d\theta}(\cos{\theta})\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=a\cos{\theta}+b\sin{\theta}\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x},\frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dy}{d\theta}=a\cos{\theta}+b\sin{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=a\cos{\theta}+b\sin{\theta}\times{\frac{1}{b\cos{\theta}-a\sin{\theta}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\therefore \frac{dy}{dx}=\frac{a\cos{\theta}+b\sin{\theta}}{b\cos{\theta}-a\sin{\theta}}\)

দেওয়া আছে,
\(x=a\sec^2{\theta}\), \(y=a\tan^3{\theta}\) যখন \(\theta=\frac{\pi}{4}\)
\(\Rightarrow x=a\sec^2{\theta}\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\sec^2{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a.2\sec{\theta}\frac{d}{d\theta}(\sec{\theta})\) ➜Note \(\sec{\theta})\) কে \(\theta\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^2)=2x\)
\(\Rightarrow \frac{dx}{d\theta}=2a\sec{\theta}\sec{\theta}\tan{\theta}\) ➜Note \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\Rightarrow \frac{dx}{d\theta}=2a\sec^2{\theta}\tan{\theta} ....(1)\)
আবার,
\(\Rightarrow y=a\tan^3{\theta}\)
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(\tan^3{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a.3\tan^2{\theta}\frac{d}{d\theta}(\tan{\theta})\) ➜Note \(\tan{\theta})\) কে \(\theta\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dy}{d\theta}=3a\tan^2{\theta}\sec^2{\theta}\) ➜Note \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\Rightarrow \frac{dy}{d\theta}=3a\sec^2{\theta}\tan^2{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=3a\sec^2{\theta}\tan^2{\theta}\times{\frac{1}{2a\sec^2{\theta}\tan{\theta}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{3}{2}\tan{\theta}\)
\(\Rightarrow \frac{dy}{dx}=\frac{3}{2}\tan{\frac{\pi}{4}}\) যখন \(\theta=\frac{\pi}{4}\)
\(\Rightarrow \frac{dy}{dx}=\frac{3}{2}.1\) ➜Note \(\because \tan{\frac{\pi}{4}}=1\)
\(\therefore \frac{dy}{dx}=\frac{3}{2}\)

দেওয়া আছে,
\(\tan{y}=\frac{2t}{1-t^2}\) এবং \(\sin{x}=\frac{2t}{1+t^2}\)
\(\Rightarrow \tan{y}=\frac{2t}{1-t^2}\)
ধরি,
\(\tan{\theta}=t\Rightarrow \theta=\tan^{-1}{t}\)
\(\Rightarrow \tan{y}=\frac{2\tan{\theta}}{1-\tan^2{\theta}}\)
\(\Rightarrow \tan{y}=\tan{2\theta}\) ➜Note \(\because \frac{2\tan{\theta}}{1-\tan^2{\theta}}=\tan{2\theta}\)
\(\Rightarrow y=2\theta\)
\(\Rightarrow y=2\tan^{-1}{t}\) ➜Note \(\because \theta=\tan^{-1}{t}\)
\(\Rightarrow \frac{dy}{dt}=2\frac{d}{dt}(\tan^{-1}{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=2\frac{1}{1+t^2}\) ➜Note \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{dy}{dt}=\frac{2}{1+t^2} ....(1)\)
আবার,
\(\sin{x}=\frac{2t}{1+t^2}\)
ধরি,
\(\tan{\theta}=t\Rightarrow \theta=\tan^{-1}{t}\)
\(\Rightarrow \sin{x}=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow \sin{x}=\sin{2\theta}\) ➜Note \(\because \frac{2\tan{\theta}}{1+\tan^2{\theta}}=\sin{2\theta}\)
\(\Rightarrow x=2\theta\)
\(\Rightarrow x=2\tan^{-1}{t}\)| Note \(\because \theta=\tan^{-1}{t}\)
\(\Rightarrow \frac{dx}{dt}=2\frac{d}{dt}(\tan^{-1}{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=2\frac{1}{1+t^2}\) ➜Note \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{dx}{dt}=\frac{2}{1+t^2} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2}{1+t^2}\times{\frac{1}{\frac{2}{1+t^2}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\therefore \frac{dy}{dx}=1\)

দেওয়া আছে,
\(x=a\cos^3{\theta}\) এবং \(y=a\sin^3{\theta}\)
\(\Rightarrow x=a\cos^3{\theta}\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{d\theta}(\cos^3{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a.3\cos^2{\theta}\frac{d}{d\theta}(\cos{\theta})\) ➜Note \(\cos{\theta})\) কে \(\theta\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dx}{d\theta}=3a\cos^2{\theta}(-\sin{\theta})\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dx}{d\theta}=-3a\cos^2{\theta}\sin{\theta} ....(1)\)
আবার,
\(\Rightarrow y=a\sin^3{\theta}\)
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{d\theta}(\sin^3{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a.3\sin^2{\theta}\frac{d}{d\theta}(\tan{\theta})\) ➜Note \(\sin{\theta})\) কে \(\theta\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dy}{d\theta}=3a\sin^2{\theta}\cos{\theta}\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{dy}{d\theta}=3a\sin^2{\theta}\cos{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=3a\sin^2{\theta}\cos{\theta}\times{\frac{1}{-3a\cos^2{\theta}\sin{\theta}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=-\frac{\sin{\theta}}{\cos{\theta}}\)
\(\therefore \frac{dy}{dx}=-\tan{\theta}\)

দেওয়া আছে,
\(x=e^t\cos{t}\) এবং \(y=e^t\sin{t}\)
\(\Rightarrow x=e^t\cos{t}\)
\(\Rightarrow \frac{dx}{dt}=\frac{d}{dt}(e^t\cos{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=e^t\frac{d}{dt}(\cos{t})+\cos{t}\frac{d}{dt}(e^t)\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dx}{dt}=-e^t\sin{t}+\cos{t}e^t\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(e^x)=e^x\)
\(\therefore \frac{dx}{dt}=e^t(\cos{t}-\sin{t}) ....(1)\)
আবার,
\(\Rightarrow y=e^t\sin{t}\)
\(\Rightarrow \frac{dy}{dt}=\frac{d}{dt}(e^t\sin{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=e^t\frac{d}{dt}(\sin{t})+\sin{t}\frac{d}{dt}(e^t)\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{dt}=e^t\cos{t}+\sin{t}e^t\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}, \frac{d}{dx}(e^x)=e^x\)
\(\therefore \frac{dy}{dt}=e^t(\cos{t}+\sin{t}) ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=e^t(\cos{t}+\sin{t})\times{\frac{1}{e^t(\cos{t}-\sin{t})}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=(\cos{t}+\sin{t})\times{\frac{1}{(\cos{t}-\sin{t})}}\)
\(\therefore \frac{dy}{dx}=\frac{\sin{t}+\cos{t}}{\cos{t}-\sin{t}}\)

দেওয়া আছে,
\(x=a\cos{t}\) এবং \(y=b\sin{t}\)
\(\Rightarrow x=\cos{t}\)
\(\Rightarrow \frac{dx}{dt}=a\frac{d}{dt}(\cos{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=-a\sin{t}\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\therefore \frac{dx}{dt}=\frac{dx}{dt}=-a\sin{t} ....(1)\)
আবার,
\(\Rightarrow y=b\sin{t}\)
\(\Rightarrow \frac{dy}{dt}=b\frac{d}{dt}(\sin{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=b\cos{t}\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dy}{dt}=b\cos{t}....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=b\cos{t}\times{\frac{1}{-a\sin{t}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=-\frac{b\cos{t}}{a\sin{t}}\)
\(\therefore \frac{dy}{dx}=-\frac{b}{a}\cot{t}\)

দেওয়া আছে,
\(x=\sqrt{t}\) এবং \(y=t-\frac{1}{\sqrt{t}}\)
\(\Rightarrow x=\sqrt{t}\)
\(\Rightarrow \frac{dx}{dt}=\frac{d}{dt}(\sqrt{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=\frac{1}{2\sqrt{t}}\) ➜Note \(\because \frac{d}{dt}(\sqrt{t})=\frac{1}{2\sqrt{t}}\)
\(\therefore \frac{dx}{dt}=\frac{1}{2\sqrt{t}} ....(1)\)
আবার,
\(\Rightarrow y=t-\frac{1}{\sqrt{t}}\)
\(\Rightarrow \frac{dy}{dt}=\frac{d}{dt}\left(t-\frac{1}{\sqrt{t}}\right)\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=\frac{d}{dt}(t)-\frac{d}{dt}\left(\frac{1}{\sqrt{t}}\right)\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dt}=1+\frac{1}{(\sqrt{t})^2}\frac{d}{dt}(\sqrt{t})\) ➜Note \(\sqrt{t}\) কে \(t\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dt}(t)=1, \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}\)
\(\Rightarrow \frac{dy}{dt}=1+\frac{1}{t}.\frac{1}{2\sqrt{t}}\) ➜Note\(\because \frac{d}{dt}(\sqrt{t})=\frac{1}{2\sqrt{t}}\)
\(\Rightarrow \frac{dy}{dt}=1+\frac{1}{2t\sqrt{t}}\)
\(\therefore \frac{dy}{dt}=\frac{2t\sqrt{t}+1}{2t\sqrt{t}}....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2t\sqrt{t}+1}{2t\sqrt{t}}\times{\frac{1}{\frac{1}{2\sqrt{t}}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{2t\sqrt{t}+1}{2t\sqrt{t}}\times{2\sqrt{t}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2t\sqrt{t}+1}{t}\)
\(\therefore \frac{dy}{dx}=2\sqrt{t}+\frac{1}{t}\)

দেওয়া আছে,
\(x=2\sin{t}\) এবং \(y=\cos{2t}\)
\(\Rightarrow x=2\sin{t}\)
\(\Rightarrow \frac{dx}{dt}=2\frac{d}{dt}(\sin{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=2\cos{t}\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dx}{dt}=2\cos{t} ....(1)\)
আবার,
\(\Rightarrow y=\cos{2t}\)
\(\Rightarrow \frac{dy}{dt}=\frac{d}{dt}(\cos{2t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=-\sin{2t}\frac{d}{dt}(2t)\) ➜Note \(2t\) কে \(t\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dy}{dt}=-\sin{2t}.2\) ➜Note\(\because \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dt}=-2\sin{2t}....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=-2\sin{2t}\times{\frac{1}{2\cos{t}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=-\frac{\sin{2t}}{\cos{t}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{2\sin{t}\cos{t}}{\cos{t}}\)
\(\therefore \frac{dy}{dx}=-2\sin{t}\)

দেওয়া আছে,
\(x=\cos^3{\theta}\) এবং \(y=\sin^3{\theta}\)
\(\Rightarrow x=\cos^3{\theta}\)
\(\Rightarrow \frac{dx}{d\theta}=\frac{d}{d\theta}(\cos^3{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=3\cos^2{\theta}\frac{d}{d\theta}(\cos{\theta})\) ➜Note \(\cos{\theta})\) কে \(\theta\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dx}{d\theta}=3\cos^2{\theta}(-\sin{\theta})\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow \frac{dx}{d\theta}=-3\cos^2{\theta}\sin{\theta} ....(1)\)
আবার,
\(\Rightarrow y=\sin^3{\theta}\)
\(\Rightarrow \frac{dy}{d\theta}=\frac{d}{d\theta}(\sin^3{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=3\sin^2{\theta}\frac{d}{d\theta}(\tan{\theta})\) ➜Note \(\sin{\theta})\) কে \(\theta\) মনে করে, সংযোজিত ফাংশনের নিয়মানুযায়ী অন্তরীকরণ করে এবং \(\because \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dy}{d\theta}=3\sin^2{\theta}\cos{\theta}\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow \frac{dy}{d\theta}=3\sin^2{\theta}\cos{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=3\sin^2{\theta}\cos{\theta}\times{\frac{1}{-3\cos^2{\theta}\sin{\theta}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=-\frac{\sin{\theta}}{\cos{\theta}}\)
\(\therefore \frac{dy}{dx}=-\tan{\theta}\)

দেওয়া আছে,
\(x=a(\cos{\phi}+\phi\sin{\phi}), y=a(\sin{\phi}-\phi\cos{\phi})\)
\(\Rightarrow x=a(\cos{\phi}+\phi\sin{\phi})\)
\(\Rightarrow \frac{dx}{d\phi}=a\frac{d}{d\phi}(\cos{\phi}+\phi\sin{\phi})\) ➜Note \(\phi\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\phi}=a\frac{d}{d\phi}(\cos{\phi})+a\frac{d}{d\phi}(\phi\sin{\phi})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\phi}=-a\sin{\phi}+a\{\phi\frac{d}{d\phi}(\sin{\phi})+\sin{\phi}\frac{d}{d\phi}(\phi)\}\)| Note\(\because \frac{d}{d\phi}(\cos{\phi})=-\sin{\phi}\), \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dx}{d\phi}=-a\sin{\phi}+a\{\phi\cos{\phi}+\sin{\phi}.1\}\) ➜Note\(\because \frac{d}{d\phi}(\sin{\phi})=\cos{\phi}, \frac{d}{d\phi}(\phi)=1\)
\(\Rightarrow \frac{dx}{d\phi}=a\{-\sin{\phi}+\phi\cos{\phi}+\sin{\phi}\}\)
\(\therefore \frac{dx}{d\phi}=a\phi\cos{\phi} ....(1)\)
আবার,
\(\Rightarrow y=a(\sin{\phi}-\phi\cos{\phi})\)
\(\Rightarrow \frac{dy}{d\phi}=a\frac{d}{d\phi}(\sin{\phi}-\phi\cos{\phi})\) ➜Note \(\phi\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\phi}=a\frac{d}{d\phi}(\sin{\phi})-a\frac{d}{d\phi}(\phi\cos{\phi})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\phi}=a\cos{\phi}-a\{\phi\frac{d}{d\phi}(\cos{\phi})+\cos{\phi}\frac{d}{d\phi}(\phi)\}\) ➜Note\(\because \frac{d}{d\phi}(\sin{\phi})=\cos{\phi}, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{d\phi}=a\cos{\phi}-a\{-\phi\sin{\phi}+\cos{\phi}.1\}\) ➜Note\(\because \frac{d}{d\phi}(\cos{\phi})=-\sin{\phi}, \frac{d}{d\phi}(\phi)=1\)
\(\Rightarrow \frac{dy}{d\phi}=a\{\cos{\phi}+\phi\sin{\phi}-\cos{\phi}\}\)
\(\therefore \frac{dy}{d\phi}=a\phi\sin{\phi} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\phi}\times{\frac{d\phi}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\phi}\times{\frac{1}{\frac{dx}{d\phi}}}\)
\(\Rightarrow \frac{dy}{dx}=a\phi\sin{\phi}\times{\frac{1}{a\phi\cos{\phi}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\phi}}{\cos{\phi}}\)
\(\therefore \frac{dy}{dx}=\tan{\phi}\)

দেওয়া আছে,
\(x=\frac{3at}{1+t^3}, y=\frac{3at^2}{1+t^3}\)
\(\Rightarrow x=\frac{3at}{1+t^3}\)
\(\Rightarrow \frac{dx}{dt}=3a\frac{d}{dt}\left(\frac{t}{1+t^3}\right)\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=3a\frac{(1+t^3)\frac{d}{dt}(t)-t\frac{d}{dt}(1+t^3)}{(1+t^3)^2}\) ➜Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dx}{dt}=3a\frac{(1+t^3).1-t(0+3t^2)}{(1+t^3)^2}\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dx}{dt}=3a\frac{1+t^3-3t^3}{(1+t^3)^2}\)
\(\therefore \frac{dx}{dt}=\frac{3a(1-2t^3)}{(1+t^3)^2} ....(1)\)
আবার,
\(\Rightarrow y=\frac{3at^2}{1+t^3}\)
\(\Rightarrow \frac{dy}{dt}=3a\frac{d}{dt}\left(\frac{t^2}{1+t^3}\right)\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=3a\frac{(1+t^3)\frac{d}{dt}(t^2)-t^2\frac{d}{dt}(1+t^3)}{(1+t^3)^2}\) ➜Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dt}=3a\frac{(1+t^3).2t-t^2(0+3t^2)}{(1+t^3)^2}\) ➜Note \(\because \frac{d}{dx}(x^2)=2x, \frac{d}{dx}(c)=0, \frac{d}{dx}(x^3)=3x^2\)
\(\Rightarrow \frac{dy}{dt}=3a\frac{2t+2t^4-3t^4}{(1+t^3)^2}\)
\(\therefore \frac{dy}{dt}=\frac{3a(2t-t^4)}{(1+t^3)^2} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{3a(2t-t^4)}{(1+t^3)^2}\times{\frac{1}{\frac{3a(1-2t^3)}{(1+t^3)^2}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{3a(2t-t^4)}{(1+t^3)^2}\times{\frac{(1+t^3)^2}{3a(1-2t^3)}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2t-t^4}{1-2t^3}\)
\(\therefore \frac{dy}{dx}=\frac{t(2-t^3)}{1-2t^3}\)

ধরি,
\(x=\frac{3at}{1+t^3} ...(1)\), \(y=\frac{3at^2}{1+t^3} ....(2)\)
\((2)\) কে \((1)\) দ্বারা ভাগ করে।
\(\frac{y}{x}=\frac{3at^2}{1+t^3}\div{\frac{3at^2}{1+t^3}}\)
\(\Rightarrow \frac{y}{x}=\frac{3at^2}{1+t^3}\times{\frac{1+t^3}{3at^2}}\)
\(\Rightarrow \frac{y}{x}=t\)
\(\Rightarrow t=\frac{y}{x}\)
\(t\)এর এই মাণ \(\) এ বসিয়ে,
\(x=\frac{3a\frac{y}{x}}{1+\left(\frac{y}{x}\right)^3}\)
\(\Rightarrow x=\frac{\frac{3ay}{x}}{1+\frac{y^3}{x^3}}\)
\(\Rightarrow x=\frac{\frac{3ay}{x}}{\frac{x^3+y^3}{x^3}}\)
\(\Rightarrow x=\frac{3ay}{x}\times{\frac{x^3}{x^3+y^3}}\)
\(\Rightarrow x=\frac{3ayx^2}{x^3+y^3}\)
\(\Rightarrow x^3+y^3=\frac{3ayx^2}{x}\)
\(\Rightarrow x^3+y^3=3axy\)
\(\Rightarrow \frac{d}{dx}(x^3+y^3)=\frac{d}{dx}(3axy)\) ➜Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(x^3)+\frac{d}{dx}(y^3)=3a\frac{d}{dx}(xy)\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow 3x^2+3y^2\frac{dy}{dx}=3a\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x)\right)\) ➜Note\(\because \frac{d}{dx}(x^3)=3x^2, \frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}\), \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow 3x^2+3y^2\frac{dy}{dx}=3a\left(x\frac{dy}{dx}+y.1\right)\) ➜Note\(\because \frac{d}{dx}(y)=\frac{dy}{dx}, \frac{d}{dx}(x)=1\)
\(\Rightarrow 3x^2+3y^2\frac{dy}{dx}=3ax\frac{dy}{dx}+3ay\)
\(\Rightarrow 3y^2\frac{dy}{dx}-3ax\frac{dy}{dx}=3ay-3x^2\)
\(\Rightarrow \frac{dy}{dx}(3y^2-3ax)=3ay-3x^2\)
\(\Rightarrow \frac{dy}{dx}=\frac{3ay-3x^2}{(3y^2-3ax)}\)
\(\Rightarrow \frac{dy}{dx}=\frac{3(ay-x^2)}{3(y^2-ax)}\)
\(\therefore \frac{dy}{dx}=\frac{ay-x^2}{y^2-ax}\)

দেওয়া আছে,
\(x=a(\theta-\sin{\theta})\) এবং \(y=a(1-\cos{\theta})\)
\(\Rightarrow x=a(\theta-\sin{\theta})\)
\(\Rightarrow \frac{dx}{d\theta}=\frac{d}{d\theta}a(\theta-\sin{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=a\{\frac{d}{d\theta}(\theta)-\frac{d}{d\theta}(\sin{\theta})\}\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{d\theta}=a(1-\cos{\theta})\) ➜Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dx}{d\theta}=a(1-\cos{\theta}) ....(1)\)
আবার,
\(y=a(1-\cos{\theta})\)
\(\Rightarrow y=a(1-\cos{\theta})\)
\(\Rightarrow \frac{dy}{d\theta}=\frac{d}{d\theta}a(1-\cos{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\{\frac{d}{d\theta}(1)-\frac{d}{d\theta}(\cos{\theta})\}\) ➜Note \(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{d\theta}=a(0+\sin{\theta})\) ➜Note \(\because \frac{d}{dx}(c)=0, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\therefore \frac{dy}{d\theta}=a\sin{\theta} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=a\sin{\theta}\times{\frac{1}{a(1-\cos{\theta})}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{\theta}}{1-\cos{\theta}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}{2\sin^2{\frac{\theta}{2}}}\) ➜Note \(\because \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}, 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}\)
\(\therefore \frac{dy}{dx}=\cot{\frac{\theta}{2}}\)

দেওয়া আছে,
\(x=\frac{a\cos{t}}{t}, y=\frac{a\sin{t}}{t}\)
\(\Rightarrow x=\frac{a\cos{t}}{t}\)
\(\Rightarrow \frac{dx}{dt}=a\frac{d}{dt}\left(\frac{\cos{t}}{t}\right)\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=a\frac{t\frac{d}{dt}(\cos{t})-\cos{t}\frac{d}{dt}(t)}{t^2}\) ➜Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dx}{dt}=a\frac{-t\sin{t}-\cos{t}.1}{t^2}\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(x)=1\)
\(\therefore \frac{dx}{dt}=-a\frac{t\sin{t}+\cos{t}}{t^2}....(1)\)
আবার,
\(\Rightarrow y=\frac{a\sin{t}}{t}\)
\(\Rightarrow \frac{dy}{dt}=a\frac{d}{dt}\left(\frac{\sin{t}}{t}\right)\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=a\frac{t\frac{d}{dt}(\sin{t})-\sin{t}\frac{d}{dt}(t)}{t^2}\) ➜Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(\Rightarrow \frac{dy}{dt}=a\frac{t\cos{t}-\sin{t}.1}{t^2}\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}, \frac{d}{dx}(x)=1\)
\(\therefore \frac{dy}{dt}=a\frac{t\cos{t}-\sin{t}}{t^2}....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=a\frac{t\cos{t}-\sin{t}}{t^2}\times{\frac{1}{-a\frac{t\sin{t}+\cos{t}}{t^2}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=a\frac{t\cos{t}-\sin{t}}{t^2}\times{-\frac{t^2}{a(t\sin{t}+\cos{t})}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{t\cos{t}-\sin{t}}{t\sin{t}+\cos{t}}\)
\(\therefore \frac{dy}{dx}=\frac{\sin{t}-t\cos{t}}{t\sin{t}+\cos{t}}\)

দেওয়া আছে,
\(x=a(\cos{t}+t\sin{t}), y=a(\sin{t}-t\cos{t})\)
\(\Rightarrow x=a(\cos{t}+t\sin{t})\)
\(\Rightarrow \frac{dx}{dt}=a\frac{d}{dt}(\cos{t}+t\sin{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=a\frac{d}{dt}(\cos{t})+a\frac{d}{dt}(t\sin{t})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{dt}=-a\sin{t}+a\{t\frac{d}{dt}(\sin{t})+\sin{t}\frac{d}{dt}(t)\}\) ➜Note\(\because \frac{d}{dt}(\cos{t})=-\sin{t}, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dx}{dt}=-a\sin{t}+a\{t\cos{t}+\sin{t}.1\}\) ➜Note\(\because \frac{d}{dt}(\sin{t})=\cos{t}, \frac{d}{dt}(t)=1\)
\(\Rightarrow \frac{dx}{dt}=a\{-\sin{t}+t\cos{t}+\sin{t}\}\)
\(\therefore \frac{dx}{dt}=at\cos{t} ....(1)\)
আবার,
\(\Rightarrow y=a(\sin{t}-t\cos{t})\)
\(\Rightarrow \frac{dy}{dt}=a\frac{d}{dt}(\sin{t}-t\cos{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=a\frac{d}{dt}(\sin{t})-a\frac{d}{dt}(t\cos{t})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dt}=a\cos{t}-a\{t\frac{d}{dt}(\cos{t})+\cos{t}\frac{d}{dt}(t)\}\) ➜Note\(\because \frac{d}{dt}(\sin{t})=\cos{t}, \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{dy}{dt}=a\cos{t}-a\{-t\sin{t}+\cos{t}.1\}\) ➜Note\(\because \frac{d}{dt}(\cos{t})=-\sin{t}, \frac{d}{dt}(t)=1\)
\(\Rightarrow \frac{dy}{dt}=a\{\cos{t}+t\sin{t}-\cos{t}\}\)
\(\therefore \frac{dy}{dt}=at\sin{t} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=at\sin{t}\times{\frac{1}{at\cos{t}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{\sin{t}}{\cos{t}}\)
\(\therefore \frac{dy}{dx}=\tan{t}\)

দেওয়া আছে,
\(x=a\cos{\theta}\) এবং \(y=a\sin{\theta}\)
\(\Rightarrow x=\cos{\theta}\)
\(\Rightarrow \frac{dx}{d\theta}=a\frac{d}{dt}(\cos{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\theta}=-a\sin{\theta}\) ➜Note \(\because \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\therefore \frac{dx}{d\theta}=-a\sin{\theta} ....(1)\)
আবার,
\(\Rightarrow y=b\sin{\theta}\)
\(\Rightarrow \frac{dy}{d\theta}=a\frac{d}{dt}(\sin{\theta})\) ➜Note \(\theta\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\theta}=a\cos{\theta}\) ➜Note \(\because \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\therefore \frac{dy}{d\theta}=a\cos{\theta}....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{d\theta}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\theta}\times{\frac{1}{\frac{dx}{d\theta}}}\)
\(\Rightarrow \frac{dy}{dx}=a\cos{\theta}\times{\frac{1}{-a\sin{\theta}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=-\frac{\cos{\theta}}{\sin{\theta}}\)
\(\therefore \frac{dy}{dx}=-\cot{\theta}\)

দেওয়া আছে,
\(x=a\sec{\phi}\) এবং \(y=b\tan{\phi}\)
\(\Rightarrow x=a\sec{\phi}\)
\(\Rightarrow \frac{dx}{d\phi}=a\frac{d}{d\phi}(\sec{\phi})\) ➜Note \(\phi\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{d\phi}=a\sec{\phi}\tan{\phi}\) ➜Note \(\because \frac{d}{dx}(\sec{x})=\sec{x}\tan{x}\)
\(\therefore \frac{dx}{d\phi}=a\sec{\phi}\tan{\phi} ....(1)\)
আবার,
\(\Rightarrow y=b\tan{\phi}\)
\(\Rightarrow \frac{dy}{d\phi}=b\frac{d}{d\phi}(\tan{\phi})\) ➜Note \(\phi\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{d\phi}=b\sec^2{\phi}\) ➜Note \(\because \frac{d}{dx}(\tan{x})=\sec^2{x}\)
\(\therefore \frac{dy}{d\phi}=b\sec^2{\phi}....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{d\phi}\times{\frac{d\phi}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{d\phi}\times{\frac{1}{\frac{dx}{d\phi}}}\)
\(\Rightarrow \frac{dy}{dx}=b\sec^2{\phi}\times{\frac{1}{a\sec{\phi}\tan{\phi}}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=\frac{b\sec{\phi}}{a\tan{\phi}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{b}{a}\sec{\phi}\cot{\phi}\)
\(\Rightarrow \frac{dy}{dx}=\frac{b}{a}\frac{1}{\cos{\phi}}.\frac{\cos{\phi}}{\sin{\phi}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{b}{a}\frac{1}{\sin{\phi}}\)
\(\therefore \frac{dy}{dx}=\frac{b}{a} cosec \ {\phi}\)

দেওয়া আছে,
\(x=a(t-\sin{t}), y=a(1+\cos{t})\)
\(\Rightarrow a(t-\sin{t})\)
\(\Rightarrow \frac{dx}{dt}=a\frac{d}{dt}(t-\sin{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dx}{dt}=a\frac{d}{dt}(t)-a\frac{d}{dt}(\sin{t})\) ➜Note\(\because \frac{d}{dx}(u-v)=\frac{d}{dx}(u)-\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dx}{dt}=a.1-a\cos{t}\) ➜Note\(\because \frac{d}{dt}(t)=1, \frac{d}{dt}(\sin{t})=\cos{t}\)
\(\therefore \frac{dx}{dt}=a(1-\cos{t}) ....(1)\)
আবার,
\(\Rightarrow y=a(1+\cos{t})\)
\(\Rightarrow \frac{dy}{dt}=a\frac{d}{dt}(1+\cos{t})\) ➜Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{dy}{dt}=a\frac{d}{dt}(1)+a\frac{d}{dt}(\cos{t})\) ➜Note\(\because \frac{d}{dx}(u+v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\)
\(\Rightarrow \frac{dy}{dt}=a.0-a\sin{t}\) ➜Note\(\because \frac{d}{dx}(c)=0, \frac{d}{dt}(\cos{t})=-\sin{t}\)
\(\therefore \frac{dy}{dt}=-a\sin{t} ....(2)\)
এখন,
\(\frac{dy}{dx}=\frac{dy}{dt}\times{\frac{dt}{dx}}\)
\(\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times{\frac{1}{\frac{dx}{dt}}}\)
\(\Rightarrow \frac{dy}{dx}=-a\sin{t}\times{\frac{1}{a(1-\cos{t})}}\) ➜Note \((1)\) ও \((2)\) এর সাহায্যে।
\(\Rightarrow \frac{dy}{dx}=-\frac{\sin{t}}{1-\cos{t}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{2\sin{\frac{t}{2}}\cos{\frac{t}{2}}}{2\sin^2{\frac{t}{2}}}\) ➜Note\(\because \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}, 1-\cos{A}=2\sin^2{\frac{A}{2}}\)
\(\Rightarrow \frac{dy}{dx}=-\frac{\cos{\frac{t}{2}}}{\sin{\frac{t}{2}}}\)
\(\Rightarrow \frac{dy}{dx}=-\cot{\frac{t}{2}}\)
\(\Rightarrow \frac{dy}{dx}=-\cot{\frac{\frac{5\pi}{3}}{2}}\) যখন \(t=\frac{5\pi}{3}\)
\(\Rightarrow \frac{dy}{dx}=-\cot{\frac{5\pi}{6}}\)
\(\Rightarrow \frac{dy}{dx}=-\cot{\left(\pi-\frac{\pi}{6}\right)}\)
\(\Rightarrow \frac{dy}{dx}=\cot{\frac{\pi}{6}}\) ➜Note দ্বিতীয় চৌকোণে \(\cot{\theta}\) ঋণাত্মক।
\(\therefore \frac{dy}{dx}=\sqrt{3}\) ➜Note\(\because \cot{\frac{\pi}{6}}=\sqrt{3}\)
(Showed)