দেওয়া আছে,
\(u=u(x)\) এবং \(v=u(x)\) উভয়ে \(x\)-এর ফাংশন।
ধরি,
\(f(x)=uv=u(x)v(x)\)
\(\therefore f(x+h)=u(x+h)v(x+h)\)
আমরা জানি,
\(\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\) ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে ।
\[\Rightarrow \frac{d}{dx}(uv)=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\left[\frac{u(x+h)v(x+h)-u(x+h)v(x)}{h}+\frac{u(x+h)v(x)-u(x)v(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}\left[u(x+h)\frac{v(x+h)-v(x)}{h}+v(x)\frac{u(x+h)-u(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}u(x+h)\lim_{h \rightarrow 0}\frac{v(x+h)-v(x)}{h}+v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}\]
\(=u(x)\frac{d}{dx}\{v(x)\}+v(x)\frac{d}{dx}\{u(x)\}\) ➜ Note \[\because \frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\(=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\therefore \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
(proved)

দেওয়া আছে,
\(u=u(x)\) এবং \(v=u(x)\) উভয়ে \(x\)-এর ফাংশন।
ধরি,
\(f(x)=\frac{u}{v}=\frac{u(x)}{v(x)}\)
\(\therefore f(x+h)=\frac{u(x+h)}{v(x+h)}\)
আমরা জানি,
\[\frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\] ➜ Note মূল নিয়মে অন্তরজের সংজ্ঞানুসারে ।
\[\Rightarrow \frac{d}{dx}(\frac{u}{v})=\lim_{h \rightarrow 0}\frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{\frac{u(x+h)v(x)-v(x+h)u(x)}{v(x+h)v(x)}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x)-v(x+h)u(x)}{hv(x+h)v(x)}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{u(x+h)v(x)-v(x+h)u(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{u(x+h)v(x)-u(x)v(x)-v(x+h)u(x)+u(x)v(x)}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \frac{\{u(x+h)v(x)-u(x)v(x)\}-\{v(x+h)u(x)-u(x)v(x)\}}{h}\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \left[\frac{u(x+h)v(x)-u(x)v(x)}{h}-\frac{v(x+h)u(x)-u(x)v(x)}{h}\right]\]
\[=\lim_{h \rightarrow 0}\frac{1}{v(x+h)v(x)}\times \left[\lim_{h \rightarrow 0}\frac{u(x+h)v(x)-u(x)v(x)}{h}-\lim_{h \rightarrow 0}\frac{v(x+h)u(x)-u(x)v(x)}{h}\right]\] \[=\frac{1}{v(x).v(x)}\times \left[v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}-u(x)\lim_{h \rightarrow 0}\frac{v(x+h)-v(x)}{h}\right]\]
\[=\frac{1}{v.v}\times \left[v(x)\frac{d}{dx}\{u(x)\}-u(x)\frac{d}{dx}\{v(x)\}\right]\] ➜ Note \[\because \frac{d}{dx}\{f(x)\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[=\frac{1}{v^2}\times \left[v\frac{d}{dx}(u)-u\frac{d}{dx}(v)\right]\]
\[=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\]
\[\therefore \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\]
(proved)

ধরি,
\(y=x^3\sin x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3\sin x)\)
\(=x^3\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(x^3)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^3\cos x+\sin x.3x^{3-1}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(x^{n})=nx^{n-1}\)
\(=x^3\cos x+\sin x.3x^{2}\)
\(=x^3\cos x+3x^{2}\sin x\)
\(=x^2(x\cos x+3\sin x)\)

ধরি,
\(y=(x^2+1)(\frac{1}{2}x+1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x^2+1)(\frac{1}{2}x+1)\}\)
\(=(x^2+1)\frac{d}{dx}\{\frac{1}{2}x+1\}+(\frac{1}{2}x+1)\frac{d}{dx}\{x^2+1\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=(x^2+1)\{\frac{1}{2}\frac{d}{dx}(x)+\frac{d}{dx}(1)\}+(\frac{1}{2}x+1)\{\frac{d}{dx}(x^2)+\frac{d}{dx}(1)\}\)
\(=(x^2+1)\{\frac{1}{2}.1+0\}+(\frac{1}{2}x+1)\{2x+0\}\) ➜ Note \(\because \frac{d}{dx}(1)=0, \frac{d}{dx}(x^{n})=nx^{n-1}\)
\(=(x^2+1).\frac{1}{2}+(\frac{1}{2}x+1).2x\)
\(=\frac{1}{2}(x^2+1)+2x(\frac{1}{2}x+1)\)
\(=\frac{1}{2}x^2+\frac{1}{2}+x^2+2x\)
\(=\frac{1}{2}x^2+x^2+2x+\frac{1}{2}\)
\(=\frac{1+2}{2}x^2+2x+\frac{1}{2}\)
\(=\frac{3}{2}x^2+2x+\frac{1}{2}\)

ধরি,
\(y=5e^x\ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5e^x\ln x)\)
\(=5e^x\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(5e^x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=5e^x\frac{1}{x}+\ln x.5\frac{d}{dx}(e^x)\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=5e^x\frac{1}{x}+\ln x.5e^x\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x\)
\(=5e^x(\frac{1}{x}+\ln x)\)

ধরি,
\(y=\frac{x^2}{x^2-4}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{x^2}{x^2-4})\)
\(=\frac{(x^2-4)\frac{d}{dx}(x^2)-(x^2)\frac{d}{dx}(x^2-4)}{(x^2-4)^2}\) ➜ Note \(\because \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(x^2-4)2x-x^2(\frac{d}{dx}x^2-\frac{d}{dx}4)}{(x^2-4)^2}\) ➜ Note \(\because \frac{d}{dx}(x^{n})=nx^{n-1}\)
\(=\frac{(x^2-4)2x-x^2(2x-0)}{(x^2-4)^2}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\)
\(=\frac{2x(x^2-4)-x^2.2x}{(x^2-4)^2}\)
\(=\frac{2x^3-8x-2x^3}{(x^2-4)^2}\)
\(=\frac{-8x}{(x^2-4)^2}\)
\(=-\frac{8x}{(x^2-4)^2}\)

ধরি,
\(y=\frac{\cos x}{\ln x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{\cos x}{\ln x})\)
\(=\frac{(\ln x)\frac{d}{dx}(\cos x)-(\cos x)\frac{d}{dx}(\ln x)}{(\ln x)^2}\) ➜ Note \(\because \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(\ln x)(-\sin x)-(\cos x)\frac{1}{x}}{(\ln x)^2}\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=\frac{-\sin x\ln x-\frac{1}{x}\cos x}{(\ln x)^2}\)
\(=-\frac{\sin x\ln x+\frac{1}{x}\cos x}{(\ln x)^2}\)
\(=-\frac{x\sin x\ln x+\cos x}{x(\ln x)^2}\) ➜ Note লব ও হর কে \(x\) দ্বারা গুণ করে।

ধরি,
\(y=\log_ax\times \ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\log_ax\times \ln x)\)
\(=\log_ax\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(\log_ax)\) ➜ Note \(\because \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\log_ax.\frac{1}{x}+\ln x.\frac{1}{x\ln a}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}\)
\(=\frac{1}{x}\log_ax+\frac{1}{x\ln a}\ln x\)
\(=\frac{1}{x\ln a}(\log_ax\ln a+\ln x)\)

ধরি,
\(y=a^xx^a\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(a^xx^a)\)
\(=a^x\frac{d}{dx}(x^a)+x^a\frac{d}{dx}(a^x)\) ➜ Note \(\because \(\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=a^xax^{a-1}+x^aa^x\ln a\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(a^x)=a^x\ln a\)
\(=a^xax^{a-1}+x.x^{a-1}a^x\ln a\)
\(=a^xx^{a-1}(a+x\ln a)\)

ধরি,
\(y=\frac{1+\sin x}{1-\sin x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{1+\sin x}{1-\sin x})\)
\(=\frac{(1-\sin x)\frac{d}{dx}(1+\sin x)-(1+\sin x)\frac{d}{dx}(1-\sin x)}{(1-\sin x)^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1-\sin x)\{\frac{d}{dx}(1)+\frac{d}{dx}(\sin x)\}-(1+\sin x)\{\frac{d}{dx}(1)-\frac{d}{dx}(\sin x)\}}{(1-\sin x)^2}\)
\(=\frac{(1-\sin x)\{0+\cos x\}-(1+\sin x)\{0-\cos x\}}{(1-\sin x)^2}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(c)=0\)
\(=\frac{(1-\sin x)\cos x+(1+\sin x)\cos x}{(1-\sin x)^2}\)
\(=\frac{\cos x(1-\sin x+1+\sin x)}{(1-\sin x)^2}\)
\(=\frac{\cos x.(2)}{(1-\sin x)^2}\)
\(=\frac{2\cos x}{(1-\sin x)^2}\)

ধরি,
\(y=3\sqrt{x}\cos x-7\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(3\sqrt{x}\cos x-7)\)
\(=\frac{d}{dx}(3\sqrt{x}\cos x)-\frac{d}{dx}(7)\)
\(=3\sqrt{x}\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(3\sqrt{x})-0\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u), \frac{d}{dx}(c)=0\)
\(=3\sqrt{x}(-\sin x)+\cos x.3\frac{d}{dx}(\sqrt{x})\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(=-3\sqrt{x}\sin x+3\cos x.\frac{1}{2\sqrt{x}}\)
\(=3\left(\frac{\cos x}{2\sqrt{x}}-\sqrt{x}\sin x\right)\)

ধরি,
\(y=\log_ax\ln x\log x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\log_ax\ln x\log x)\)
\(=\ln x\log x\frac{d}{dx}(\log_ax)+\log_ax\log x\frac{d}{dx}(\ln x)+\log_ax\ln x\frac{d}{dx}(\log x)\) ➜ Note \(\because \frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
\(=\ln x\log x\frac{1}{x\ln a}+\log_ax\log x\frac{1}{x}+\log_ax\ln x\frac{1}{x\ln 10}\) ➜ Note \(\because \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=\frac{1}{x}\left(\frac{\ln x\log x}{\ln a}+\log_ax\log x+\frac{\log_ax\ln x}{\ln 10}\right)\)

ধরি,
\(y=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{\sqrt{x}+1}{\sqrt{x}-1})\)
\(=\frac{(\sqrt{x}-1)\frac{d}{dx}(\sqrt{x}+1)-(\sqrt{x}+1)\frac{d}{dx}(\sqrt{x}-1)}{(\sqrt{x}-1)^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(\sqrt{x}-1)\frac{d}{dx}(\sqrt{x}+1)-(\sqrt{x}+1)\frac{d}{dx}(\sqrt{x}-1)}{(\sqrt{x}-1)^2}\)
\(=\frac{(\sqrt{x}-1)\{\frac{d}{dx}(\sqrt{x})+\frac{d}{dx}(1)\}-(\sqrt{x}+1)\{\frac{d}{dx}(\sqrt{x})-\frac{d}{dx}(1)\}}{(\sqrt{x}-1)^2}\)
\(=\frac{(\sqrt{x}-1)\{\frac{1}{2\sqrt{x}}+0\}-(\sqrt{x}+1)\{\frac{1}{2\sqrt{x}}-0\}}{(\sqrt{x}-1)^2}\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0\)
\(=\frac{(\sqrt{x}-1)\frac{1}{2\sqrt{x}}-(\sqrt{x}+1)\frac{1}{2\sqrt{x}}}{(\sqrt{x}-1)^2}\)
\(=\frac{\frac{1}{2\sqrt{x}}(\sqrt{x}-1-\sqrt{x}-1)}{(\sqrt{x}-1)^2}\)
\(=\frac{\frac{1}{2\sqrt{x}}(-2)}{(\sqrt{x}-1)^2}\)
\(=\frac{-\frac{1}{\sqrt{x}}}{(\sqrt{x}-1)^2}\)
\(=-\frac{1}{\sqrt{x}(\sqrt{x}-1)^2}\)

ধরি,
\(y=\frac{e^x+\ln x}{\log_ax}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{e^x+\ln x}{\log_ax})\)
\(=\frac{(\log_ax)\frac{d}{dx}(e^x+\ln x)-(e^x+\ln x)\frac{d}{dx}(\log_ax)}{(\log_ax)^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\log_ax\{\frac{d}{dx}(e^x)+\frac{d}{dx}(\ln x)\}-(e^x+\ln x)\frac{d}{dx}(\log_ax)}{(\log_ax)^2}\)
\(=\frac{\log_ax\{e^x+\frac{1}{x}\}-(e^x+\ln x)\frac{1}{x\ln a}}{(\log_ax)^2}\) ➜ Note \(\because \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(\ln x)=\frac{1}{x}; \frac{d}{dx}(e^x)=e^x\)
\(=\frac{\frac{1}{x\ln a}\{x\ln a\log_ax(e^x+\frac{1}{x})-(e^x+\ln x)\}}{(\log_ax)^2}\)
\(=\frac{\frac{1}{x\ln a}\{\ln a\log_ax(xe^x+1)-(e^x+\ln x)\}}{(\log_ax)^2}\)
\(=\frac{\ln a\log_ax(xe^x+1)-(e^x+\ln x)}{x\ln a(\log_ax)^2}\)

ধরি,
\(y=\frac{\ln x}{\sqrt{x}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\frac{\ln x}{\sqrt{x}})\)
\(=\frac{\sqrt{x}\frac{d}{dx}(\ln x)-\ln x\frac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\sqrt{x}\frac{1}{x}-\ln x\frac{1}{2\sqrt{x}}}{x}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}; \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(=\frac{\frac{1}{\sqrt{x}}-\ln x\frac{1}{2\sqrt{x}}}{x}\)
\(=\frac{\frac{1}{\sqrt{x}}-\ln x\frac{1}{2}\times \frac{1}{\sqrt{x}}}{x}\)
\(=\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}\ln x\times \frac{1}{\sqrt{x}}}{x}\)
\(=\frac{\frac{1}{\sqrt{x}}-\ln x^{\frac{1}{2}}\times \frac{1}{\sqrt{x}}}{x}\)
\(=\frac{\frac{1}{\sqrt{x}}(1-\ln \sqrt{x})}{x}\)
\(=\frac{1-\ln \sqrt{x}}{\sqrt{x}.x}\)
\(=\frac{1-\ln \sqrt{x}}{x^{\frac{1}{2}}.x}\)
\(=\frac{1-\ln \sqrt{x}}{x^{\frac{1}{2}+1}}\)
\(=\frac{1-\ln \sqrt{x}}{x^{\frac{1+2}{2}}}\)
\(=\frac{1-\ln \sqrt{x}}{x^{\frac{3}{2}}}\)

ধরি,
\(y=x^2\ln x-8e^x\cos x+7\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2\ln x-8e^x\cos x+7)\)
\(=\frac{d}{dx}(x^2\ln x)-\frac{d}{dx}(8e^x\cos x)+\frac{d}{dx}(7)\)
\(=\{x^2\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(x^2)\}-\{8e^x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(8e^x)\}+\frac{d}{dx}(7)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\{x^2\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(x^2)\}-\{8e^x\frac{d}{dx}(\cos x)+8\cos x\frac{d}{dx}(e^x)\}+\frac{d}{dx}(7)\)
\(=\{x^2\frac{1}{x}+\ln x.2x\}-\{8e^x(-\sin x)+8\cos xe^x\}+0\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\cos x)=-\sin x\), \(\frac{d}{dx}(c)=0, \frac{d}{dx}(e^x)=e^x\)
\(=\{x+2x\ln x\}-\{-8e^x\sin x+8e^x\cos x\}\)
\(=x+2x\ln x+8e^x\sin x-8e^x\cos x\)
\(=x(1+2\ln x)+8e^x(\sin x-\cos x)\)

ধরি,
\(y=(x^2+3)(2x^2-1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(x^2+3)(2x^2-1)\}\)
\(=(x^2+3)\frac{d}{dx}(2x^2-1)+(2x^2-1)\frac{d}{dx}(x^2+3)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=(x^2+3)\{\frac{d}{dx}(2x^2)-\frac{d}{dx}(1)\}+(2x^2-1)\{\frac{d}{dx}(x^2)+\frac{d}{dx}(3)\}\)
\(=(x^2+3)\{2\frac{d}{dx}(x^2)-\frac{d}{dx}(1)\}+(2x^2-1)\{\frac{d}{dx}(x^2)+\frac{d}{dx}(3)\}\)
\(=(x^2+3)\{2.2x-0\}+(2x^2-1)\{2x+0\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\)
\(=(x^2+3).4x+(2x^2-1).2x\)
\(=(x^2+3).4x+(2x^2-1).2x\)
\(=4x^3+12x+4x^3-2x\)
\(=8x^3+10x\)

ধরি,
\(y=(3x+4)(5x^2-1)\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}\{(3x+4)(5x^2-1)\}\)
\(=(3x+4)\frac{d}{dx}(5x^2-1)+(5x^2-1)\frac{d}{dx}(3x+4)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=(3x+4)\{\frac{d}{dx}(5x^2)-\frac{d}{dx}(1)\}+(5x^2-1)\{\frac{d}{dx}(3x)+\frac{d}{dx}(4)\}\)
\(=(3x+4)\{5\frac{d}{dx}(x^2)-\frac{d}{dx}(1)\}+(5x^2-1)\{3\frac{d}{dx}(x)+\frac{d}{dx}(4)\}\)
\(=(3x+4)\{5.2x-0\}+(5x^2-1)\{3.1+0\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\)
\(=(3x+4).10x+(5x^2-1).3\)
\(=30x^2+40x+15x^2-3\)
\(=45x^2+40x-3\)

ধরি,
\(y=x^2\ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2\ln x)\)
\(=x^2\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(x^2)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^2.\frac{1}{x}+\ln x.2x\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=x+2x\ln x\)
\(=x(1+2\ln x)\)
\(=x(1+\ln x^2)\)

ধরি,
\(y=x^3e^x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3e^x)\)
\(=x^3\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(x^3)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^3.e^x+e^x.3x^2\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=x^3e^x+3x^2e^x\)
\(=x^2e^x(x+3)\)

ধরি,
\(y=5e^x\ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5e^x\ln x)\)
\(=5e^x\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(5e^x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=5e^x\frac{d}{dx}(\ln x)+5\ln x\frac{d}{dx}(e^x)\)
\(=5e^x.\frac{1}{x}+5\ln x.e^x\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=5e^x.\frac{1}{x}+5e^x\ln x\)
\(=5e^x\left(\frac{1}{x}+\ln x\right)\)

ধরি,
\(y=x^3\log_ax\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3\log_ax)\)
\(=x^3\frac{d}{dx}(\log_ax)+\log_ax\frac{d}{dx}(x^3)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^3\frac{1}{x\ln a}+\log_ax.3x^2\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}\)
\(=x^2\frac{1}{\ln a}+3x^2\log_ax\)
\(=x^2\left(\frac{1}{\ln a}+3\log_ax\right)\)

ধরি,
\(y=5e^x\log_ax\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5e^x\log_ax)\)
\(=5e^x\frac{d}{dx}(\log_ax)+\log_ax\frac{d}{dx}(5e^x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=5e^x\frac{d}{dx}(\log_ax)+5\log_ax\frac{d}{dx}(e^x)\)
\(=5e^x\frac{1}{x\ln a}+5\log_ax.e^x\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}\)
\(=5e^x\frac{1}{x\ln a}+5e^x\log_ax\)
\(=5e^x\left(\frac{1}{x\ln a}+\log_ax\right)\)

ধরি,
\(y=e^{x}\cos x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^{x}\cos x)\)
\(=e^{x}\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(e^{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=e^{x}(-\sin x)+\cos x.e^{x}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\cos x)=-\sin x\)
\(=-e^{x}\sin x+e^x\cos x\)
\(=e^{x}(\cos x-\sin x)\)

ধরি,
\(y=x^ne^x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^ne^x)\)
\(=x^n\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(x^n)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^n.e^x+e^x.nx^{n-1}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=e^xx^n+nx^{n-1}e^x\)
\(=e^xx^n+nx^nx^{-1}e^x\)
\(=e^xx^n+\frac{nx^n}{x}e^x\)
\(=e^xx^n(1+\frac{n}{x})\)

ধরি,
\(y=x^{10}10^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^{10}10^{x})\)
\(=x^{10}\frac{d}{dx}(10^{x})+10^{x}\frac{d}{dx}(x^{10})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^{10}10^{x}\ln {10}+10^{x}.10x^{10-1}\) ➜ Note \(\because \frac{d}{dx}(10^x)=10^x\ln {10}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=x^{10}10^{x}\ln {10}+10x^{9}10^{x}\)
\(= x^{10}10^{x}\left(\ln 10+\frac{10}{x}\right)\)

ধরি,
\(y=x^{10}10^{x}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3\ln x)\)
\(=x^3\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(x^3)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^3.\frac{1}{x}+\ln x.3x^2\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=x^2+3x^2\ln x\)
\(=x^2(1+3\ln x)\)
\(=x^2(1+\ln x^3)\)

ধরি,
\(y=2^{x}\sin x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(2^{x}\sin x)\)
\(=2^{x}\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(2^{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=2^{x}\cos x+\sin x.2^{x}\ln 2\) ➜ Note \(\because \frac{d}{dx}(2^x)=2^x\ln a, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=2^{x}\cos x+2^{x}\sin x\ln 2\)
\(= 2^{x}(\cos x+\ln 2\sin x)\)

ধরি, \(y=ax\ln x+be^x\sin x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(ax\ln x+be^x\sin x)\)
\(=\frac{d}{dx}(ax\ln x)+\frac{d}{dx}(be^x\sin x)\)
\(=a\frac{d}{dx}(x\ln x)+b\frac{d}{dx}(e^x\sin x)\)
\(=a\{x\frac{d}{dx}(\ln x)\}+a\{\ln x\frac{d}{dx}(x)\}+b\{e^x\frac{d}{dx}(\sin x)\}+b\{\sin x\frac{d}{dx}(e^x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=a\{x.\frac{1}{x}\}+a\{\ln x.1\}+b\{e^x\cos x\}+b\{\sin xe^x\}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(x)=1\),\(\frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\sin x)=\cos x\)
\(=a.1+a\ln x+be^x\cos x+be^x\sin x\)
\(=a(1+\ln x)+be^x(\sin x+\cos x)\)

ধরি,
\(y=ax^m\ln x+bx\sin x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(ax^m\ln x+bx\sin x)\)
\(=\frac{d}{dx}(ax^m\ln x)+\frac{d}{dx}(bx\sin x)\)
\(=a\frac{d}{dx}(x^m\ln x)+b\frac{d}{dx}(x\sin x)\)
\(=a\{x^m\frac{d}{dx}(\ln x)\}+a\{\ln x\frac{d}{dx}(x^m)\}+b\{x\frac{d}{dx}(\sin x)\}+b\{\sin x\frac{d}{dx}(x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=a\{x^m.\frac{1}{x}\}+a\{\ln x.mx^{m-1}\}+b\{x\cos x\}+b\{\sin x.1\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\),\(\frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\sin x)=\cos x\)
\(=a\{x^m.\frac{1}{x}\}+a\{\ln x.mx^{m-1}\}+b\{x\cos x\}+b\{\sin x.1\}\)
\(=a\frac{x^m}{x}+amx^{m-1}\ln x+bx\cos x+b\sin x\)
\(=ax^{m-1}+amx^{m-1}\ln x+bx\cos x+b\sin x\)
\(=ax^{m-1}(1+\ln x^{m})+b(x\cos x+\sin x)\)

ধরি,
\(y=x^3\log_ax-5e^x\cos x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3\log_ax-5e^x\cos x)\)
\(=\frac{d}{dx}(x^3\log_ax)-\frac{d}{dx}(5e^x\cos x)\)
\(=\frac{d}{dx}(x^3\log_ax)-5\frac{d}{dx}(e^x\cos x)\)
\(=x^3\frac{d}{dx}(\log_ax)+\log_ax\frac{d}{dx}(x^3)-5\{e^x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(e^x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^3\frac{1}{x\ln a}+\log_ax.3x^2-5\{e^x(-\sin x)+\cos x.e^x\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x\),\(\frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(\cos x)=-\sin x\)
\(=x^2\frac{1}{\ln a}+3x^2\log_ax-5\{-e^x\sin x+e^x\cos x\}\)
\(=x^2\frac{1}{\ln a}+3x^2\log_ax+5e^x\sin x-5e^x\cos x\)
\(=x^2\left(\frac{1}{\ln a}+3\log_ax\right)+5e^x(\sin x-\cos x)\)

ধরি,
\(y=x^2\log_ax+7e^x\cos x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^2\log_ax+7e^x\cos x)\)
\(=\frac{d}{dx}(x^2\log_ax)+\frac{d}{dx}(7e^x\cos x)\)
\(=\frac{d}{dx}(x^2\log_ax)+7\frac{d}{dx}(e^x\cos x)\)
\(=x^2\frac{d}{dx}(\log_ax)+\log_ax\frac{d}{dx}(x^2)+7\{e^x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(e^x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^2\frac{1}{x\ln a}+\log_ax.2x+7\{e^x(-\sin x)+\cos x.e^x\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x\),\(\frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(\cos x)=-\sin x\)
\(=x\frac{1}{\ln a}+2x\log_ax+7\{-e^x\sin x+e^x\cos x\}\)
\(=x\frac{1}{\ln a}+2x\log_ax-7e^x\sin x+7e^x\cos x\)
\(=x\left(\frac{1}{\ln a}+2\log_ax\right)+7e^x(\cos x-\sin x)\)

ধরি,
\(y=\sqrt{x}\sin x-8\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{x}\sin x-8)\)
\(=\frac{d}{dx}(\sqrt{x}\sin x)-\frac{d}{dx}(8)\)
\(=\frac{d}{dx}(\sqrt{x}\sin x)-0\)
\(=\sqrt{x}\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(\sqrt{x})\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\sqrt{x}\cos x+\sin x\frac{1}{2\sqrt{x}}\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(\sin x)=\cos x\)\)
\(=\sqrt{x}\cos x+\frac{\sin x}{2\sqrt{x}}\)

ধরি,
\(y=5\log_ax\times \ln x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(5\log_ax\times \ln x)\)
\(=5\{\log_ax\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(\log_ax)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=5\{\log_ax.\frac{1}{x}+\ln x.\frac{1}{x\ln a}\}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}\)
\(=5\{\frac{1}{x}\log_ax+\frac{1}{x\ln a}\ln x\}\)
\(=\frac{5}{x\ln a}(\log_ax\ln a+\ln x)\)

ধরি,
\(y=e^x\sin x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(e^x\sin x)\)
\(=e^x\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(e^x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=e^x\cos x+\sin xe^x\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(e^x)=e^x\)
\(=e^x\cos x+e^x\sin x\)
\(=e^x(\sin x+\cos x)\)

ধরি,
\(y=7\sqrt{x}\cos x+e^x\sin x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(7\sqrt{x}\cos x+e^x\sin x)\)
\(=\frac{d}{dx}(7\sqrt{x}\cos x)+\frac{d}{dx}(e^x\sin x)\)
\(=7\frac{d}{dx}(\sqrt{x}\cos x)+\frac{d}{dx}(e^x\sin x)\)
\(=7\{\sqrt{x}\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(\sqrt{x})\}+e^x\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(e^x)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=7\{\sqrt{x}(-\sin x)+\cos x\frac{1}{2\sqrt{x}}\}+e^x\cos x+\sin xe^x\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(e^x)=e^x\),\( \frac{d}{dx}(\cos)=-\sin x, \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\)
\(=7\{-\sqrt{x}\sin x+\cos x\frac{1}{2\sqrt{x}}\}+e^x\cos x+\sin xe^x\)
\(=\frac{7\cos x}{2\sqrt{x}}-7\sqrt{x}\sin x+e^x(\sin x+\cos x)\)

ধরি,
\(y=x^3\log_ax+9e^x\cos x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(x^3\log_ax+9e^x\cos x)\)
\(=\frac{d}{dx}(x^3\log_ax)+\frac{d}{dx}(9e^x\cos x)\)
\(=\frac{d}{dx}(x^3\log_ax)+9\frac{d}{dx}(e^x\cos x)\)
\(=x^3\frac{d}{dx}(\log_ax)+\log_ax\frac{d}{dx}(x^3)+9\{e^x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(e^x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=x^3\frac{1}{x\ln a}+\log_ax.3x^2+9\{e^x(-\sin x)+\cos x.e^x\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x\),\(\frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(\cos x)=-\sin x\)
\(=\frac{x^2}{\ln a}+3x^2\log_ax+9\{-e^x\sin x+e^x\cos x\}\)
\(=\frac{x^2}{\ln a}+3x^2\log_ax-9e^x\sin x+9e^x\cos x\)
\(=3x^2\log_ax+\frac{x^2}{\ln a}+9e^x(\cos x-\sin x)\)

ধরি,
\(y=7x^3\log_ax+8e^x\sec x\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(7x^3\log_ax+8e^x\sec x)\)
\(=\frac{d}{dx}(7x^3\log_ax)+\frac{d}{dx}(8e^x\sec x)\)
\(=7\frac{d}{dx}(x^3\log_ax)+8\frac{d}{dx}(e^x\sec x)\)
\(=7\{x^3\frac{d}{dx}(\log_ax)+\log_ax\frac{d}{dx}(x^3)\}+8\{e^x\frac{d}{dx}(\sec x)+\sec x\frac{d}{dx}(e^x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=7\{x^3\frac{1}{x\ln a}+\log_ax.3x^2\}+8\{e^x\sec x\tan x+\sec xe^x\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x\),\(\frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(\sec x)=\sec x\tan x\)
\(=7\{\frac{x^2}{\ln a}+3x^2\log_ax\}+8\{e^x\sec x\tan x+e^x\sec x\}\)
\(=\frac{7x^2}{\ln a}+21x^2\log_ax+8e^x\sec x(\tan x+1)\)

মনে করি,
\(y=\frac{x}{x^2+a^2}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x}{x^2+a^2}\right)\)
\(=\frac{(x^2+a^2)\frac{d}{dx}(x)-x\frac{d}{dx}(x^2+a^2)}{(x^2+a^2)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(x^2+a^2)\frac{d}{dx}(x)-x\{\frac{d}{dx}(x^2)+\frac{d}{dx}(a^2)\}}{(x^2+a^2)^2}\)
\(=\frac{(x^2+a^2).1-x\{2x+0\}}{(x^2+a^2)^2}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=\frac{x^2+a^2-x.2x}{(x^2+a^2)^2}\)
\(=\frac{x^2+a^2-2x^2}{(x^2+a^2)^2}\)
\(=\frac{a^2-x^2}{(x^2+a^2)^2}\)

মনে করি,
\(y=\frac{1+\sqrt{x}}{1-\sqrt{x}}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)\)
\(=\frac{(1-\sqrt{x})\frac{d}{dx}(1+\sqrt{x})-(1+\sqrt{x})\frac{d}{dx}(1-\sqrt{x})}{(1-\sqrt{x})^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1-\sqrt{x})\{\frac{d}{dx}(1)+\frac{d}{dx}(\sqrt{x})\}-(1+\sqrt{x})\{\frac{d}{dx}(1)-\frac{d}{dx}(\sqrt{x})\}}{(1-\sqrt{x})^2}\)
\(=\frac{(1-\sqrt{x})\{0+\frac{1}{2\sqrt{x}}\}-(1+\sqrt{x})\{0-\frac{1}{2\sqrt{x}}\}}{(1-\sqrt{x})^2}\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0\)
\(=\frac{(1-\sqrt{x})\frac{1}{2\sqrt{x}}+(1+\sqrt{x})\frac{1}{2\sqrt{x}}}{(1-\sqrt{x})^2}\)
\(=\frac{\frac{1}{2\sqrt{x}}(1-\sqrt{x}+1+\sqrt{x})}{(1-\sqrt{x})^2}\)
\(=\frac{\frac{1}{2\sqrt{x}}.2}{(1-\sqrt{x})^2}\)
\(=\frac{\frac{1}{\sqrt{x}}}{(1-\sqrt{x})^2}\)
\(=\frac{1}{\sqrt{x}(1-\sqrt{x})^2}\)

মনে করি,
\(y=\frac{x^2}{x^2-4}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^2}{x^2-4}\right)\)
\(=\frac{(x^2-4)\frac{d}{dx}(x^2)-(x^2)\frac{d}{dx}(x^2-4)}{(x^2-4)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(x^2-4)\frac{d}{dx}(x^2)-x^2\{\frac{d}{dx}(x^2)-\frac{d}{dx}(4)\}}{(x^2-4)^2}\)
\(=\frac{(x^2-4).2x-x^2\{2x-0\}}{(x^2-4)^2}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(c)=0\)
\(=\frac{2x^3-8x-x^2.2x}{(x^2-4)^2}\)
\(=\frac{2x^3-8x-2x^3}{(x^2-4)^2}\)
\(=\frac{-8x}{(x^2-4)^2}\)
\(=-\frac{8x}{(x^2-4)^2}\)

মনে করি,
\(y=\frac{x^2+x+1}{x^2-x+1}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^2+x+1}{x^2-x+1}\right)\)
\(=\frac{(x^2-x+1)\frac{d}{dx}(x^2+x+1)-(x^2+x+1)\frac{d}{dx}(x^2-x+1)}{(x^2-x+1)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(x^2-x+1)\{\frac{d}{dx}(x^2)+\frac{d}{dx}(x)+\frac{d}{dx}(1)\}-(x^2+x+1)\{\frac{d}{dx}(x^2)-\frac{d}{dx}(x)+\frac{d}{dx}(1)\}}{(x^2-x+1)^2}\)
\(=\frac{(x^2-x+1)\{2x+1+0\}-(x^2+x+1)\{2x-1+0\}}{(x^2-x+1)^2}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=\frac{(x^2-x+1)(2x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}\)
\(=\frac{(2x^3-2x^2+2x+x^2-x+1)-(2x^3+2x^2+2x-x^2-x-1)}{(x^2-x+1)^2}\)
\(=\frac{(2x^3-x^2+x+1)-(2x^3+x^2+x-1)}{(x^2-x+1)^2}\)
\(=\frac{(2x^3-x^2+x+1-2x^3-x^2-x+1)}{(x^2-x+1)^2}\)
\(=\frac{(-2x^2+2)}{(x^2-x+1)^2}\)
\(=\frac{2(1-x^2)}{(x^2-x+1)^2}\)

মনে করি,
\(y=\frac{1+x+x^2}{x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1+x+x^2}{x}\right)\)
\(=\frac{x\frac{d}{dx}(x^2+x+1)-(x^2+x+1)\frac{d}{dx}(x)}{x^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{x\frac{d}{dx}(x^2+x+1)-(x^2+x+1)\frac{d}{dx}(x)}{x^2}\)
\(=\frac{x\{\frac{d}{dx}(x^2)+\frac{d}{dx}(x)+\frac{d}{dx}(1)\}-(x^2+x+1)\frac{d}{dx}(x)}{x^2}\)
\(=\frac{x\{2x+1+0\}-(x^2+x+1).1}{x^2}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=\frac{x(2x+1)-(x^2+x+1)}{x^2}\)
\(=\frac{2x^2+x-x^2-x-1}{x^2}\)
\(=\frac{x^2-1}{x^2}\)

মনে করি,
\(y=\frac{x}{e^x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x}{e^x}\right)\)
\(=\frac{e^x\frac{d}{dx}(x)-x\frac{d}{dx}(e^x)}{(e^x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{e^x\frac{d}{dx}(x)-x\frac{d}{dx}(e^x)}{(e^x)^2}\)
\(=\frac{e^x.1-xe^x}{(e^x)^2}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(x)=1\)
\(=\frac{e^x(1-x)}{(e^x)^2}\)
\(=\frac{1-x}{e^x}\)

মনে করি,
\(y=\frac{1-\tan x}{1+\tan x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1-\tan x}{1+\tan x}\right)\)
\(=\frac{(1+\tan x)\frac{d}{dx}(1-\tan x)-(1-\tan x)\frac{d}{dx}(1+\tan x)}{(1+\tan x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\tan x)\{\frac{d}{dx}(1)-\frac{d}{dx}(\tan x)\}-(1-\tan x)\{\frac{d}{dx}(1)+\frac{d}{dx}(\tan x)\}}{(1+\tan x)^2}\)
\(=\frac{(1+\tan x)\{0-\sec^2 x\}-(1-\tan x)\{0+\sec^2 x\}}{(1+\tan x)^2}\) ➜ Note \(\because \frac{d}{dx}(\tan x)=\sec^2 x, \frac{d}{dx}(c)=0\)
\(=\frac{(1+\tan x)(-\sec^2 x)-(1-\tan x)\sec^2 x}{(1+\tan x)^2}\)
\(=\frac{-(1+\tan x)\sec^2 x-(1-\tan x)\sec^2 x}{(1+\tan x)^2}\)
\(=\frac{\sec^2 x(-1-\tan x-1+\tan x)}{(1+\tan x)^2}\)
\(=\frac{\sec^2 x(-2)}{(1+\tan x)^2}\)
\(=\frac{-2\sec^2 x}{(1+\tan x)^2}\)

মনে করি,
\(y=\frac{x^n+\tan x}{e^x-\cot x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^n+\tan x}{e^x-\cot x}\right)\)
\(=\frac{(e^x-\cot x)\frac{d}{dx}(x^n+\tan x)-(x^n+\tan x)\frac{d}{dx}(e^x-\cot x)}{(e^x-\cot x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(e^x-\cot x)\{\frac{d}{dx}(x^n)+\frac{d}{dx}(\tan x)\}-(x^n+\tan x)\{\frac{d}{dx}(e^x)-\frac{d}{dx}(\cot x)\}}{(e^x-\cot x)^2}\)
\(=\frac{(e^x-\cot x)\{nx^{n-1}+\sec^2 x\}-(x^n+\tan x)\{e^x-(-cosec^2 \ x)\}}{(e^x-\cot x)^2}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(\tan x)=\sec^2 x\), \(\frac{d}{dx}(\cot x)=-cosec^2 \ x, \frac{d}{dx}(e^x)=e^x\)
\(=\frac{(e^x-\cot x)(nx^{n-1}+\sec^2 x)-(x^n+\tan x)(e^x+cosec^2 \ x)}{(e^x-\cot x)^2}\)

মনে করি,
\(y=\frac{\tan x+\cot x}{3e^x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\tan x+\cot x}{3e^x}\right)\)
\(=\frac{3e^x\frac{d}{dx}(\tan x+\cot x)-(\tan x+\cot x)\frac{d}{dx}(3e^x)}{(3e^x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{3e^x\{\frac{d}{dx}(\tan x)+\frac{d}{dx}(\cot x)\}-(\tan x+\cot x).3\frac{d}{dx}(e^x)}{(3e^x)^2}\)
\(=\frac{3e^x\{\sec^ x-cosec^2 \ x\}-(\tan x+\cot x).3e^x}{(3e^x)^2}\) ➜ Note \(\because \frac{d}{dx}(\cot x)=-cosec^2 \ x, \frac{d}{dx}(\tan x)=\sec^2 x\), \(\frac{d}{dx}(e^x)=e^x\)
\(=\frac{3e^x(\sec^2 x-cosec^2 \ x)-3e^x(\tan x+\cot x)}{(3e^x)^2}\)
\(=\frac{3e^x(\sec^2 x-cosec^2 \ x-\tan x-\cot x)}{(3e^x)^2}\)
\(=\frac{(\sec^2 x-cosec^2 \ x)-(\tan x+\cot x)}{3e^x}\)
\(=\frac{(1+\tan^2 x-1-\cot^2 x)-(\tan x+\cot x)}{3e^x}\) ➜ Note \(\because \sec^2 x=1+\tan^2 x, cosec^2 \ x=1+\cot^2 x \)
\(=\frac{(\tan^2 x-\cot^2 x)-(\tan x+\cot x)}{3e^x}\)
\(=\frac{(\tan x+\cot x)\{(\tan x-\cot x)-1\}}{3e^x}\)
\(=\frac{(\tan x+\cot x)(\tan x-\cot x-1)}{3e^x}\)

মনে করি,
\(y=\frac{\sin x}{x^2+\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\sin x}{x^2+\cos x}\right)\)
\(=\frac{(x^2+\cos x)\frac{d}{dx}(\sin x)-\sin x\frac{d}{dx}(x^2+\cos x)}{(x^2+\cos x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(x^2+\cos x)\frac{d}{dx}(\sin x)-\sin x\{\frac{d}{dx}(x^2)+\frac{d}{dx}(\cos x)\}}{(x^2+\cos x)^2}\)
\(=\frac{(x^2+\cos x)\cos x-\sin x\{2x+(-\sin x)\}}{(x^2+\cos x)^2}\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(\sin x)=\cos x\), \(\frac{d}{dx}(x^n)=nx^{n-1}\)
\(=\frac{(x^2+\cos x)\cos x-\sin x(2x-\sin x)}{(x^2+\cos x)^2}\)
\(=\frac{x^2\cos x+\cos^2 x-2x\sin x+\sin^2 x}{(x^2+\cos x)^2}\)
\(=\frac{x^2\cos x-2x\sin x+\sin^2 x+\cos^2 x}{(x^2+\cos x)^2}\)
\(=\frac{x^2\cos x-2x\sin x+1}{(x^2+\cos x)^2}\) ➜ Note \(\because \sin^2 x+\cos^2 x=1\)
\(=\frac{1+x(x\cos x-2\sin x)}{(x^2+\cos x)^2}\)

মনে করি,
\(y=\frac{\log_ax}{x^2}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\log_ax}{x^2}\right)\)
\(=\frac{x^2\frac{d}{dx}(\log_ax)-\log_ax\frac{d}{dx}(x^2)}{(x^2)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{x^2\frac{1}{x\ln a}-\log_ax.2x}{x^4}\) ➜ Note \(\because \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=\frac{\frac{x}{\ln a}-2x\log_ax}{x^4}\)
\(=\frac{\frac{x}{\ln a}(1-2\ln a\log_ax)}{x^4}\)
\(=\frac{x(1-2\ln a\log_ax)}{x^4\ln a}\)
\(=\frac{1-\ln x^2}{x^3\ln a}\) ➜ Note \(\because \ln a\log_ax=\ln x\)

মনে করি,
\(y=\frac{x^{n}}{\log_ax}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^{n}}{\log_ax}\right)\)
\(=\frac{\log_ax\frac{d}{dx}(x^{n})-x^{n}\frac{d}{dx}(\log_ax)}{(\log_ax)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\log_ax.nx^{n-1}-x^{n}\frac{1}{x\ln a}}{(\log_ax)^2}\) ➜ Note \(\because \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=\frac{nx^{n-1}\log_ax-x^{n-1}\frac{1}{\ln a}}{(\log_ax)^2}\)
\(=\frac{x^{n-1}\frac{1}{\ln a}(n\ln a\log_ax-1)}{(\log_ax)^2}\)
\(=\frac{x^{n-1}(n\ln x-1)}{\ln a(\log_ax)^2}\) ➜ Note \(\because \ln a\log_ax=\ln x\)

মনে করি,
\(y=\frac{1-\ln x}{1+\ln x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1-\ln x}{1+\ln x}\right)\)
\(=\frac{(1+\ln x)\frac{d}{dx}(1-\ln x)-(1-\ln x)\frac{d}{dx}(1+\ln x)}{(1+\ln x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\ln x)\{\frac{d}{dx}(1)-\frac{d}{dx}(\ln x)\}-(1-\ln x)\{\frac{d}{dx}(1)+\frac{d}{dx}(\ln x)\}}{(1+\ln x)^2}\)
\(=\frac{(1+\ln x)\{0-\frac{1}{x}\}-(1-\ln x)\{0+\frac{1}{x}\}}{(1+\ln x)^2}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(c)=0\)
\(=\frac{-(1+\ln x)\frac{1}{x}-(1-\ln x)\frac{1}{x}}{(1+\ln x)^2}\)
\(=\frac{\frac{1}{x}(-1-\ln x-1+\ln x)}{(1+\ln x)^2}\)
\(=\frac{\frac{1}{x}.(-2)}{(1+\ln x)^2}\)
\(=-\frac{2}{x(1+\ln x)^2}\)

মনে করি,
\(y=\frac{\sec x}{1+\sec x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\sec x}{1+\sec x}\right)\)
\(=\frac{(1+\sec x)\frac{d}{dx}(\sec x)-\sec x\frac{d}{dx}(1+\sec x)}{(1+\sec x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\sec x)\frac{d}{dx}(\sec x)-\sec x\{\frac{d}{dx}(1)+\frac{d}{dx}(\sec x)\}}{(1+\sec x)^2}\)
\(=\frac{(1+\sec x)\sec x\tan x-\sec x\{0+\sec x\tan x\}}{(1+\sec x)^2}\) ➜ Note \(\because \frac{d}{dx}(\sec x)=\sec x\tan x, \frac{d}{dx}(c)=0\)
\(=\frac{(1+\sec x)\sec x\tan x-\sec^2 x\tan x}{(1+\sec x)^2}\)
\(=\frac{\sec x\tan x(1+\sec x-\sec x)}{(1+\sec x)^2}\)
\(=\frac{\frac{1}{\cos x}\times \frac{\sin x}{\cos x}.(1)}{\left(1+\frac{1}{\cos x}\right)^2}\) ➜ Note \(\because \sec x=\frac{1}{\cos x}, tan x=\frac{\sin x}{\cos x}\)
\(=\frac{\frac{\sin x}{\cos^2 x}}{\left(\frac{\cos x+1}{\cos x}\right)^2}\)
\(=\frac{\sin x}{\cos^2 x\frac{(\cos x+1)^2}{\cos^2 x}}\)
\(=\frac{\sin x}{(\cos x+1)^2}\)
\(=\frac{\sin x}{(1+\cos x)^2}\)

মনে করি,
\(y=\frac{e^x-\tan x}{\ln x-\cot x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{e^x-\tan x}{\ln x-\cot x}\right)\)
\(=\frac{(\ln x-\cot x)\frac{d}{dx}(e^x-\tan x)-(e^x-\tan x)\frac{d}{dx}(\ln x-\cot x)}{(\ln x-\cot x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(\ln x-\cot x)\{\frac{d}{dx}(e^x)-\frac{d}{dx}(\tan x)\}-(e^x-\tan x)\{\frac{d}{dx}(\ln x)-\frac{d}{dx}(\cot x)\}}{(\ln x-\cot x)^2}\)
\(=\frac{(\ln x-\cot x)\{e^x-\sec^2 x\}-(e^x-\tan x)\{\frac{1}{x}-(-cosec^2 \ x)\}}{(\ln x-\cot x)^2}\) ➜ Note \(\because \frac{d}{dx}(\tan x)=\sec^2 x, \frac{d}{dx}(\ln x)=\frac{1}{x}\),\(\frac{d}{dx}(\cot x)=-cosec^2 \ x, \frac{d}{dx}(e^x)=e^x\)
\(=\frac{(\ln x-\cot x)(e^x-\sec^2 x)-(e^x-\tan x)\left(\frac{1}{x}+cosec^2 \ x\right)}{(\ln x-\cot x)^2}\)

মনে করি,
\(y=\frac{\ln x}{\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\ln x}{\cos x}\right)\)
\(=\frac{\cos x\frac{d}{dx}(\ln x)-\ln x\frac{d}{dx}(\cos x)}{(\cos x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\cos x\frac{1}{x}-\ln x(-\sin x)}{(\cos x)^2}\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=\frac{\cos x\frac{1}{x}+\ln x\sin x}{(\cos x)^2}\)
\(=\frac{\cos x+x\ln x\sin x}{x(\cos x)^2}\)

মনে করি,
\(y=\frac{\sin x+\cos x}{\sin 2x}\)
\(=\frac{\sin x+\cos x}{\sin 2x}\)
\(=\frac{\cos x\left(\frac{\sin x}{\cos x}+1\right)}{\sin 2x}\)
\(=\frac{\cos x(\tan x+1)}{2\sin x\cos x}\) ➜ Note \(\because \tan x=\frac{\sin x}{\cos x}, \sin 2x=2\sin x\cos x\)
\(=\frac{\tan x+1}{2\sin x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\tan x+1}{2\sin x}\right)\)
\(=\frac{2\sin x\frac{d}{dx}(\tan x+1)-(\tan x+1)\frac{d}{dx}(2\sin x)}{(2\sin x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{2\sin x\{\frac{d}{dx}(\tan x)+\frac{d}{dx}(1)\}-2(\tan x+1)\frac{d}{dx}(\sin x)}{(2\sin x)^2}\)
\(=\frac{2\sin x\{\sec^2 x+0\}-2(\tan x+1)\cos x}{4\sin^2 x}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(\tan x)=\sec^2 x, \frac{d}{dx}(c)=0\)
\(=\frac{2\sin x\sec^2 x-2\cos x(\tan x+1)}{4\sin^2 x}\)
\(=\frac{2\sin x\sec^2 x-2\cos x\left(\frac{\sin x}{\cos x}+1\right)}{4\sin^2 x}\) ➜ Note \(\because \tan x=\frac{\sin x}{\cos x}\)
\(=\frac{2\sin x\sec^2 x-2\sin x-2\cos x}{4\sin^2 x}\)
\(=\frac{2\sin x\left(\sec^2 x-1-\frac{\cos x}{\sin x}\right)}{4\sin^2 x}\)
\(=\frac{\sec^2 x-1-\cot x}{2\sin x}\)
\(=\frac{1+\tan^2 x-1-\cot x}{2\sin x}\)
\(=\frac{\tan^2 x-\cot x}{2\sin x}\)
\(=\frac{1}{2}\frac{\tan^2 x-\cot x}{\sin x}\)
\(=\frac{1}{2}\left(\frac{\tan^2 x}{\sin x}-\frac{\cot x}{\sin x}\right)\)
\(=\frac{1}{2}\left(\frac{\frac{\sin^2 x}{\cos^2 x}}{\sin x}-\frac{\frac{\cos x}{\sin x}}{\sin x}\right)\) ➜ Note \(\because \tan x=\frac{\sin x}{\cos x}, \cot x=\frac{\cos x}{\sin x}\)
\(=\frac{1}{2}\left(\frac{\sin^2 x}{\cos^2 x\sin x}-\frac{\cos x}{\sin^2 x}\right)\)
\(=\frac{1}{2}\left(\frac{\sin x}{\cos^2 x}-\frac{\cos x}{\sin^2 x}\right)\)
\(=\frac{1}{2}\left(\frac{1}{\cos x}\times \frac{\sin x}{\cos x}-\frac{1}{\sin x}\times \frac{\cos x}{\sin x}\right)\)
\(=\frac{1}{2}(\sec x\tan x-cosec \ x\cot x)\) ➜ Note \(\because \tan x=\frac{\sin x}{\cos x},\cot x=\frac{\cos x}{\sin x}\), \(\sec x=\frac{1}{\cos x}, cosec \ x=\frac{1}{\sin x}\)

মনে করি,
\(y=\frac{1+\sin x}{1+\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1+\sin x}{1+\cos x}\right)\)
\(=\frac{(1+\cos x)\frac{d}{dx}(1+\sin x)-(1+\sin x)\frac{d}{dx}(1+\cos x)}{(1+\cos x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\cos x)\{\frac{d}{dx}(1)+\frac{d}{dx}(\sin x)\}-(1+\sin x)\{\frac{d}{dx}(1)+\frac{d}{dx}(\cos x)\}}{(1+\cos x)^2}\)
\(=\frac{(1+\cos x)\{0+\cos x\}-(1+\sin x)\{0+(-\sin x)\}}{(1+\cos x)^2}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(\cos x)=-\sin x\)
\(=\frac{(1+\cos x)\cos x-(1+\sin x)\{-\sin x\}}{(1+\cos x)^2}\)
\(=\frac{(1+\cos x)\cos x+\sin x(1+\sin x)}{(1+\cos x)^2}\)
\(=\frac{\cos x+\cos^2 x+\sin x+\sin^2 x}{(1+\cos x)^2}\)
\(=\frac{\sin x+\cos x+\sin^2 x+\cos^2 x}{(1+\cos x)^2}\)
\(=\frac{\sin x+\cos x+1}{(1+\cos x)^2}\) ➜ Note \(\because \sin^2 x+\cos^2 x=1\)

মনে করি,
\(y=\frac{\sin x}{x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\sin x}{x}\right)\)
\(=\frac{x\frac{d}{dx}(\sin x)-\sin x\frac{d}{dx}(x)}{x^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{x\cos x-\sin x.1}{x^2}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(x)=1\)
\(=\frac{x\cos x-\sin x}{x^2}\)

মনে করি,
\(y=\frac{x^4}{\ln x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^4}{\ln x}\right)\)
\(=\frac{\ln x\frac{d}{dx}(x^4)-x^4\frac{d}{dx}(\ln x)}{(\ln x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\ln x.4x^3-x^4\frac{1}{x}}{(\ln x)^2}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=\frac{4x^3\ln x-x^3}{(\ln x)^2}\)
\(=\frac{x^3(4\ln x-1)}{(\ln x)^2}\)
\(=\frac{x^3(\ln x^4-1)}{(\ln x)^2}\)

মনে করি,
\(y=\frac{\tan x-\cot x}{\tan x+\cot x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\tan x-\cot x}{\tan x+\cot x}\right)\)
\(=\frac{(\tan x+\cot x)\frac{d}{dx}(\tan x-\cot x)-(\tan x-\cot x)\frac{d}{dx}(\tan x+\cot x)}{(\tan x+\cot x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(\tan x+\cot x)\{\frac{d}{dx}(\tan x)-\frac{d}{dx}(\cot x)\}-(\tan x-\cot x)\{\frac{d}{dx}(\tan x)+\frac{d}{dx}(\cot x)\}}{(\tan x+\cot x)^2}\)
\(=\frac{(\tan x+\cot x)\{\sec^2 x-(-cosec^2 \ x)\}-(\tan x-\cot x)\{\sec^2 x+(-cosec^2 \ x)\}}{(\tan x+\cot x)^2}\) ➜ Note \(\because \frac{d}{dx}(\tan x)=\sec^2 x, \frac{d}{dx}(\cot x)=-cosec^2 \ x\)
\(=\frac{(\tan x+\cot x)\{\sec^2 x+cosec^2 \ x\}-(\tan x-\cot x)\{\sec^2 x-cosec^2 \ x\}}{(\tan x+\cot x)^2}\)
\(=\frac{(\sec^2 x\tan x+\sec^2 x\cot x+cosec^2 \ x\tan x+cosec^2 \ x\cot x)-(\sec^2 x\tan x-\sec^2 x\cot x-cosec^2 \ x\tan x+cosec^2 \ x\cot x)}{(\tan x+\cot x)^2}\)
\(=\frac{\sec^2 x\tan x+\sec^2 x\cot x+cosec^2 \ x\tan x+cosec^2 \ x\cot x-\sec^2 x\tan x+\sec^2 x\cot x+cosec^2 \ x\tan x-cosec^2 \ x\cot x}{(\tan x+\cot x)^2}\)
\(=\frac{2\sec^2 x\cot x+2cosec^2 \ x\tan x}{(\tan x+\cot x)^2}\)
\(=\frac{2\frac{1}{\cos^2 x}\times \frac{\cos x}{\sin x}+2\frac{1}{\sin^2 x}\times \frac{\sin x}{\cos x}}{\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)^2}\) ➜ Note \(\because \tan x=\frac{\sin x}{\cos x}, \cot x=\frac{\cos x}{\sin x}\),\(\sec x=\frac{1}{\cos x}, cosec \ x=\frac{1}{\sin x}\)
\(=\frac{2\frac{1}{\cos x\sin x}+2\frac{1}{\cos x\sin x}}{\left(\frac{\sin^2 x+\cos^2 x}{\sin x\cos x}\right)^2}\)
\(=\frac{4\frac{1}{\cos x\sin x}}{\frac{1}{\sin^2 x\cos^2 x}}\) ➜ Note \(\because \sin^2 x+\cos^2 x=1\)
\(=4\frac{1}{\cos x\sin x}\times \frac{\sin^2 x\cos^2 x}{1}\)
\(=4\sin x\cos x\)
\(=2.2\sin x\cos x\)
\(=2\sin 2x\) ➜ Note \(\because 2\sin x\cos x=\sin 2x\)

মনে করি,
\(y=\frac{a\sin x+b\cos x}{a\cos x-b\sin x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{a\sin x+b\cos x}{a\cos x-b\sin x}\right)\)
\(=\frac{(a\cos x-b\sin x)\frac{d}{dx}(a\sin x+b\cos x)-(a\sin x+b\cos x)\frac{d}{dx}(a\cos x-b\sin x)}{(a\cos x-b\sin x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(a\cos x-b\sin x)\{\frac{d}{dx}(a\sin x)+\frac{d}{dx}(b\cos x)\}-(a\sin x+b\cos x)\{\frac{d}{dx}(a\cos x)-\frac{d}{dx}(b\sin x)\}}{(a\cos x-b\sin x)^2}\)
\(=\frac{(a\cos x-b\sin x)\{a\frac{d}{dx}(\sin x)+b\frac{d}{dx}(\cos x)\}-(a\sin x+b\cos x)\{a\frac{d}{dx}(\cos x)-b\frac{d}{dx}(\sin x)\}}{(a\cos x-b\sin x)^2}\)
\(=\frac{(a\cos x-b\sin x)\{a\cos x+b(-\sin x)\}-(a\sin x+b\cos x)\{a(-\sin x)-b\cos x\}}{(a\cos x-b\sin x)^2}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(\cos x)=-\sin x\)
\(=\frac{(a\cos x-b\sin x)\{a\cos x-b\sin x\}+(a\sin x+b\cos x)\{a\sin x+b\cos x\}}{(a\cos x-b\sin x)^2}\)
\(=\frac{(a\cos x-b\sin x)^2+(a\sin x+b\cos x)^2}{(a\cos x-b\sin x)^2}\)
\(=\frac{a^2\cos^2 x-2ab\sin x\cos x+b^2\sin^2 x+a^2\sin^2 x+2ab\sin x\cos x+b^2\cos^2 x}{(a\cos x-b\sin x)^2}\)
\(=\frac{a^2\cos^2 x+b^2\sin^2 x+a^2\sin^2 x+b^2\cos^2 x}{(a\cos x-b\sin x)^2}\)
\(=\frac{a^2(\sin^2 x+\cos^2 x)+b^2(\sin^2 x+\cos^2 x)}{(a\cos x-b\sin x)^2}\)
\(=\frac{a^2.(1)+b^2.(1)}{(a\cos x-b\sin x)^2}\) ➜ Note \(\because \sin^2 x+\cos^2 x=1\)
\(=\frac{a^2+b^2}{(a\cos x-b\sin x)^2}\)

মনে করি,
\(y=\frac{cosec \ x+\cot x}{cosec \ x-\cot x}\)
\(=\frac{\frac{1}{\sin x}+\frac{\cos x}{\sin x}}{\frac{1}{\sin x}-\frac{\cos x}{\sin x}}\) ➜ Note \(\because cosec \ x=\frac{1}{\sin x}, \cot x=\frac{\cos x}{\sin x}\)
\(=\frac{\frac{1+\cos x}{\sin x}}{\frac{1-\cos x}{\sin x}}\)
\(=\frac{1+\cos x}{\sin x}\times \frac{\sin x}{1-\cos x}\)
\(=\frac{1+\cos x}{1-\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1+\cos x}{1-\cos x}\right)\)
\(=\frac{(1-\cos x)\frac{d}{dx}(1+\cos x)-(1+\cos x)\frac{d}{dx}(1-\cos x)}{(1-\cos x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1-\cos x)\{\frac{d}{dx}(1+\cos x)\}-(1+\cos x)\{\frac{d}{dx}(1-\cos x)\}}{(1-\cos x)^2}\)
\(=\frac{(1-\cos x)\{\frac{d}{dx}(1)+\frac{d}{dx}(\cos x)\}-(1+\cos x)\{\frac{d}{dx}(1)-\frac{d}{dx}(\cos x)\}}{(1-\cos x)^2}\)
\(=\frac{(1-\cos x)\{0+(-\sin x)\}-(1+\cos x)\{0-(-\sin x)\}}{(1-\cos x)^2}\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x\)
\(=\frac{-(1-\cos x)\sin x-(1+\cos x)\sin x}{(1-\cos x)^2}\)
\(=\frac{\sin x(-1+\cos x-1-\cos x)}{(1-\cos x)^2}\)
\(=\frac{\sin x.(-2)}{(1-\cos x)^2}\)
\(=-\frac{2\sin x}{(1-\cos x)^2}\)

মনে করি,
\(y=\frac{e^x}{\ln x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{e^x}{\ln x}\right)\)
\(=\frac{\ln x\frac{d}{dx}(e^x)-e^x\frac{d}{dx}(\ln x)}{(\ln x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\ln x.e^x-e^x\frac{1}{x}}{(\ln x)^2}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(e^x)=e^x\)
\(=\frac{\ln x.e^x-\frac{e^x}{x}}{(\ln x)^2}\)
\(=\frac{\frac{e^x}{x}(x\ln x-1)}{(\ln x)^2}\)
\(=\frac{e^x(x\ln x-1)}{x(\ln x)^2}\)

মনে করি,
\(y=\frac{\ln x}{1+\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{\ln x}{1+\cos x}\right)\)
\(=\frac{(1+\cos x)\frac{d}{dx}(\ln x)-\ln x\frac{d}{dx}(1+\cos x)}{(1+\cos x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\cos x)\frac{d}{dx}(\ln x)-\ln x\{\frac{d}{dx}(1)+\frac{d}{dx}(\cos x)\}}{(1+\cos x)^2}\)
\(=\frac{(1+\cos x)\frac{1}{x}-\ln x\{0+(-\sin x)\}}{(1+\cos x)^2}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(c)=0\)
\(=\frac{(1+\cos x)\frac{1}{x}+\ln x\sin x}{(1+\cos x)^2}\)
\(=\frac{1+\cos x+x\ln x\sin x}{x(1+\cos x)^2}\) ➜ Note লব ও হরের সহিত \(x\) গুণ করে।

মনে করি,
\(y=\frac{e^x}{\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{e^x}{\cos x}\right)\)
\(=\frac{\cos x\frac{d}{dx}(e^x)-e^x\frac{d}{dx}(\cos x)}{\cos^2 x}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\cos x.e^x-e^x(-\sin x)}{\cos^2 x}\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(e^x)=e^x\)
\(=\frac{e^x\cos x+e^x\sin x}{\cos^2 x}\)
\(=\frac{e^x(\sin x+\cos x)}{\cos^2 x}\)

মনে করি,
\(y=\frac{1-\cot x}{1+\cot x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1-\cot x}{1+\cot x}\right)\)
\(=\frac{(1+\cot x)\frac{d}{dx}(1-\cot x)-(1-\cot x)\frac{d}{dx}(1+\cot x)}{(1+\cot x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\cot x)\{\frac{d}{dx}(1)-\frac{d}{dx}(\cot x)\}-(1-\cot x)\{\frac{d}{dx}(1)+\frac{d}{dx}(\cot x)\}}{(1+\cot x)^2}\)
\(=\frac{(1+\cot x)\{0-(-cosec^2 \ x)\}-(1-\cot x)\{0+(-cosec^2 \ x)\}}{(1+\cot x)^2}\) ➜ Note \(\because \frac{d}{dx}(\cot x)=-cosec^2 \ x, \frac{d}{dx}(c)=0\)
\(=\frac{(1+\cot x)cosec^2 \ x+(1-\cot x)cosec^2 \ x}{(1+\cot x)^2}\)
\(=\frac{cosec^2 \ x(1+\cot x+1-\cot x)}{(1+\cot x)^2}\)
\(=\frac{cosec^2 \ x.(2)}{(1+\cot x)^2}\)
\(=\frac{2cosec^2 \ x}{(1+\cot x)^2}\)

মনে করি,
\(y=\frac{x^n+\cot x}{e^x-\tan x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x^n+\cot x}{e^x-\tan x}\right)\)
\(=\frac{(e^x-\tan x)\frac{d}{dx}(x^n+\cot x)-(x^n+\cot x)\frac{d}{dx}(e^x-\tan x)}{(e^x-\tan x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(e^x-\tan x)\{\frac{d}{dx}(x^n)+\frac{d}{dx}(\cot x)\}-(x^n+\cot x)\{\frac{d}{dx}(e^x)-\frac{d}{dx}(\tan x)\}}{(e^x-\tan x)^2}\)
\(=\frac{(e^x-\tan x)\{nx^{n-1}+(-cosec^2 \ x)\}-(x^n+\cot x)\{e^x-\sec^2 x\}}{(e^x-\tan x)^2}\) ➜ Note \(\because \frac{d}{dx}(\cot x)=-cosec^2 \ x, \frac{d}{dx}(x^n)=nx^{n-1}\), \(\frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\tan x)=\sec^2 x \)
\(=\frac{(e^x-\tan x)(nx^{n-1}-cosec^2 \ x)-(x^n+\cot x)(e^x-\sec^2 x)}{(e^x-\tan x)^2}\)

মনে করি,
\(y=\frac{x\sin x}{1+\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x\sin x}{1+\cos x}\right)\)
\(=\frac{(1+\cos x)\frac{d}{dx}(x\sin x)-x\sin x\frac{d}{dx}(1+\cos x)}{(1+\cos x)^2}\) ➜ Note \(\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\cos x)\{x\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(x)\}-x\sin x\{\frac{d}{dx}(1)+\frac{d}{dx}(\cos x)\}}{(1+\cos x)^2}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\frac{(1+\cos x)\{x\cos x+\sin x.1\}-x\sin x\{0+(-\sin x)\}}{(1+\cos x)^2}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(c)=0\)
\(=\frac{(1+\cos x)\{x\cos x+\sin x\}+x\sin^2 x}{(1+\cos x)^2}\)
\(=\frac{x\cos x+\sin x+\sin x\cos x+x(\sin^2 x+\cos^2 x)}{(1+\cos x)^2}\)
\(=\frac{x\cos x+\sin x+\sin x\cos x+x.(1)}{(1+\cos x)^2}\)
\(=\frac{x\cos x+\sin x+\sin x\cos x+x}{(1+\cos x)^2}\)
\(=\frac{x+x\cos x+\sin x+\sin x\cos x}{(1+\cos x)^2}\)
\(=\frac{x(1+\cos x)+\sin x(1+\cos x)}{(1+\cos x)^2}\)
\(=\frac{(1+\cos x)(x+\sin x)}{(1+\cos x)^2}\)
\(=\frac{x+\sin x}{1+\cos x}\)

মনে করি,
\(y=e^{x}\cos x\ln x\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(e^{x}\cos x\ln x)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\cos x\ln x\frac{d}{dx}(e^x)+e^{x}\ln x\frac{d}{dx}(\cos x)+e^{x}\cos x\frac{d}{dx}(\ln x)\) ➜ Note \(\because \frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
\(=\cos x\ln x.e^x+e^{x}\ln x.(-\sin x)+e^{x}\cos x.\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(e^x)=e^x\)
\(=e^x\cos x\ln x-e^{x}\ln x\sin x+e^{x}\cos x.\frac{1}{x}\)
\(=\frac{e^x}{x}(x\cos x\ln x-x\ln x\sin x+\cos x)\)

মনে করি,
\(y=\frac{e^x}{x^2\ln x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{e^x}{x^2\ln x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{x^2\ln x\frac{d}{dx}(e^x)-e^x\frac{d}{dx}(x^2\ln x)}{(x^2\ln x)^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{x^2\ln x\frac{d}{dx}(e^x)-e^x\{\ln x\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(\ln x)\}}{(x^2\ln x)^2}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\frac{x^2\ln x.e^x-e^x\{\ln x.2x+x^2.\frac{1}{x}\}}{(x^2\ln x)^2}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=\frac{e^xx^2\ln x-e^x(2x\ln x+x)}{(x^2\ln x)^2}\)
\(=\frac{e^xx^2\ln x-2xe^x\ln x+xe^x}{(x^2\ln x)^2}\)
\(=\frac{xe^x(x\ln x-2\ln x+1)}{(x^2\ln x)^2}\)
\(=\frac{xe^x(x\ln x-\ln x^2+1)}{x^4(\ln x)^2}\)
\(=\frac{e^x(x\ln x-\ln x^2+1)}{x^3(\ln x)^2}\)

মনে করি,
\(y=\frac{e^{x}\ln x}{\cos x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{e^{x}\ln x}{\cos x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{\cos x\frac{d}{dx}(e^{x}\ln x)-e^{x}\ln x\frac{d}{dx}(\cos x)}{\cos^2 x}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\cos x\{e^{x}\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(e^x)\}-e^{x}\ln x\frac{d}{dx}(\cos x)}{\cos^2 x}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\frac{\cos x\{e^{x}.\frac{1}{x}+\ln x.e^x\}-e^{x}\ln x(-\sin x)}{\cos^2 x}\) ➜ Note \(\because \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(\cos x)=-\sin x\)
\(=\frac{\cos x(\frac{e^{x}}{x}+e^{x}\ln x)+e^{x}\ln x\sin x}{\cos^2 x}\)
\(=\frac{\frac{e^{x}}{x}\cos x+e^{x}\ln x\cos x+e^{x}\ln x\sin x}{\cos^2 x}\)
\(=\frac{\frac{e^{x}}{x}\cos x+e^{x}\ln x(\cos x+\sin x)}{\cos^2 x}\)
\(=\frac{e^{x}\cos x+xe^{x}\ln x(\sin x+\cos x)}{x\cos^2 x}\) ➜ Note লব ও হরের সহিত \(x\) গুণ করে।

মনে করি,
\(y=x^2e^x\ln x\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^2e^x\ln x)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=e^x\ln x\frac{d}{dx}(x^2)+x^2\ln x\frac{d}{dx}(e^x)+x^2e^x\frac{d}{dx}(\ln x)\) ➜ Note \(\because \frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
\(=e^x\ln x.2x+x^2\ln x.e^x+x^2e^x.\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=2xe^x\ln x+x^2e^x\ln x+xe^x\)
\(=xe^x(2\ln x+x\ln x+1)\)
\(=xe^x(\ln x^2+x\ln x+1)\)

মনে করি,
\(y=x^2\sin x\ln x\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^2\sin x\ln x)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\sin x\ln x\frac{d}{dx}(x^2)+x^2\ln x\frac{d}{dx}(\sin x)+x^2\sin x\frac{d}{dx}(\ln x)\) ➜ Note \(\because \frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
\(=\sin x\ln x.2x+x^2\ln x.\cos x+x^2\sin x.\frac{1}{x}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=2x\sin x\ln x+x^2\ln x\cos x+x\sin x\)
\(=x\sin x+x\ln x(2\sin x+x\cos x)\)
\(=x\sin x+x\ln x(x\cos x+2\sin x)\)

মনে করি,
\(y=x\ln x\log_ax\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x\ln x\log_ax)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\ln x\log_ax\frac{d}{dx}(x)+x\log_ax\frac{d}{dx}(\ln x)+x\ln x\frac{d}{dx}(\log_ax)\) ➜ Note \(\because \frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
\(=\ln x\log_ax.1+x\log_ax.\frac{1}{x}+x\ln x.\frac{1}{x\ln a}\) ➜ Note \(\because \frac{d}{dx}(x)=1, \frac{d}{dx}(\log_ax)=\frac{1}{x\ln a}, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=\ln x\log_ax+\log_ax+\frac{1}{\ln a}\ln x\)
\(=\ln x\log_ax+\log_ax+\ln_ae\times \ln_ex\)
\(=\ln x\log_ax+\log_ax+\log_ax\) ➜ Note \(\because \ln_ae\times \ln_ex=\log_ax\)
\(=\ln x\log_ax+2\log_ax\)
\(=\log_ax(\ln x+2)\)

মনে করি,
\(y=e^x\ln x(x^3+x^4)\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\{e^x\ln x(x^3+x^4)\}\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\ln x(x^3+x^4)\frac{d}{dx}(e^x)+e^x(x^3+x^4)\frac{d}{dx}(\ln x)+e^x\ln x\frac{d}{dx}(x^3+x^4)\) ➜ Note \(\because \frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
\(=\ln x(x^3+x^4)\frac{d}{dx}(e^x)+e^x(x^3+x^4)\frac{d}{dx}(\ln x)+e^x\ln x\{\frac{d}{dx}(x^3)+\frac{d}{dx}(x^4)\}\)
\(=\ln x(x^3+x^4).e^x+e^x(x^3+x^4).\frac{1}{x}+e^x\ln x\{3x^2+4x^3\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=e^x\ln x(x^3+x^4)+e^x.x(x^2+x^3).\frac{1}{x}+e^x\ln x(3x^2+4x^3)\)
\(=e^x\ln x(x^3+x^4)+e^x(x^2+x^3)+e^x\ln x(3x^2+4x^3)\)
\(=e^x(x^2+x^3)+e^x\ln x(x^3+x^4)+e^x\ln x(3x^2+4x^3)\)
\(=e^x(x^2+x^3)+e^x\ln x(x^3+x^4+3x^2+4x^3)\)
\(=e^x(x^2+x^3)+e^x\ln x(x^4+5x^3+3x^2)\)

মনে করি,
\(y=x^2\log_ax-x^3\ln a^x+6xe^x\ln x\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^2\log_ax-x^3\ln a^x+6xe^x\ln x)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{d}{dx}(x^2\log_ax)-\frac{d}{dx}(x^3\ln a^x)+\frac{d}{dx}(6xe^x\ln x)\)
\(=\frac{d}{dx}(x^2\log_ax)-\frac{d}{dx}(x^3\ln a^x)+6\frac{d}{dx}(xe^x\ln x)\)
\(=x^2\frac{d}{dx}(\log_ax)+\log_ax\frac{d}{dx}(x^2)\)-\(\{x^3\frac{d}{dx}(\ln a^x)+\ln a^x\frac{d}{dx}(x^3)\}\)+\(6\{e^x\ln x\frac{d}{dx}(x)+x\ln x\frac{d}{dx}(e^x)+xe^x\frac{d}{dx}(\ln x)\}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\), \(\frac{d}{dx}(uvw)=vw\frac{d}{dx}(u)+uw\frac{d}{dx}(v)+uv\frac{d}{dx}(w)\)
\(=x^2.\frac{1}{x\ln a}+\log_ax.2x\)-\(\{x^3\frac{d}{dx}(x\ln a)+\ln a^x.3x^2\}\)+\(6\{e^x\ln x.1+x\ln x.e^x+xe^x.\frac{1}{x}\}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=x\frac{1}{\ln a}+2x\log_ax\)-\(\{x^3\ln a\frac{d}{dx}(x)+3x^2\ln a^x\}\)+\(6\{e^x\ln x+xe^x\ln x+e^x\}\)
\(=x\frac{1}{\ln a}+2x\log_ax\)-\(\{x^3\ln a.1+3x^2\ln a^x\}\)+\(6e^x(\ln x+x\ln x+1)\)
\(=x\frac{1}{\ln a}+2x\log_ax\)-\((x^3\ln a+3x^2.x\ln a)\)+\(6e^x(\ln x+x\ln x+1)\)
\(=x\frac{1}{\ln a}+2x\log_ax\)-\((x^3\ln a+3x^3\ln a)\)+\(6e^x(1+\ln x+x\ln x)\)
\(=x(\frac{1}{\ln a}+2\log_ax)-4x^3\ln a\)+\(6e^x(1+\ln x+x\ln x)\)

মনে করি,
\(y=\frac{x\ln x}{1+\sin x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x\ln x}{1+\sin x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{(1+\sin x)\frac{d}{dx}(x\ln x)-x\ln x\frac{d}{dx}(1+\sin x)}{(1+\sin x)^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(1+\sin x)\{x\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}(x)\}-x\ln x\{\frac{d}{dx}(1)+\frac{d}{dx}(\sin x)\}}{(1+\sin x)^2}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\frac{(1+\sin x)\{x.\frac{1}{x}+\ln x.1\}-x\ln x\{0+\cos x\}}{(1+\sin x)^2}\) ➜ Note \(\because \frac{d}{dx}(\ln x)=\frac{1}{x}, \frac{d}{dx}(c)=0, \frac{d}{dx}(\sin x)=\cos x\)
\(=\frac{(1+\sin x)(1+\ln x)-x\ln x\cos x}{(1+\sin x)^2}\)

মনে করি,
\(y=4e^t\sin t\)
\(\therefore \frac{d}{dt}(y)=\frac{d}{dt}(4e^t\sin t)\) ➜ Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=4e^t\frac{d}{dt}(\sin t)+\sin t\frac{d}{dt}(4e^t)\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=4e^t\frac{d}{dt}(\sin t)+\sin t.4\frac{d}{dt}(e^t)\)
\(=4e^t.\cos t+\sin t.4e^t\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(e^x)=e^x\)
\(=4e^t\cos t+4e^t\sin t\)
\(=4e^t(\sin t+\cos t)\)

মনে করি,
\(y=\frac{e^t+\ln t}{\sin t}\)
\(\therefore \frac{d}{dt}(y)=\frac{d}{dt}\left(\frac{e^t+\ln t}{\sin t}\right)\) ➜ Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{\sin t\frac{d}{dt}(e^t+\ln t)-(e^t+\ln t)\frac{d}{dt}(\sin t)}{\sin^2 t}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{\sin t\{\frac{d}{dt}(e^t)+\frac{d}{dt}(\ln t)\}-(e^t+\ln t)\frac{d}{dt}(\sin t)}{\sin^2 t}\)
\(=\frac{\sin t\{e^t+\frac{1}{t}\}-(e^t+\ln t).\cos t}{\sin^2 t}\) ➜ Note \(\because \frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(e^x)=e^x, \frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(=\frac{\sin t\left(\frac{te^t+1}{t}\right)-\cos t(e^t+\ln t)}{\sin^2 t}\)
\(=\frac{\sin t(te^t+1)-t\cos t(e^t+\ln t)}{t\sin^2 t}\) ➜ Note লব ও হরের সহিত \(t\) গুণ করে।

মনে করি,
\(y=\frac{x\cos x}{(x+1)\sin x}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{x\cos x}{(x+1)\sin x}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{(x+1)\sin x\frac{d}{dx}(x\cos x)-x\cos x\frac{d}{dx}\{(x+1)\sin x\}}{\{(x+1)\sin x\}^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(x+1)\sin x\{x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(x)\}-x\cos x\{(x+1)\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(x+1)\}}{(x+1)^2\sin^2 x}\) ➜ Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(=\frac{(x+1)\sin x\{x.(-\sin x)+\cos x.1\}-x\cos x\{(x+1).\cos x+\sin x.1\}}{(x+1)^2\sin^2 x}\) ➜ Note \(\because \frac{d}{dx}(\cos x)=-\sin x, \frac{d}{dx}(x)=1\),\(\frac{d}{dx}(\sin x)=\cos x, \frac{d}{dx}(c)=0\)
\(=\frac{(x+1)\sin x(\cos x-x\sin x)-x\cos x\{(x+1)\cos x+\sin x\}}{(x+1)^2\sin^2 x}\)
\(=\frac{(x+1)\sin x\cos x-x(x+1)\sin^2 x-x(x+1)\cos^2 x-x\sin x\cos x}{(x+1)^2\sin^2 x}\)
\(=\frac{x\sin x\cos x+\sin x\cos x-x(x+1)\sin^2 x-x(x+1)\cos^2 x-x\sin x\cos x}{(x+1)^2\sin^2 x}\)
\(=\frac{\sin x\cos x-x(x+1)(\sin^2 x+\cos^2 x)}{(x+1)^2\sin^2 x}\)
\(=\frac{\sin x\cos x-x(x+1).(1)}{(x+1)^2\sin^2 x}\) ➜ Note \(\because \sin^2 x+\cos^2 x=1\)
\(=\frac{\sin x\cos x-x(x+1)}{(x+1)^2\sin^2 x}\)

দেওয়া আছে,
\(y=\frac{1-2x}{x^3+5}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}\left(\frac{1-2x}{x^3+5}\right)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{(x^3+5)\frac{d}{dx}(1-2x)-(1-2x)\frac{d}{dx}(x^3+5)}{(x^3+5)^2}\) ➜ Note \(\because \frac{d}{dx}(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}\)
\(=\frac{(x^3+5)\{\frac{d}{dx}(1)-\frac{d}{dx}(2x)\}-(1-2x)\{\frac{d}{dx}(x^3)+\frac{d}{dx}(5)\}}{(x^3+5)^2}\)
\(=\frac{(x^3+5)\{0-2.1\}-(1-2x)\{3x^2+0\}}{(x^3+5)^2}\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1, \frac{d}{dx}(c)=0\)
\(=\frac{-2(x^3+5)-3x^2(1-2x)}{(x^3+5)^2}\)
\(=\frac{-2x^3-10-3x^2+6x^3}{(x^3+5)^2}\)
\(=\frac{4x^3-3x^2-10}{(x^3+5)^2}\)
\(x=1\)-এর জন্য,
\(\frac{d}{dx}(y)=\frac{4.1^3-3.1^2-10}{(1^3+5)^2} \)
\(=\frac{4.1-3.1-10}{(1+5)^2} \)
\(=\frac{4-3-10}{(6)^2} \)
\(=\frac{4-13}{36} \)
\(=\frac{-9}{36} \)
\(=-\frac{1}{4} \)

দেওয়া আছে,
\(s=ut+\frac{1}{2}ft^2\)
\(\therefore \frac{d}{dt}(s)=\frac{d}{dx}(ut+\frac{1}{2}ft^2)\) ➜ Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{d}{dt}(ut)+\frac{d}{dt}(\frac{1}{2}ft^2)\)
\(=u\frac{d}{dt}(t)+\frac{1}{2}f\frac{d}{dt}(t^2)\)
\(=u.1+\frac{1}{2}f.2t\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\)
\(=u+ft\)
\(\therefore \frac{d}{dt}(s)=u+ft\)
( proved )

দেওয়া আছে,
\(f(x)=80x-16x^2\)
\(\therefore \acute{f}(x)=\frac{d}{dx}\{f(x)\}=\frac{d}{dx}(80x-16x^2)\) ➜ Note \(\because \acute{f}(x)=\frac{d}{dx}\{f(x)\}\)
\(=\frac{d}{dx}(80x)-\frac{d}{dx}(16x^2)\)
\(=80\frac{d}{dx}(x)-16\frac{d}{dx}(x^2)\)
\(=80.1-16.2x\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\)
\(=80-32x\)
আবার,
দেওয়া আছে,
\(\acute{f}(x)=16\)
\(\Rightarrow 80-32x=16\)
\(\Rightarrow 16=80-32x\)
\(\Rightarrow 32x=80-16\)
\(\Rightarrow 32x=64\)
\(\Rightarrow x=\frac{64}{32}\)
\(\therefore x=2\)

দেওয়া আছে,
\(y=x(x^2-12)\)
\(=x^3-12x\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(x^3-12x)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{d}{dx}(x^3)-\frac{d}{dx}(12x)\)
\(=\frac{d}{dx}(x^3)-12\frac{d}{dx}(x)\)
\(=3x^2-12.1\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}, \frac{d}{dx}(x)=1\)
\(=3x^2-12\)
আবার,
দেওয়া আছে,
\(\frac{d}{dx}(y)=0\)
\(\Rightarrow 3x^2-12=0\)
\(\Rightarrow 3x^2=12\)
\(\Rightarrow x^2=\frac{12}{3}\)
\(\Rightarrow x^2=4\)
\(\Rightarrow x=\pm \sqrt{4}\)
\(\therefore x=\pm 2\)

দেওয়া আছে,
\(y=4x^2\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(4x^2)\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=4\frac{d}{dx}(x^2)\)
\(=4.2x\) ➜ Note \(\because \frac{d}{dx}(x^n)=nx^{n-1}\)
\(=8x\)
\(x=2\) বিন্দুতে,
\(\frac{d}{dx}(y)=8.2\)
\(=16\)

দেওয়া আছে,
\(s=\sqrt{t}+7\)
\(\therefore \frac{d}{dt}(s)=\frac{d}{dt}(\sqrt{t}+7)\) ➜ Note \(t\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(=\frac{d}{dt}(\sqrt{t})+\frac{d}{dt}(7)\)
\(=\frac{1}{2\sqrt{t}}+0\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(c)=0\)
\(=\frac{1}{2\sqrt{t}}\)
যখন, \(t=9\)
\(\frac{d}{dt}(s)=\frac{1}{2\sqrt{9}}\)
\(=\frac{1}{2.3}\)
\(=\frac{1}{6}\)

দেওয়া আছে,
\(y=\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(=\sqrt{x}+\frac{1}{x^{\frac{1}{2}}}\)
\(=\sqrt{x}+x^{-\frac{1}{2}}\)
\(\therefore \frac{d}{dx}(y)=\frac{d}{dx}(\sqrt{x}+x^{-\frac{1}{2}})\) ➜ Note \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}(y)=\frac{d}{dt}(\sqrt{x})+\frac{d}{dt}(x^{-\frac{1}{2}})\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{-\frac{1}{2}-1}\) ➜ Note \(\because \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}, \frac{d}{dx}(x^n)=nx^{n-1}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-1-2}{2}}\)
\(\Rightarrow \frac{d}{dx}(y)=\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-3}{2}}\)
\(\Rightarrow 2x\frac{d}{dx}(y)=\frac{2x}{2\sqrt{x}}-\frac{1}{2}x^{\frac{-3}{2}}.2x\) ➜ Note উভয় পার্শে \(2x\) গুণ করে।
\(\Rightarrow 2x\frac{d}{dx}(y)=\frac{x}{\sqrt{x}}-x^{\frac{-3}{2}}.x\)
\(\Rightarrow 2x\frac{d}{dx}(y)=\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}}-x^{\frac{-3}{2}+1}\)
\(\Rightarrow 2x\frac{d}{dx}(y)=\sqrt{x}-x^{\frac{-3+2}{2}}\)
\(\Rightarrow 2x\frac{d}{dx}(y)=\sqrt{x}-x^{\frac{-1}{2}}\)
\(\Rightarrow 2x\frac{d}{dx}(y)=\sqrt{x}-\frac{1}{x^{\frac{1}{2}}}\)
\(\Rightarrow 2x\frac{d}{dx}(y)=\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Rightarrow 2x\frac{d}{dx}(y)+y=\sqrt{x}-\frac{1}{\sqrt{x}}+y\) ➜ Note উভয় পার্শে \(y\) যোগ করে।
\(\Rightarrow 2x\frac{d}{dx}(y)+y=\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}\) ➜ Note \(\because y=\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(\Rightarrow 2x\frac{d}{dx}(y)+y=2\sqrt{x}\)
( proved )