\displaystyle \text{Differentiate the following functions with respect to } x \ (1 - 57):

\displaystyle \textbf{Question 1.}\ \sin(3x+5)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin(3x+5)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\{\sin(3x+5)\}

\displaystyle =\cos(3x+5)\cdot \frac{d}{dx}(3x+5)\quad[\text{using chain rule}]

\displaystyle =\cos(3x+5)\cdot 3

\displaystyle =3\cos(3x+5)

\displaystyle \text{So, } \frac{d}{dx}\{\sin(3x+5)\}=3\cos(3x+5)

\displaystyle \textbf{Question 2. }\ \tan^2 x

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\tan^{2}x

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\{\tan^{2}x\}

\displaystyle =2\tan x\cdot \frac{d}{dx}(\tan x)\quad[\text{using chain rule}]

\displaystyle =2\tan x\cdot \sec^{2}x

\displaystyle \text{So, } \frac{d}{dx}\{\tan^{2}x\}=2\tan x\sec^{2}x

\displaystyle \textbf{Question 3. }\ \tan(x^\circ+45^\circ)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\tan(x^\circ+45^\circ)

\displaystyle \Rightarrow\ y=\tan\!\left\{(x+45)\frac{\pi}{180}\right\}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\tan\!\left\{(x+45)\frac{\pi}{180}\right\}\right]

\displaystyle =\sec^{2}\!\left\{(x+45)\frac{\pi}{180}\right\}\cdot \frac{d}{dx}\left\{(x+45)\frac{\pi}{180}\right\}\quad[\text{using chain rule}]

\displaystyle =\frac{\pi}{180}\sec^{2}(x^\circ+45^\circ)

\displaystyle \text{So, } \frac{d}{dx}\{\tan(x^\circ+45^\circ)\}=\frac{\pi}{180}\sec^{2}(x^\circ+45^\circ)

\displaystyle \textbf{Question 4. }\ \sin(\log x)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin(\log x)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\{\sin(\log x)\}

\displaystyle =\cos(\log x)\cdot \frac{d}{dx}(\log x)\quad[\text{using chain rule}]

\displaystyle =\frac{1}{x}\cos(\log x)

\displaystyle \text{So, } \frac{d}{dx}\{\sin(\log x)\}=\frac{1}{x}\cos(\log x)

\displaystyle \textbf{Question 5. }\ e^{\sin\sqrt{x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{\sin\sqrt{x}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{\sin\sqrt{x}}\right)

\displaystyle =e^{\sin\sqrt{x}}\cdot \frac{d}{dx}(\sin\sqrt{x})\quad[\text{using chain rule}]

\displaystyle =e^{\sin\sqrt{x}}\cdot \cos\sqrt{x}\cdot \frac{d}{dx}(\sqrt{x})\quad[\text{using chain rule}]

\displaystyle =e^{\sin\sqrt{x}}\cdot \cos\sqrt{x}\cdot \frac{1}{2\sqrt{x}}

\displaystyle =\frac{\cos\sqrt{x}\,e^{\sin\sqrt{x}}}{2\sqrt{x}}

\displaystyle \text{So, } \frac{d}{dx}\left(e^{\sin\sqrt{x}}\right)=\frac{\cos\sqrt{x}\,e^{\sin\sqrt{x}}}{2\sqrt{x}}

\displaystyle \textbf{Question 6. }\ e^{\tan x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{\tan x}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{\tan x}\right)

\displaystyle =e^{\tan x}\cdot \frac{d}{dx}(\tan x)\quad[\text{using chain rule}]

\displaystyle =e^{\tan x}\cdot \sec^{2}x

\displaystyle \text{So, } \frac{d}{dx}\left(e^{\tan x}\right)=\sec^{2}x\,e^{\tan x}

\displaystyle \textbf{Question 7. }\ \sin^2(2x+1)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin^{2}(2x+1)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\{\sin^{2}(2x+1)\}

\displaystyle =2\sin(2x+1)\cdot \frac{d}{dx}\{\sin(2x+1)\}\quad[\text{using chain rule}]

\displaystyle =2\sin(2x+1)\cdot \cos(2x+1)\cdot \frac{d}{dx}(2x+1)\quad[\text{using chain rule}]

\displaystyle =4\sin(2x+1)\cos(2x+1)

\displaystyle =2\sin\{2(2x+1)\}\quad[\because\ \sin 2A=2\sin A\cos A]

\displaystyle =2\sin(4x+2)

\displaystyle \text{So, } \frac{d}{dx}\{\sin^{2}(2x+1)\}=2\sin(4x+2)

\displaystyle \textbf{Question 8. }\ \log_7(2x-3)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log_{7}(2x-3)

\displaystyle \Rightarrow\ y=\frac{\log(2x-3)}{\log 7}\quad\left[\because\ \log_{a}b=\frac{\log b}{\log a}\right]

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{1}{\log 7}\cdot \frac{d}{dx}\{\log(2x-3)\}

\displaystyle =\frac{1}{\log 7}\cdot \frac{1}{(2x-3)}\cdot \frac{d}{dx}(2x-3)\quad[\text{using chain rule}]

\displaystyle =\frac{2}{(2x-3)\log 7}

\displaystyle \text{Hence, } \frac{d}{dx}\{\log_{7}(2x-3)\}=\frac{2}{(2x-3)\log 7}

\displaystyle \textbf{Question 9. }\ \tan 5x^\circ

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\tan 5x^\circ

\displaystyle \Rightarrow\ y=\tan\!\left(5x\times\frac{\pi}{180}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\tan\!\left(5x\times\frac{\pi}{180}\right)\right]

\displaystyle =\sec^{2}\!\left(5x\times\frac{\pi}{180}\right)\cdot \frac{d}{dx}\left(5x\times\frac{\pi}{180}\right)\quad[\text{using chain rule}]

\displaystyle =\frac{5\pi}{180}\sec^{2}\!\left(5x\times\frac{\pi}{180}\right)

\displaystyle =\frac{5\pi}{180}\sec^{2}(5x^\circ)

\displaystyle \text{Hence, } \frac{d}{dx}\{\tan 5x^\circ\}=\frac{5\pi}{180}\sec^{2}(5x^\circ)

\displaystyle \textbf{Question 10. }\ 2^{x^3}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=2^{x^{3}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(2^{x^{3}}\right)

\displaystyle =2^{x^{3}}\cdot \log_{e}2\cdot \frac{d}{dx}(x^{3})\quad[\text{using chain rule}]

\displaystyle =2^{x^{3}}\cdot \log_{e}2\cdot 3x^{2}

\displaystyle =3x^{2}\cdot 2^{x^{3}}\log_{e}2

\displaystyle \text{Hence, } \frac{d}{dx}\left(2^{x^{3}}\right)=3x^{2}\cdot 2^{x^{3}}\log_{e}2

\displaystyle \textbf{Question 11. }\ 3^{e^x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=3^{e^{x}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(3^{e^{x}}\right)

\displaystyle =3^{e^{x}}\cdot \log_{e}3\cdot \frac{d}{dx}(e^{x})\quad[\text{using chain rule}]

\displaystyle =3^{e^{x}}\cdot \log_{e}3\cdot e^{x}

\displaystyle =e^{x}\cdot 3^{e^{x}}\log_{e}3

\displaystyle \text{So, } \frac{d}{dx}\left(3^{e^{x}}\right)=e^{x}\cdot 3^{e^{x}}\log_{e}3

\displaystyle \textbf{Question 12. }\ \log_x 3

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log_{x}3

\displaystyle \Rightarrow\ y=\frac{\log 3}{\log x}\quad\left[\because\ \log_{a}b=\frac{\log b}{\log a}\right]

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\log 3}{\log x}\right)

\displaystyle =\log 3\cdot \frac{d}{dx}\left((\log x)^{-1}\right)

\displaystyle =\log 3\cdot\left[-(\log x)^{-2}\right]\cdot \frac{d}{dx}(\log x)\quad[\text{using chain rule}]

\displaystyle =-\frac{\log 3}{(\log x)^{2}}\cdot \frac{1}{x}

\displaystyle =-\frac{1}{x\log 3\,(\log_{3}x)^{2}}\quad\left[\because\ \frac{\log b}{\log a}=\log_{a}b\right]

\displaystyle \text{So, } \frac{d}{dx}(\log_{x}3)=-\frac{1}{x\log 3\,(\log_{3}x)^{2}}

\displaystyle \textbf{Question 13. }\ 3^{x^2+2x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=3^{x^{2}+2x}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(3^{x^{2}+2x}\right)

\displaystyle =3^{x^{2}+2x}\cdot \log_{e}3\cdot \frac{d}{dx}(x^{2}+2x)\quad[\text{using chain rule}]

\displaystyle =3^{x^{2}+2x}\cdot \log_{e}3\cdot (2x+2)

\displaystyle =(2x+2)\,3^{x^{2}+2x}\log_{e}3

\displaystyle \text{So, } \frac{d}{dx}\left(3^{x^{2}+2x}\right)=(2x+2)\,3^{x^{2}+2x}\log_{e}3

\displaystyle \textbf{Question 14. }\ \sqrt{\frac{a^2-x^2}{a^2+x^2}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}

\displaystyle \Rightarrow\ y=\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{\tfrac12}

\displaystyle \text{Differentiate } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{\tfrac12}

\displaystyle =\frac12\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{\tfrac12-1}\cdot \frac{d}{dx}\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)\quad[\text{Using chain rule}]

\displaystyle =\frac12\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{-\tfrac12}\cdot \left\{\frac{(a^{2}+x^{2})\frac{d}{dx}(a^{2}-x^{2})-(a^{2}-x^{2})\frac{d}{dx}(a^{2}+x^{2})}{(a^{2}+x^{2})^{2}}\right\}

\displaystyle =\frac12\left(\frac{a^{2}+x^{2}}{a^{2}-x^{2}}\right)^{\tfrac12}\cdot \left\{\frac{(a^{2}+x^{2})(-2x)-(a^{2}-x^{2})(2x)}{(a^{2}+x^{2})^{2}}\right\}

\displaystyle =\frac12\left(\frac{a^{2}+x^{2}}{a^{2}-x^{2}}\right)^{\tfrac12}\cdot \left\{\frac{-2x(a^{2}+x^{2})-2x(a^{2}-x^{2})}{(a^{2}+x^{2})^{2}}\right\}

\displaystyle =\frac12\left(\frac{a^{2}+x^{2}}{a^{2}-x^{2}}\right)^{\tfrac12}\cdot \left\{\frac{-2xa^{2}-2x^{3}-2xa^{2}+2x^{3}}{(a^{2}+x^{2})^{2}}\right\}

\displaystyle =\frac12\left(\frac{a^{2}+x^{2}}{a^{2}-x^{2}}\right)^{\tfrac12}\cdot \left(\frac{-4xa^{2}}{(a^{2}+x^{2})^{2}}\right)

\displaystyle =\frac{-2xa^{2}}{\sqrt{a^{2}-x^{2}}\,(a^{2}+x^{2})^{\tfrac32}}

\displaystyle \text{So, }\frac{d}{dx}\left(\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}\right)=\frac{-2a^{2}x}{\sqrt{a^{2}-x^{2}}\,(a^{2}+x^{2})^{\tfrac32}}

\displaystyle \textbf{Question 15. }\ 3^{x\log x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=3^{x\log x}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(3^{x\log x}\right)

\displaystyle =3^{x\log x}\cdot \log_{e}3\cdot \frac{d}{dx}(x\log x)\quad[\text{Using chain rule}]

\displaystyle =3^{x\log x}\cdot \log_{e}3\left[x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x)\right]

\displaystyle =3^{x\log x}\cdot \log_{e}3\left(\frac{x}{x}+\log x\right)

\displaystyle =3^{x\log x}(1+\log x)\log_{e}3

\displaystyle \text{So, } \frac{d}{dx}\left(3^{x\log x}\right)=3^{x\log x}(1+\log x)\log_{e}3

\displaystyle \textbf{Question 16. }\ \sqrt{\frac{1+\sin x}{1-\sin x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sqrt{\frac{1+\sin x}{1-\sin x}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right)^{\tfrac12}

\displaystyle =\frac12\left(\frac{1+\sin x}{1-\sin x}\right)^{\tfrac12-1}\cdot \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right)

\displaystyle =\frac12\left(\frac{1-\sin x}{1+\sin x}\right)^{\tfrac12} \left[\frac{(1-\sin x)(\cos x)-(1+\sin x)(-\cos x)}{(1-\sin x)^2}\right]

\displaystyle =\frac12\left(\frac{1-\sin x}{1+\sin x}\right)^{\tfrac12} \left[\frac{\cos x-\cos x\sin x+\cos x+\sin x\cos x}{(1-\sin x)^2}\right]

\displaystyle =\frac12\cdot\frac{2\cos x}{\sqrt{1+\sin x}(1-\sin x)^{\tfrac32}}

\displaystyle =\frac{\cos x}{\sqrt{1+\sin x}(1-\sin x)^{\tfrac32}}

\displaystyle =\frac{\cos x}{\sqrt{1-\sin^2 x}(1-\sin x)} \quad[\because\ 1-\sin^2 x=\cos^2 x]

\displaystyle =\frac{\cos x}{\cos x(1-\sin x)}

\displaystyle =\frac{1}{1-\sin x}

\displaystyle =\frac{1+\sin x}{1-\sin^2 x}

\displaystyle =\frac{1+\sin x}{\cos^2 x}

\displaystyle =\sec x(\sec x+\tan x)

\displaystyle \text{Hence, } \frac{dy}{dx}=\sec x(\sec x+\tan x)

\displaystyle \textbf{Question 17. }\ \sqrt{\frac{1-x^2}{1+x^2}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sqrt{\frac{1-x^{2}}{1+x^{2}}}

\displaystyle \Rightarrow\ y=\left(\frac{1-x^{2}}{1+x^{2}}\right)^{\tfrac12}

\displaystyle \text{Differentiate } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1-x^{2}}{1+x^{2}}\right)^{\tfrac12}

\displaystyle =\frac12\left(\frac{1-x^{2}}{1+x^{2}}\right)^{\tfrac12-1} \cdot \frac{d}{dx}\left(\frac{1-x^{2}}{1+x^{2}}\right)\quad[\text{Using chain rule}]

\displaystyle =\frac12\left(\frac{1-x^{2}}{1+x^{2}}\right)^{-\tfrac12} \left\{\frac{(1+x^{2})\frac{d}{dx}(1-x^{2})-(1-x^{2})\frac{d}{dx}(1+x^{2})}{(1+x^{2})^{2}}\right\}

\displaystyle =\frac12\left(\frac{1+x^{2}}{1-x^{2}}\right)^{\tfrac12} \left\{\frac{(1+x^{2})(-2x)-(1-x^{2})(2x)}{(1+x^{2})^{2}}\right\}

\displaystyle =\frac12\left(\frac{1+x^{2}}{1-x^{2}}\right)^{\tfrac12} \left\{\frac{-2x-2x^{3}-2x+2x^{3}}{(1+x^{2})^{2}}\right\}

\displaystyle =\frac12\left(\frac{1+x^{2}}{1-x^{2}}\right)^{\tfrac12} \left(\frac{-4x}{(1+x^{2})^{2}}\right)

\displaystyle =\frac{-2x}{\sqrt{1-x^{2}}\,(1+x^{2})^{\tfrac32}}

\displaystyle \text{So, }\frac{d}{dx}\left(\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right) =\frac{-2x}{\sqrt{1-x^{2}}\,(1+x^{2})^{\tfrac32}}

\displaystyle \textbf{Question 18. }\ (\log \sin x)^2

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=(\log\sin x)^{2}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\{(\log\sin x)^{2}\}

\displaystyle =2(\log\sin x)\cdot \frac{d}{dx}(\log\sin x)

\displaystyle =2(\log\sin x)\cdot \frac{1}{\sin x}\cdot \frac{d}{dx}(\sin x)

\displaystyle =2(\log\sin x)\cdot \frac{1}{\sin x}\cdot \cos x

\displaystyle =2(\log\sin x)\cot x

\displaystyle \text{So, } \frac{d}{dx}\{(\log\sin x)^{2}\}=2(\log\sin x)\cot x

\displaystyle \textbf{Question 19. }\ \sqrt{\frac{1+x}{1-x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sqrt{\frac{1+x}{1-x}}

\displaystyle \Rightarrow\ y=\left(\frac{1+x}{1-x}\right)^{\tfrac12}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{1+x}{1-x}\right)^{\tfrac12}

\displaystyle =\frac12\left(\frac{1+x}{1-x}\right)^{\tfrac12-1} \cdot \frac{d}{dx}\left(\frac{1+x}{1-x}\right)\quad[\text{Using chain rule}]

\displaystyle =\frac12\left(\frac{1+x}{1-x}\right)^{-\tfrac12} \left\{\frac{(1-x)\frac{d}{dx}(1+x)-(1+x)\frac{d}{dx}(1-x)}{(1-x)^2}\right\} \quad[\text{Using quotient rule}]

\displaystyle =\frac12\left(\frac{1-x}{1+x}\right)^{\tfrac12} \left\{\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2}\right\}

\displaystyle =\frac12\left(\frac{1-x}{1+x}\right)^{\tfrac12} \left(\frac{1-x+1+x}{(1-x)^2}\right)

\displaystyle =\frac12\left(\frac{1-x}{1+x}\right)^{\tfrac12} \left(\frac{2}{(1-x)^2}\right)

\displaystyle =\frac{1}{\sqrt{1+x}\,(1-x)^{\tfrac32}}

\displaystyle \text{So, }\frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right) =\frac{1}{\sqrt{1+x}\,(1-x)^{\tfrac32}}

\displaystyle \textbf{Question 20. }\ \sin\!\left(\frac{1+x^2}{1-x^2}\right)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin\!\left(\frac{1+x^{2}}{1-x^{2}}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\sin\!\left(\frac{1+x^{2}}{1-x^{2}}\right)\right]

\displaystyle =\cos\!\left(\frac{1+x^{2}}{1-x^{2}}\right)\cdot \frac{d}{dx}\left(\frac{1+x^{2}}{1-x^{2}}\right)\quad[\text{Using chain rule}]

\displaystyle =\cos\!\left(\frac{1+x^{2}}{1-x^{2}}\right) \left[\frac{(1-x^{2})\frac{d}{dx}(1+x^{2})-(1+x^{2})\frac{d}{dx}(1-x^{2})}{(1-x^{2})^{2}}\right] \quad[\text{Using quotient rule}]

\displaystyle =\cos\!\left(\frac{1+x^{2}}{1-x^{2}}\right) \left[\frac{(1-x^{2})(2x)-(1+x^{2})(-2x)}{(1-x^{2})^{2}}\right]

\displaystyle =\cos\!\left(\frac{1+x^{2}}{1-x^{2}}\right) \left[\frac{2x-2x^{3}+2x+2x^{3}}{(1-x^{2})^{2}}\right]

\displaystyle =\frac{4x}{(1-x^{2})^{2}}\cos\!\left(\frac{1+x^{2}}{1-x^{2}}\right)

\displaystyle \text{So, }\frac{d}{dx}\left\{\sin\!\left(\frac{1+x^{2}}{1-x^{2}}\right)\right\} =\frac{4x}{(1-x^{2})^{2}}\cos\!\left(\frac{1+x^{2}}{1-x^{2}}\right)

\displaystyle \textbf{Question 21. }\ e^{3x}\cos 2x

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{3x}\cos 2x

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{3x}\cos 2x\right)

\displaystyle =e^{3x}\cdot \frac{d}{dx}(\cos 2x)+\cos 2x\cdot \frac{d}{dx}(e^{3x}) \quad[\text{Using product rule}]

\displaystyle =e^{3x}\cdot(-\sin 2x)\cdot \frac{d}{dx}(2x) +\cos 2x\cdot e^{3x}\cdot \frac{d}{dx}(3x) \quad[\text{Using chain rule}]

\displaystyle =-2e^{3x}\sin 2x+3e^{3x}\cos 2x

\displaystyle =e^{3x}(3\cos 2x-2\sin 2x)

\displaystyle \text{So, } \frac{d}{dx}\left(e^{3x}\cos 2x\right) =e^{3x}(3\cos 2x-2\sin 2x)

\displaystyle \textbf{Question 22. }\ \sin(\log \sin x)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin(\log\sin x)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\{\sin(\log\sin x)\}

\displaystyle =\cos(\log\sin x)\cdot \frac{d}{dx}(\log\sin x)\quad[\text{Using chain rule}]

\displaystyle =\cos(\log\sin x)\cdot \frac{1}{\sin x}\cdot \frac{d}{dx}(\sin x) \quad[\text{Using chain rule}]

\displaystyle =\cos(\log\sin x)\cdot \frac{\cos x}{\sin x}

\displaystyle =\cos(\log\sin x)\cot x

\displaystyle \text{Hence, } \frac{d}{dx}\{\sin(\log\sin x)\} =\cos(\log\sin x)\cot x

\displaystyle \textbf{Question 23. }\ e^{\tan 3x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{\tan 3x}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{\tan 3x}\right)

\displaystyle =e^{\tan 3x}\cdot \frac{d}{dx}(\tan 3x)\quad[\text{using chain rule}]

\displaystyle =e^{\tan 3x}\cdot \sec^{2}3x\cdot \frac{d}{dx}(3x)\quad[\text{using chain rule}]

\displaystyle =e^{\tan 3x}\cdot \sec^{2}3x\cdot 3

\displaystyle =3e^{\tan 3x}\sec^{2}3x

\displaystyle \text{So, } \frac{d}{dx}\left(e^{\tan 3x}\right)=3e^{\tan 3x}\sec^{2}3x

\displaystyle \textbf{Question 24. }\ e^{\sqrt{\cot x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{\sqrt{\cot x}}

\displaystyle \Rightarrow\ y=e^{(\cot x)^{\tfrac12}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{(\cot x)^{\tfrac12}}\right)

\displaystyle =e^{(\cot x)^{\tfrac12}}\cdot \frac{d}{dx}\left((\cot x)^{\tfrac12}\right) \quad[\text{using chain rule}]

\displaystyle =e^{\sqrt{\cot x}}\cdot \frac12(\cot x)^{\tfrac12-1}\cdot \frac{d}{dx}(\cot x) \quad[\text{using chain rule}]

\displaystyle =e^{\sqrt{\cot x}}\cdot \frac12(\cot x)^{-\tfrac12}\cdot (-\mathrm{cosec}^{2}x)

\displaystyle =-\frac{e^{\sqrt{\cot x}}\mathrm{cosec}^{2}x}{2\sqrt{\cot x}}

\displaystyle \text{So, } \frac{d}{dx}\left(e^{\sqrt{\cot x}}\right) =-\frac{e^{\sqrt{\cot x}}\mathrm{cosec}^{2}x}{2\sqrt{\cot x}}

\displaystyle \textbf{Question 25. }\ \log\!\left(\frac{\sin x}{1+\cos x}\right)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log\!\left(\frac{\sin x}{1+\cos x}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[\log\!\left(\frac{\sin x}{1+\cos x}\right)\right]

\displaystyle =\frac{1}{\left(\dfrac{\sin x}{1+\cos x}\right)} \cdot \frac{d}{dx}\left(\frac{\sin x}{1+\cos x}\right) \quad[\text{Using chain rule}]

\displaystyle =\left(\frac{1+\cos x}{\sin x}\right) \left[ \frac{(1+\cos x)\dfrac{d}{dx}(\sin x)-\sin x\dfrac{d}{dx}(1+\cos x)} {(1+\cos x)^{2}} \right] \quad[\text{Using quotient rule}]

\displaystyle =\left(\frac{1+\cos x}{\sin x}\right) \left[ \frac{(1+\cos x)\cos x-\sin x(-\sin x)} {(1+\cos x)^{2}} \right]

\displaystyle =\left(\frac{1+\cos x}{\sin x}\right) \left[ \frac{\cos x+\cos^{2}x+\sin^{2}x} {(1+\cos x)^{2}} \right]

\displaystyle =\left(\frac{1+\cos x}{\sin x}\right) \left[ \frac{1+\cos x}{(1+\cos x)^{2}} \right]

\displaystyle =\frac{1}{\sin x}

\displaystyle =\mathrm{cosec} x

\displaystyle \text{So, } \frac{d}{dx}\left\{\log\!\left(\frac{\sin x}{1+\cos x}\right)\right\} =\mathrm{cosec} x

\displaystyle \textbf{Question 26. }\ \log\sqrt{\frac{1-\cos x}{1+\cos x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log\sqrt{\frac{1-\cos x}{1+\cos x}}

\displaystyle \Rightarrow\ y=\log\left(\frac{1-\cos x}{1+\cos x}\right)^{\tfrac12}

\displaystyle \Rightarrow\ y=\frac12\log\left(\frac{1-\cos x}{1+\cos x}\right) \quad[\text{using }\log a^{b}=b\log a]

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left\{\frac12\log\left(\frac{1-\cos x}{1+\cos x}\right)\right\}

\displaystyle =\frac12\cdot \frac{1}{\left(\dfrac{1-\cos x}{1+\cos x}\right)} \cdot \frac{d}{dx}\left(\frac{1-\cos x}{1+\cos x}\right) \quad[\text{Using chain rule}]

\displaystyle =\frac12\left(\frac{1+\cos x}{1-\cos x}\right) \left[ \frac{(1+\cos x)\dfrac{d}{dx}(1-\cos x)-(1-\cos x)\dfrac{d}{dx}(1+\cos x)} {(1+\cos x)^{2}} \right] \quad[\text{Using quotient rule}]

\displaystyle =\frac12\left(\frac{1+\cos x}{1-\cos x}\right) \left[ \frac{(1+\cos x)\sin x-(1-\cos x)(-\sin x)} {(1+\cos x)^{2}} \right]

\displaystyle =\frac12\left(\frac{1+\cos x}{1-\cos x}\right) \left[ \frac{\sin x+\sin x\cos x+\sin x-\sin x\cos x} {(1+\cos x)^{2}} \right]

\displaystyle =\frac12\left(\frac{1+\cos x}{1-\cos x}\right) \left[ \frac{2\sin x}{(1+\cos x)^{2}} \right]

\displaystyle =\frac{\sin x}{(1-\cos x)(1+\cos x)}

\displaystyle =\frac{\sin x}{1-\cos^{2}x}

\displaystyle =\frac{\sin x}{\sin^{2}x}

\displaystyle =\frac{1}{\sin x}

\displaystyle =\mathrm{cosec} x

\displaystyle \text{So, }\frac{d}{dx}\left(\log\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\mathrm{cosec} x

\displaystyle \textbf{Question 27. }\ \tan(e^{\sin x})

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\tan\!\left(e^{\sin x}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[\tan\!\left(e^{\sin x}\right)\right]

\displaystyle =\sec^{2}\!\left(e^{\sin x}\right)\cdot \frac{d}{dx}\left(e^{\sin x}\right) \quad[\text{using chain rule}]

\displaystyle =\sec^{2}\!\left(e^{\sin x}\right)\cdot e^{\sin x}\cdot \frac{d}{dx}(\sin x) \quad[\text{using chain rule}]

\displaystyle =\sec^{2}\!\left(e^{\sin x}\right)\cdot e^{\sin x}\cdot \cos x

\displaystyle =\cos x\,\sec^{2}\!\left(e^{\sin x}\right)e^{\sin x}

\displaystyle \text{So, } \frac{d}{dx}\left\{\tan\!\left(e^{\sin x}\right)\right\} =\cos x\,\sec^{2}\!\left(e^{\sin x}\right)e^{\sin x}

\displaystyle \textbf{Question 28. }\ \log\!\left(x+\sqrt{x^{2}+1}\right)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log\!\left(x+\sqrt{x^{2}+1}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[\log\!\left(x+\sqrt{x^{2}+1}\right)\right]

\displaystyle =\frac{1}{x+\sqrt{x^{2}+1}}\cdot \frac{d}{dx}\left(x+(x^{2}+1)^{\tfrac12}\right) \quad[\text{Using chain rule}]

\displaystyle =\frac{1}{x+\sqrt{x^{2}+1}} \left[1+\frac12(x^{2}+1)^{\tfrac12-1}\cdot \frac{d}{dx}(x^{2}+1)\right]

\displaystyle =\frac{1}{x+\sqrt{x^{2}+1}} \left[1+\frac{1}{2\sqrt{x^{2}+1}}\cdot 2x\right]

\displaystyle =\frac{1}{x+\sqrt{x^{2}+1}} \left[\frac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}}\right]

\displaystyle =\frac{1}{\sqrt{x^{2}+1}}

\displaystyle \text{So, } \frac{d}{dx}\left\{\log\!\left(x+\sqrt{x^{2}+1}\right)\right\} =\frac{1}{\sqrt{x^{2}+1}}

\displaystyle \textbf{Question 29. }\ \frac{e^{x}\log x}{x^{2}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\frac{e^{x}\log x}{x^{2}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{x^{2}\dfrac{d}{dx}(e^{x}\log x)-(e^{x}\log x)\dfrac{d}{dx}(x^{2})}{(x^{2})^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{x^{2}\left\{e^{x}\dfrac{d}{dx}(\log x)+\log x\dfrac{d}{dx}(e^{x})\right\} -e^{x}\log x\cdot 2x}{x^{4}} \quad[\text{Using product rule}]

\displaystyle =\frac{x^{2}\left(\frac{e^{x}}{x}+e^{x}\log x\right)-2xe^{x}\log x}{x^{4}}

\displaystyle =\frac{\dfrac{x^{2}e^{x}(1+x\log x)}{x}-2xe^{x}\log x}{x^{4}}

\displaystyle =\frac{xe^{x}\left(1+x\log x-2\log x\right)}{x^{4}}

\displaystyle =\frac{e^{x}}{x^{3}}\left(1+x\log x-2\log x\right)

\displaystyle =e^{x}x^{-2}\left(\frac{1}{x}+\log x-\frac{2\log x}{x}\right)

\displaystyle \text{So, } \frac{d}{dx}\left(\frac{e^{x}\log x}{x^{2}}\right) =e^{x}x^{-2}\left(\frac{1}{x}+\log x-\frac{2\log x}{x}\right)

\displaystyle \textbf{Question 30. }\ \log(\mathrm{cosec} x-\cot x)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log(\mathrm{cosec}\,x-\cot x)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\{\log(\mathrm{cosec}\,x-\cot x)\}

\displaystyle =\frac{1}{(\mathrm{cosec}\,x-\cot x)}\cdot \frac{d}{dx}(\mathrm{cosec}\,x-\cot x)

\displaystyle =\frac{1}{(\mathrm{cosec}\,x-\cot x)} \left(-\mathrm{cosec}\,x\cot x+\mathrm{cosec}^{2}x\right)

\displaystyle =\frac{\mathrm{cosec}\,x(\mathrm{cosec}\,x-\cot x)}{(\mathrm{cosec}\,x-\cot x)}

\displaystyle =\mathrm{cosec}\,x

\displaystyle \text{So, } \frac{d}{dx}\{\log(\mathrm{cosec}\,x-\cot x)\} =\mathrm{cosec}\,x

\displaystyle \textbf{Question 31. }\ \frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right]

\displaystyle =\frac{(e^{2x}-e^{-2x})\dfrac{d}{dx}(e^{2x}+e^{-2x}) -(e^{2x}+e^{-2x})\dfrac{d}{dx}(e^{2x}-e^{-2x})} {(e^{2x}-e^{-2x})^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{(e^{2x}-e^{-2x})\left(2e^{2x}-2e^{-2x}\right) -(e^{2x}+e^{-2x})\left(2e^{2x}+2e^{-2x}\right)} {(e^{2x}-e^{-2x})^{2}} \quad[\text{Using chain rule}]

\displaystyle =\frac{2(e^{2x}-e^{-2x})^{2}-2(e^{2x}+e^{-2x})^{2}} {(e^{2x}-e^{-2x})^{2}}

\displaystyle =\frac{2\left(e^{4x}-2+e^{-4x}\right) -2\left(e^{4x}+2+e^{-4x}\right)} {(e^{2x}-e^{-2x})^{2}}

\displaystyle =\frac{-8}{(e^{2x}-e^{-2x})^{2}}

\displaystyle \text{So, } \frac{d}{dx}\left(\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right) =\frac{-8}{(e^{2x}-e^{-2x})^{2}}

\displaystyle \textbf{Question 32. }\ \log\!\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log\!\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[\log\!\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)\right]

\displaystyle =\frac{1}{\left(\dfrac{x^{2}+x+1}{x^{2}-x+1}\right)}\cdot \frac{d}{dx}\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right) \quad[\text{Using chain rule}]

\displaystyle =\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right) \left[ \frac{(x^{2}-x+1)\dfrac{d}{dx}(x^{2}+x+1)-(x^{2}+x+1)\dfrac{d}{dx}(x^{2}-x+1)} {(x^{2}-x+1)^{2}} \right] \quad[\text{Using quotient rule}]

\displaystyle =\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right) \left[ \frac{(x^{2}-x+1)(2x+1)-(x^{2}+x+1)(2x-1)} {(x^{2}-x+1)^{2}} \right]

\displaystyle =\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right) \left[ \frac{(2x^{3}-x^{2}+x+1)-(2x^{3}+x^{2}+x-1)} {(x^{2}-x+1)^{2}} \right]

\displaystyle =\left(\frac{x^{2}-x+1}{x^{2}+x+1}\right) \left[\frac{-2x^{2}+2}{(x^{2}-x+1)^{2}}\right]

\displaystyle =\frac{-2(x^{2}-1)}{(x^{2}+x+1)(x^{2}-x+1)}

\displaystyle =\frac{-2(x^{2}-1)}{x^{4}+x^{2}+1}

\displaystyle \text{So, }\frac{d}{dx}\left\{\log\!\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)\right\} =\frac{-2(x^{2}-1)}{x^{4}+x^{2}+1}

\displaystyle \textbf{Question 33. }\ \tan^{-1}(e^{x})

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\tan^{-1}(e^{x})

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(\tan^{-1} e^{x}\right)

\displaystyle =\frac{1}{1+(e^{x})^{2}}\cdot \frac{d}{dx}(e^{x}) \quad[\text{Using chain rule}]

\displaystyle =\frac{1}{1+e^{2x}}\cdot e^{x}

\displaystyle =\frac{e^{x}}{1+e^{2x}}

\displaystyle \text{So, } \frac{d}{dx}\left(\tan^{-1} e^{x}\right) =\frac{e^{x}}{1+e^{2x}}

\displaystyle \textbf{Question 34. }\ e^{\sin^{-1}(2x)}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{\sin^{-1}2x}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(e^{\sin^{-1}2x}\right)

\displaystyle =e^{\sin^{-1}2x}\cdot \frac{d}{dx}(\sin^{-1}2x) \quad[\text{using chain rule}]

\displaystyle =e^{\sin^{-1}2x}\cdot \frac{1}{\sqrt{1-(2x)^{2}}}\cdot \frac{d}{dx}(2x)

\displaystyle =\frac{2e^{\sin^{-1}2x}}{\sqrt{1-4x^{2}}}

\displaystyle \text{So, } \frac{d}{dx}\left(e^{\sin^{-1}2x}\right) =\frac{2e^{\sin^{-1}2x}}{\sqrt{1-4x^{2}}}

\displaystyle \textbf{Question 35. }\ \sin(2\sin^{-1}x)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin\!\left(2\sin^{-1}x\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[\sin\!\left(2\sin^{-1}x\right)\right]

\displaystyle =\cos\!\left(2\sin^{-1}x\right)\cdot \frac{d}{dx}\left(2\sin^{-1}x\right) \quad[\text{using chain rule}]

\displaystyle =\cos\!\left(2\sin^{-1}x\right)\cdot 2\cdot \frac{1}{\sqrt{1-x^{2}}}

\displaystyle =\frac{2\cos\!\left(2\sin^{-1}x\right)}{\sqrt{1-x^{2}}}

\displaystyle \text{So, } \frac{d}{dx}\left\{\sin\!\left(2\sin^{-1}x\right)\right\} =\frac{2\cos\!\left(2\sin^{-1}x\right)}{\sqrt{1-x^{2}}}

\displaystyle \textbf{Question 36. }\ e^{\tan^{-1}\sqrt{x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{\tan^{-1}\sqrt{x}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(e^{\tan^{-1}\sqrt{x}}\right)

\displaystyle =e^{\tan^{-1}\sqrt{x}}\cdot \frac{d}{dx}\left(\tan^{-1}\sqrt{x}\right) \quad[\text{Using chain rule}]

\displaystyle =e^{\tan^{-1}\sqrt{x}}\cdot \frac{1}{1+(\sqrt{x})^{2}}\cdot \frac{d}{dx}(\sqrt{x}) \quad[\text{Using chain rule}]

\displaystyle =e^{\tan^{-1}\sqrt{x}}\cdot \frac{1}{1+x}\cdot \frac{1}{2\sqrt{x}}

\displaystyle =\frac{e^{\tan^{-1}\sqrt{x}}}{2\sqrt{x}(1+x)}

\displaystyle \text{So, } \frac{d}{dx}\left(e^{\tan^{-1}\sqrt{x}}\right) =\frac{e^{\tan^{-1}\sqrt{x}}}{2\sqrt{x}(1+x)}

\displaystyle \textbf{Question 37. }\ \sqrt{\tan^{-1}\!\left(\frac{x}{2}\right)}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sqrt{\tan^{-1}\!\left(\frac{x}{2}\right)}

\displaystyle \Rightarrow\ y=\left\{\tan^{-1}\!\left(\frac{x}{2}\right)\right\}^{\tfrac12}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left\{\tan^{-1}\!\left(\frac{x}{2}\right)\right\}^{\tfrac12}

\displaystyle =\frac12\left\{\tan^{-1}\!\left(\frac{x}{2}\right)\right\}^{\tfrac12-1} \cdot \frac{d}{dx}\left[\tan^{-1}\!\left(\frac{x}{2}\right)\right] \quad[\text{Using chain rule}]

\displaystyle =\frac12\left\{\tan^{-1}\!\left(\frac{x}{2}\right)\right\}^{-\tfrac12} \cdot \frac{1}{1+\left(\tfrac{x}{2}\right)^{2}} \cdot \frac{d}{dx}\left(\frac{x}{2}\right)

\displaystyle =\frac12\left\{\tan^{-1}\!\left(\frac{x}{2}\right)\right\}^{-\tfrac12} \cdot \frac{1}{1+\tfrac{x^{2}}{4}} \cdot \frac12

\displaystyle =\frac{1}{(4+x^{2})\sqrt{\tan^{-1}\!\left(\frac{x}{2}\right)}}

\displaystyle \text{So, } \frac{d}{dx}\left\{\sqrt{\tan^{-1}\!\left(\frac{x}{2}\right)}\right\} =\frac{1}{(4+x^{2})\sqrt{\tan^{-1}\!\left(\frac{x}{2}\right)}}

\displaystyle \textbf{Question 38. }\ \log(\tan^{-1}x)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log(\tan^{-1}x)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\{\log(\tan^{-1}x)\}

\displaystyle =\frac{1}{\tan^{-1}x}\cdot \frac{d}{dx}(\tan^{-1}x) \quad[\text{Using chain rule}]

\displaystyle =\frac{1}{\tan^{-1}x}\cdot \frac{1}{1+x^{2}}

\displaystyle =\frac{1}{(1+x^{2})\tan^{-1}x}

\displaystyle \text{So, } \frac{d}{dx}\{\log(\tan^{-1}x)\} =\frac{1}{(1+x^{2})\tan^{-1}x}

\displaystyle \textbf{Question 39. }\ \frac{2^{x}\cos x}{(x^{2}+3)^{2}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\frac{2^{x}\cos x}{(x^{2}+3)^{2}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\frac{2^{x}\cos x}{(x^{2}+3)^{2}}\right]

\displaystyle =\frac{(x^{2}+3)^{2}\dfrac{d}{dx}(2^{x}\cos x)-(2^{x}\cos x)\dfrac{d}{dx}(x^{2}+3)^{2}} {\left[(x^{2}+3)^{2}\right]^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{(x^{2}+3)^{2}\left\{2^{x}\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(2^{x})\right\} -(2^{x}\cos x)\cdot 2(x^{2}+3)\frac{d}{dx}(x^{2}+3)} {(x^{2}+3)^{4}} \quad[\text{Using product rule and chain rule}]

\displaystyle =\frac{(x^{2}+3)^{2}\left\{2^{x}(-\sin x)+\cos x\cdot 2^{x}\log_{e}2\right\} -(2^{x}\cos x)\cdot 2(x^{2}+3)\cdot (2x)} {(x^{2}+3)^{4}}

\displaystyle =\frac{2^{x}(x^{2}+3)\left\{(x^{2}+3)(\cos x\log_{e}2-\sin x)-4x\cos x\right\}} {(x^{2}+3)^{4}}

\displaystyle =\frac{2^{x}}{(x^{2}+3)^{2}}\left[\cos x\log_{e}2-\sin x-\frac{4x\cos x}{(x^{2}+3)}\right]

\displaystyle \text{So, } \frac{d}{dx}\left[\frac{2^{x}\cos x}{(x^{2}+3)^{2}}\right] =\frac{2^{x}}{(x^{2}+3)^{2}}\left[\cos x\log_{e}2-\sin x-\frac{4x\cos x}{(x^{2}+3)}\right]

\displaystyle \textbf{Question 40. }\ x\sin 2x+5^{x}+k^{k}+(\tan^{2}x)^{3}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=x\sin 2x+5^{x}+k^{k}+(\tan^{2}x)^{3}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[x\sin 2x+5^{x}+k^{k}+\tan^{6}x\right]

\displaystyle =\frac{d}{dx}(x\sin 2x)+\frac{d}{dx}(5^{x})+\frac{d}{dx}(k^{k})+\frac{d}{dx}(\tan^{6}x)

\displaystyle =\left[x\frac{d}{dx}(\sin 2x)+\sin 2x\frac{d}{dx}(x)\right] +5^{x}\log_{e}5+0+6\tan^{5}x\frac{d}{dx}(\tan x) \quad[\text{Using product rule and chain rule}]

\displaystyle =\left[x\cos 2x\cdot \frac{d}{dx}(2x)+\sin 2x\right] +5^{x}\log_{e}5+6\tan^{5}x\sec^{2}x

\displaystyle =2x\cos 2x+\sin 2x+5^{x}\log_{e}5+6\tan^{5}x\sec^{2}x

\displaystyle \text{So, } \frac{d}{dx}\left\{x\sin 2x+5^{x}+k^{k}+(\tan^{2}x)^{3}\right\} =2x\cos 2x+\sin 2x+5^{x}\log_{e}5+6\tan^{5}x\sec^{2}x

\displaystyle \textbf{Question 41. }\ \log(3x+2)-x^{2}\log(2x-1)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log(3x+2)-x^{2}\log(2x-1)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[\log(3x+2)-x^{2}\log(2x-1)\right]

\displaystyle =\frac{d}{dx}\{\log(3x+2)\}-\frac{d}{dx}\{x^{2}\log(2x-1)\}

\displaystyle =\frac{1}{(3x+2)}\cdot \frac{d}{dx}(3x+2) -\left[x^{2}\frac{d}{dx}\{\log(2x-1)\}+\log(2x-1)\frac{d}{dx}(x^{2})\right] \quad[\text{Using product rule and chain rule}]

\displaystyle =\frac{3}{3x+2} -\left[x^{2}\cdot \frac{1}{2x-1}\cdot \frac{d}{dx}(2x-1)+\log(2x-1)\cdot 2x\right]

\displaystyle =\frac{3}{3x+2}-\frac{2x^{2}}{2x-1}-2x\log(2x-1)

\displaystyle \text{So, } \frac{d}{dx}\left[\log(3x+2)-x^{2}\log(2x-1)\right] =\frac{3}{3x+2}-\frac{2x^{2}}{2x-1}-2x\log(2x-1)

\displaystyle \textbf{Question 42. }\ \frac{3x^{2}\sin x}{\sqrt{7-x^{2}}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\frac{3x^{2}\sin x}{\sqrt{7-x^{2}}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left\{\frac{3x^{2}\sin x}{(7-x^{2})^{\tfrac12}}\right\}

\displaystyle =\frac{(7-x^{2})^{\tfrac12}\dfrac{d}{dx}(3x^{2}\sin x) -(3x^{2}\sin x)\dfrac{d}{dx}(7-x^{2})^{\tfrac12}} {\left[(7-x^{2})^{\tfrac12}\right]^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{(7-x^{2})^{\tfrac12}\cdot 3\left[x^{2}\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}(x^{2})\right] -(3x^{2}\sin x)\cdot \frac12(7-x^{2})^{-\tfrac12}\frac{d}{dx}(7-x^{2})} {7-x^{2}} \quad[\text{Using product rule and chain rule}]

\displaystyle =\frac{(7-x^{2})^{\tfrac12}\cdot 3(x^{2}\cos x+2x\sin x) -(3x^{2}\sin x)\cdot \frac12(7-x^{2})^{-\tfrac12}(-2x)} {7-x^{2}}

\displaystyle =\frac{(7-x^{2})^{\tfrac12}\cdot 3(x^{2}\cos x+2x\sin x) +3x^{3}\sin x(7-x^{2})^{-\tfrac12}} {7-x^{2}}

\displaystyle =\frac{3(x^{2}\cos x+2x\sin x)}{\sqrt{7-x^{2}}} +\frac{3x^{3}\sin x}{(7-x^{2})^{\tfrac32}}

\displaystyle =\frac{6x\sin x+3x^{2}\cos x}{\sqrt{7-x^{2}}} +\frac{3x^{3}\sin x}{(7-x^{2})^{\tfrac32}}

\displaystyle \text{So, } \frac{d}{dx}\left(\frac{3x^{2}\sin x}{\sqrt{7-x^{2}}}\right) =\frac{6x\sin x+3x^{2}\cos x}{\sqrt{7-x^{2}}} +\frac{3x^{3}\sin x}{(7-x^{2})^{\tfrac32}}

\displaystyle \textbf{Question 43. }\ \sin^{2}\!\big(\log(2x+3)\big)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin^{2}\{\log(2x+3)\}

\displaystyle \Rightarrow\ \frac{dy}{dx} =\frac{d}{dx}\left[\sin^{2}\{\log(2x+3)\}\right]

\displaystyle =2\sin\{\log(2x+3)\}\cdot \frac{d}{dx}\{\sin(\log(2x+3))\} \quad[\text{using chain rule}]

\displaystyle =2\sin\{\log(2x+3)\}\cos\{\log(2x+3)\}\cdot \frac{d}{dx}\{\log(2x+3)\}

\displaystyle =\sin\{2\log(2x+3)\}\cdot \frac{1}{2x+3}\cdot \frac{d}{dx}(2x+3) \quad[\because\ 2\sin A\cos A=\sin 2A]

\displaystyle =\sin\{2\log(2x+3)\}\cdot \frac{2}{2x+3}

\displaystyle \text{So, } \frac{d}{dx}\left[\sin^{2}\{\log(2x+3)\}\right] =\sin\{2\log(2x+3)\}\left(\frac{2}{2x+3}\right)

\displaystyle \textbf{Question 44. }\ e^{x}\log(\sin 2x)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{x}\log(\sin 2x)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left[e^{x}\log(\sin 2x)\right]

\displaystyle =e^{x}\cdot \frac{d}{dx}\{\log(\sin 2x)\} +\log(\sin 2x)\cdot \frac{d}{dx}(e^{x}) \quad[\text{Using product rule}]

\displaystyle =e^{x}\cdot \frac{1}{\sin 2x}\cdot \frac{d}{dx}(\sin 2x) +e^{x}\log(\sin 2x) \quad[\text{Using chain rule}]

\displaystyle =e^{x}\cdot \frac{1}{\sin 2x}\cdot \cos 2x\cdot \frac{d}{dx}(2x) +e^{x}\log(\sin 2x)

\displaystyle =\frac{2e^{x}\cos 2x}{\sin 2x}+e^{x}\log(\sin 2x)

\displaystyle =2e^{x}\cot 2x+e^{x}\log(\sin 2x)

\displaystyle \text{So, } \frac{d}{dx}\left[e^{x}\log(\sin 2x)\right] =2e^{x}\cot 2x+e^{x}\log(\sin 2x)

\displaystyle \textbf{Question 45. }\ \frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}}

\displaystyle \text{Answer:}

\displaystyle \text{We have, }\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}}

\displaystyle \text{By rationalising, we get}

\displaystyle \frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}-\sqrt{x^{2}-1}} \times \frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}

\displaystyle =\frac{(\sqrt{x^{2}+1}+\sqrt{x^{2}-1})^{2}} {(\sqrt{x^{2}+1})^{2}-(\sqrt{x^{2}-1})^{2}}

\displaystyle =\frac{(\sqrt{x^{2}+1})^{2}+(\sqrt{x^{2}-1})^{2} +2\sqrt{x^{2}+1}\sqrt{x^{2}-1}} {x^{2}+1-(x^{2}-1)}

\displaystyle =\frac{x^{2}+1+x^{2}-1+2\sqrt{x^{4}-1}}{2}

\displaystyle =\frac{2x^{2}+2\sqrt{x^{4}-1}}{2}

\displaystyle =x^{2}+\sqrt{x^{4}-1}

\displaystyle \text{Now, let } y=x^{2}+\sqrt{x^{4}-1}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(x^{2}+\sqrt{x^{4}-1}\right)

\displaystyle =2x+\frac{1}{2\sqrt{x^{4}-1}}\cdot \frac{d}{dx}(x^{4}-1)

\displaystyle =2x+\frac{1}{2\sqrt{x^{4}-1}}\cdot 4x^{3}

\displaystyle =2x+\frac{2x^{3}}{\sqrt{x^{4}-1}}

\displaystyle \textbf{Question 46. }\ \log\!\left(x+2+\sqrt{x^{2}+4x+1}\right)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log\!\left[x+2+\sqrt{x^{2}+4x+1}\right]

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left\{\log\!\left[x+2+\sqrt{x^{2}+4x+1}\right]\right\}

\displaystyle =\frac{1}{x+2+\sqrt{x^{2}+4x+1}} \cdot \frac{d}{dx}\left[x+2+(x^{2}+4x+1)^{\tfrac12}\right] \quad[\text{Using chain rule}]

\displaystyle =\frac{1}{x+2+\sqrt{x^{2}+4x+1}} \left[1+0+\frac12(x^{2}+4x+1)^{-\tfrac12}\frac{d}{dx}(x^{2}+4x+1)\right]

\displaystyle =\frac{1}{x+2+\sqrt{x^{2}+4x+1}} \left[1+\frac{2x+4}{2\sqrt{x^{2}+4x+1}}\right]

\displaystyle =\frac{1}{x+2+\sqrt{x^{2}+4x+1}} \left[\frac{\sqrt{x^{2}+4x+1}+x+2}{\sqrt{x^{2}+4x+1}}\right]

\displaystyle =\frac{1}{\sqrt{x^{2}+4x+1}}

\displaystyle \text{So, } \frac{d}{dx}\left\{\log\!\left[x+2+\sqrt{x^{2}+4x+1}\right]\right\} =\frac{1}{\sqrt{x^{2}+4x+1}}

\displaystyle \textbf{Question 47. }\ (\sin^{-1}x^{4})^{4}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\left(\sin^{-1}x^{4}\right)^{4}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(\sin^{-1}x^{4}\right)^{4}

\displaystyle =4\left(\sin^{-1}x^{4}\right)^{3}\cdot \frac{d}{dx}\left(\sin^{-1}x^{4}\right) \quad[\text{Using chain rule}]

\displaystyle =4\left(\sin^{-1}x^{4}\right)^{3}\cdot \frac{1}{\sqrt{1-(x^{4})^{2}}}\cdot \frac{d}{dx}(x^{4}) \quad[\text{Using chain rule}]

\displaystyle =4\left(\sin^{-1}x^{4}\right)^{3}\cdot \frac{1}{\sqrt{1-x^{8}}}\cdot 4x^{3}

\displaystyle =\frac{16x^{3}\left(\sin^{-1}x^{4}\right)^{3}}{\sqrt{1-x^{8}}}

\displaystyle \text{So, } \frac{d}{dx}\left(\sin^{-1}x^{4}\right)^{4} =\frac{16x^{3}\left(\sin^{-1}x^{4}\right)^{3}}{\sqrt{1-x^{8}}}

\displaystyle \textbf{Question 48. }\ \sin^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\sin^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left\{\sin^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)\right\}

\displaystyle =\frac{1}{\sqrt{1-\left(\dfrac{x}{\sqrt{x^{2}+a^{2}}}\right)^{2}}} \cdot \frac{d}{dx}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right) \quad[\text{Using chain rule}]

\displaystyle =\frac{1}{\sqrt{\dfrac{a^{2}}{x^{2}+a^{2}}}} \cdot \frac{(\sqrt{x^{2}+a^{2}})\dfrac{d}{dx}(x) - x \dfrac{d}{dx}\!\left(\sqrt{x^{2}+a^{2}}\right)} {x^{2}+a^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{\sqrt{x^{2}+a^{2}}}{a} \cdot \frac{\sqrt{x^{2}+a^{2}}-\dfrac{x}{2\sqrt{x^{2}+a^{2}}}\cdot 2x} {x^{2}+a^{2}}

\displaystyle =\frac{\sqrt{x^{2}+a^{2}}}{a} \cdot \frac{x^{2}+a^{2}-x^{2}}{\sqrt{x^{2}+a^{2}}(x^{2}+a^{2})}

\displaystyle =\frac{a}{x^{2}+a^{2}}

\displaystyle \text{So, } \frac{d}{dx}\left\{\sin^{-1}\!\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)\right\} =\frac{a}{x^{2}+a^{2}}

\displaystyle \textbf{Question 49. }\ \frac{e^{x}\sin x}{(x^{2}+2)^{3}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\frac{e^{x}\sin x}{(x^{2}+2)^{3}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{(x^{2}+2)^{3}\dfrac{d}{dx}(e^{x}\sin x)-(e^{x}\sin x)\dfrac{d}{dx}(x^{2}+2)^{3}} {\left[(x^{2}+2)^{3}\right]^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{(x^{2}+2)^{3}\left[e^{x}\cos x+e^{x}\sin x\right]-(e^{x}\sin x)\cdot 3(x^{2}+2)^{2}\cdot 2x} {(x^{2}+2)^{6}} \quad[\text{Using product rule and chain rule}]

\displaystyle =\frac{(x^{2}+2)^{3}\left[e^{x}\cos x+e^{x}\sin x\right]-6xe^{x}\sin x\,(x^{2}+2)^{2}} {(x^{2}+2)^{6}}

\displaystyle =\frac{(x^{2}+2)^{2}\left[(x^{2}+2)(e^{x}\cos x+e^{x}\sin x)-6xe^{x}\sin x\right]} {(x^{2}+2)^{6}}

\displaystyle =\frac{(x^{2}+2)(e^{x}\cos x+e^{x}\sin x)-6xe^{x}\sin x}{(x^{2}+2)^{4}}

\displaystyle =\frac{e^{x}\sin x+e^{x}\cos x}{(x^{2}+2)^{3}}-\frac{6xe^{x}\sin x}{(x^{2}+2)^{4}}

\displaystyle \text{So, } \frac{dy}{dx} =\frac{e^{x}\sin x+e^{x}\cos x}{(x^{2}+2)^{3}}-\frac{6xe^{x}\sin x}{(x^{2}+2)^{4}}

\displaystyle \textbf{Question 50. }\ 3e^{-3x}\log(1+x)

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=3e^{-3x}\log(1+x)

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =3\frac{d}{dx}\left[e^{-3x}\log(1+x)\right]

\displaystyle =3\left[e^{-3x}\frac{d}{dx}\{\log(1+x)\} +\log(1+x)\frac{d}{dx}(e^{-3x})\right] \quad[\text{Using product rule and chain rule}]

\displaystyle =3\left[e^{-3x}\cdot \frac{1}{1+x} +\log(1+x)\cdot (-3e^{-3x})\right]

\displaystyle =3\left\{\frac{e^{-3x}}{1+x}-3e^{-3x}\log(1+x)\right\}

\displaystyle =3e^{-3x}\left\{\frac{1}{1+x}-3\log(1+x)\right\}

\displaystyle \text{So, } \frac{d}{dx}\left[3e^{-3x}\log(1+x)\right] =3e^{-3x}\left\{\frac{1}{1+x}-3\log(1+x)\right\}

\displaystyle \textbf{Question 51. }\ \frac{x^{2}+2}{\sqrt{\cos x}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\frac{x^{2}+2}{\sqrt{\cos x}}

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{\sqrt{\cos x}\,\frac{d}{dx}(x^{2}+2)-(x^{2}+2)\frac{d}{dx}(\sqrt{\cos x})}{(\sqrt{\cos x})^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{\sqrt{\cos x}\cdot 2x-(x^{2}+2)\left(\frac12(\cos x)^{-\tfrac12}\cdot (-\sin x)\right)}{\cos x} \quad[\text{Using chain rule}]

\displaystyle =\frac{2x\sqrt{\cos x}+\frac{(x^{2}+2)\sin x}{2\sqrt{\cos x}}}{\cos x}

\displaystyle =\frac{4x\cos x+(x^{2}+2)\sin x}{2(\cos x)^{\tfrac32}}

\displaystyle =\frac{2x}{\sqrt{\cos x}}+\frac{(x^{2}+2)\sin x}{2(\cos x)^{\tfrac32}}

\displaystyle =\frac{1}{\sqrt{\cos x}}\left\{2x+\frac12\cdot \frac{(x^{2}+2)\sin x}{\cos x}\right\}

\displaystyle =\frac{1}{\sqrt{\cos x}}\left\{2x+\frac{(x^{2}+2)\tan x}{2}\right\}

\displaystyle \text{So, } \frac{d}{dx}\left(\frac{x^{2}+2}{\sqrt{\cos x}}\right) =\frac{1}{\sqrt{\cos x}}\left\{2x+\frac{(x^{2}+2)\tan x}{2}\right\}

\displaystyle \textbf{Question 52. }\ \frac{x^{2}(1-x^{2})^{3}}{\cos 2x}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\frac{x^{2}(1-x^{2})^{3}}{\cos 2x}

\displaystyle \Rightarrow\ \frac{dy}{dx} =\frac{\cos 2x\,\dfrac{d}{dx}\{x^{2}(1-x^{2})^{3}\}-x^{2}(1-x^{2})^{3}\dfrac{d}{dx}(\cos 2x)}{\cos^{2}2x} \quad[\text{Using quotient rule}]

\displaystyle =\frac{\cos 2x\left[x^{2}\frac{d}{dx}(1-x^{2})^{3}+(1-x^{2})^{3}\frac{d}{dx}(x^{2})\right] -x^{2}(1-x^{2})^{3}(-2\sin 2x)}{\cos^{2}2x} \quad[\text{Using product rule and chain rule}]

\displaystyle =\frac{\cos 2x\left[x^{2}\cdot 3(1-x^{2})^{2}\frac{d}{dx}(1-x^{2})+(1-x^{2})^{3}\cdot 2x\right] +2x^{2}(1-x^{2})^{3}\sin 2x}{\cos^{2}2x}

\displaystyle =\frac{\cos 2x\left[x^{2}\cdot 3(1-x^{2})^{2}(-2x)+2x(1-x^{2})^{3}\right] +2x^{2}(1-x^{2})^{3}\sin 2x}{\cos^{2}2x}

\displaystyle =\frac{\cos 2x\left[-6x^{3}(1-x^{2})^{2}+2x(1-x^{2})^{3}\right] +2x^{2}(1-x^{2})^{3}\sin 2x}{\cos^{2}2x}

\displaystyle =\frac{2x(1-x^{2})^{2}\cos 2x-6x^{3}(1-x^{2})^{2}\cos 2x+2x^{2}(1-x^{2})^{3}\sin 2x}{\cos^{2}2x}

\displaystyle =\frac{2x(1-x^{2})^{2}}{\cos 2x}-\frac{6x^{3}(1-x^{2})^{2}}{\cos 2x} +\frac{2x^{2}(1-x^{2})^{3}\sin 2x}{\cos^{2}2x}

\displaystyle =2x(1-x^{2})^{2}\sec 2x\left[1-3x^{2}+x(1-x^{2})\tan 2x\right]

\displaystyle \text{So, } \frac{d}{dx}\left\{\frac{x^{2}(1-x^{2})^{3}}{\cos 2x}\right\} =2x(1-x^{2})^{2}\sec 2x\left\{1-3x^{2}+x(1-x^{2})\tan 2x\right\}

\displaystyle \textbf{Question 53. }\ \log\!\left(\cot\!\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)

\displaystyle \text{Answer:}

\displaystyle \log\!\left\{\cot\!\left(\frac{\pi}{4}+\frac{x}{2}\right)\right\} =\frac{-\mathrm{cosec}^{2}\!\left(\frac{\pi}{4}+\frac{x}{2}\right)} {2\cot\!\left(\frac{\pi}{4}+\frac{x}{2}\right)}

\displaystyle =\frac{-1}{2\cos\!\left(\frac{\pi}{4}+\frac{x}{2}\right) \sin\!\left(\frac{\pi}{4}+\frac{x}{2}\right)}

\displaystyle =\frac{-1}{\sin\!\left(\frac{\pi}{2}+x\right)}

\displaystyle =\frac{-1}{\cos x}

\displaystyle =-\sec x

\displaystyle \textbf{Question 54. }\ e^{ax}\sec x\tan 2x

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=e^{ax}\sec x\tan 2x

\displaystyle \text{Differentiate } y \text{ with respect to } x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(e^{ax}\sec x\tan 2x\right)

\displaystyle =e^{ax}\frac{d}{dx}(\sec x\tan 2x) +\sec x\tan 2x\frac{d}{dx}(e^{ax}) \quad[\text{Using product rule}]

\displaystyle =e^{ax}\left[\sec x\tan x\tan 2x +\sec x\cdot \frac{d}{dx}(\tan 2x)\right] +ae^{ax}\sec x\tan 2x

\displaystyle =e^{ax}\left[\sec x\tan x\tan 2x +\sec x\cdot 2\sec^{2}2x\right] +ae^{ax}\sec x\tan 2x

\displaystyle =ae^{ax}\sec x\tan 2x +e^{ax}\sec x\tan x\tan 2x +2e^{ax}\sec x\sec^{2}2x

\displaystyle =e^{ax}\sec x\left\{a\tan 2x+\tan x\tan 2x+2\sec^{2}2x\right\}

\displaystyle \text{So, } \frac{d}{dx}\left(e^{ax}\sec x\tan 2x\right) =e^{ax}\sec x\left\{a\tan 2x+\tan x\tan 2x+2\sec^{2}2x\right\}

\displaystyle \textbf{Question 55. }\ \log(\cos x^{2})

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log(\cos x^{2})

\displaystyle \text{Differentiating } y \text{ with respect to } x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\{\log(\cos x^{2})\}

\displaystyle =\frac{1}{\cos x^{2}}\cdot \frac{d}{dx}(\cos x^{2}) \quad[\text{using chain rule}]

\displaystyle =\frac{1}{\cos x^{2}}\cdot (-\sin x^{2})\cdot \frac{d}{dx}(x^{2})

\displaystyle =\frac{-2x\sin x^{2}}{\cos x^{2}}

\displaystyle =-2x\tan x^{2}

\displaystyle \text{So, } \frac{d}{dx}\{\log(\cos x^{2})\}=-2x\tan x^{2}

\displaystyle \textbf{Question 56. }\ \cos(\log x)^{2}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\cos\!\left((\log x)^{2}\right)

\displaystyle \text{Differentiating } y \text{ with respect to } x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left\{\cos\!\left((\log x)^{2}\right)\right\}

\displaystyle =-\sin\!\left((\log x)^{2}\right)\cdot \frac{d}{dx}\left((\log x)^{2}\right) \quad[\text{using chain rule}]

\displaystyle =-\sin\!\left((\log x)^{2}\right)\cdot 2\log x \cdot \frac{1}{x}

\displaystyle =\frac{-2\log x\,\sin\!\left((\log x)^{2}\right)}{x}

\displaystyle \text{So, } \frac{d}{dx}\left\{\cos\!\left((\log x)^{2}\right)\right\} =\frac{-2\log x\,\sin\!\left((\log x)^{2}\right)}{x}

\displaystyle \textbf{Question 57. }\ \log\sqrt{\frac{x-1}{x+1}}

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log\sqrt{\frac{x-1}{x+1}}

\displaystyle \Rightarrow\ y=\log\left(\frac{x-1}{x+1}\right)^{\tfrac12}

\displaystyle \Rightarrow\ y=\frac12\log\left(\frac{x-1}{x+1}\right)

\displaystyle \Rightarrow\ y=\frac12\,[\log(x-1)-\log(x+1)]

\displaystyle \text{Differentiating } y \text{ with respect to } x,

\displaystyle \frac{dy}{dx} =\frac12\left[\frac{d}{dx}\{\log(x-1)\}-\frac{d}{dx}\{\log(x+1)\}\right]

\displaystyle =\frac12\left(\frac{1}{x-1}-\frac{1}{x+1}\right)

\displaystyle =\frac12\left(\frac{(x+1)-(x-1)}{x^{2}-1}\right)

\displaystyle =\frac12\left(\frac{2}{x^{2}-1}\right)

\displaystyle =\frac{1}{x^{2}-1}

\displaystyle \text{So, } \frac{dy}{dx}=\frac{1}{x^{2}-1}

\displaystyle \textbf{Question 58. }\ \text{If } y=\log\left(\sqrt{x-1}-\sqrt{x+1}\right),\ \text{show that }\frac{dy}{dx}=\frac{-1}{2\sqrt{x^{2}-1}}

\displaystyle \text{Here, } y=\log(\sqrt{x-1}-\sqrt{x+1})

\displaystyle \text{Differentiating } y \text{ with respect to } x \text{, we get}

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\{\log(\sqrt{x-1}-\sqrt{x+1})\}

\displaystyle =\frac{1}{\sqrt{x-1}-\sqrt{x+1}}\, \frac{d}{dx}(\sqrt{x-1}-\sqrt{x+1}) \quad[\text{using chain rule}]

\displaystyle =\frac{1}{\sqrt{x-1}-\sqrt{x+1}} \left(\frac{d}{dx}\sqrt{x-1}-\frac{d}{dx}\sqrt{x+1}\right)

\displaystyle =\frac{1}{\sqrt{x-1}-\sqrt{x+1}} \left(\frac{1}{2\sqrt{x-1}}-\frac{1}{2\sqrt{x+1}}\right)

\displaystyle =\frac{1}{2(\sqrt{x-1}-\sqrt{x+1})} \left(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x-1}\sqrt{x+1}}\right)

\displaystyle =-\frac{1}{2}\cdot \frac{1}{\sqrt{x-1}\sqrt{x+1}}

\displaystyle =-\frac{1}{2\sqrt{x^{2}-1}}

\displaystyle \text{So, } \frac{dy}{dx}=-\frac{1}{2\sqrt{x^{2}-1}}

\displaystyle \text{Answer:}

\displaystyle \textbf{Question 59. }\ \text{If } y=\sqrt{x+1}+\sqrt{x-1},\ \text{prove that }\sqrt{x^{2}-1}\,\frac{dy}{dx}=\frac{1}{2}\,y 

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\sqrt{x+1}+\sqrt{x-1}

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}(\sqrt{x+1})+\frac{d}{dx}(\sqrt{x-1})

\displaystyle =\frac12(x+1)^{-\tfrac12}+\frac12(x-1)^{-\tfrac12}

\displaystyle =\frac12\left(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{x-1}}\right)

\displaystyle =\frac12\left(\frac{\sqrt{x-1}+\sqrt{x+1}}{\sqrt{x+1}\sqrt{x-1}}\right)

\displaystyle =\frac12\left(\frac{y}{\sqrt{x^{2}-1}}\right)

\displaystyle \Rightarrow\ (\sqrt{x^{2}-1})\frac{dy}{dx}=\frac12\,y

\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 60. }\ \text{If } y=\frac{x}{x+2},\ \text{prove that }x\frac{dy}{dx}=(1-y)\,y

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\frac{x}{x+2}

\displaystyle \text{Differentiating with respect to } x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(\frac{x}{x+2}\right)

\displaystyle =\frac{(x+2)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(x+2)}{(x+2)^{2}} \quad[\text{Using quotient rule}]

\displaystyle =\frac{(x+2)-x}{(x+2)^{2}}

\displaystyle =\frac{1}{x+2}-\frac{x}{(x+2)^{2}}

\displaystyle =\frac{y}{x}-\frac{y^{2}}{x} \quad[\because\ x+2=\tfrac{x}{y}]

\displaystyle =\frac{1}{x}\,y(1-y)

\displaystyle \Rightarrow\ x\frac{dy}{dx}=y(1-y)

\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 61. }\ \text{If } y=\log\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right),\ \text{prove that }\frac{dy}{dx}=\frac{x-1}{2x(x+1)}

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\log\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)

\displaystyle \text{Differentiate it with respect to } x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left\{\log\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\right\}

\displaystyle =\frac{1}{\sqrt{x}+\frac{1}{\sqrt{x}}}\cdot \frac{d}{dx}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) \quad[\text{Using chain rule}]

\displaystyle =\frac{1}{\sqrt{x}+\frac{1}{\sqrt{x}}}\left(\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}\right)

\displaystyle =\frac{\sqrt{x}}{x+1}\left(\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}\right)

\displaystyle =\frac{1}{2}\cdot \frac{\sqrt{x}}{x+1}\left(\frac{x-1}{x\sqrt{x}}\right)

\displaystyle =\frac{x-1}{2x(x+1)}

\displaystyle \text{So, } \frac{dy}{dx}=\frac{x-1}{2x(x+1)}

\displaystyle \textbf{Question 62. }\ \text{If } y=\log\sqrt{\frac{1+\tan x}{1-\tan x}},\ \text{prove that }\frac{dy}{dx}=\sec 2x

\displaystyle \text{Answer:}

\displaystyle \text{Let } y=\log\sqrt{\frac{1+\tan x}{1-\tan x}}

\displaystyle \Rightarrow\ y=\log\left(\frac{1+\tan x}{1-\tan x}\right)^{\tfrac12}

\displaystyle \Rightarrow\ y=\frac12\log\left(\frac{1+\tan x}{1-\tan x}\right)

\displaystyle \Rightarrow\ y=\frac12\{\log(1+\tan x)-\log(1-\tan x)\}

\displaystyle \Rightarrow\ \frac{dy}{dx} =\frac12\left\{\frac{d}{dx}\log(1+\tan x)-\frac{d}{dx}\log(1-\tan x)\right\}

\displaystyle =\frac12\left\{\frac{1}{1+\tan x}\frac{d}{dx}(1+\tan x)-\frac{1}{1-\tan x}\frac{d}{dx}(1-\tan x)\right\} \quad[\text{Using chain rule}]

\displaystyle =\frac12\left\{\frac{1}{1+\tan x}\sec^{2}x-\frac{1}{1-\tan x}(-\sec^{2}x)\right\}

\displaystyle =\frac12\left\{\frac{\sec^{2}x}{1+\tan x}+\frac{\sec^{2}x}{1-\tan x}\right\}

\displaystyle =\frac12\sec^{2}x\left\{\frac{(1-\tan x)+(1+\tan x)}{1-\tan^{2}x}\right\}

\displaystyle =\frac12\sec^{2}x\left(\frac{2}{1-\tan^{2}x}\right)

\displaystyle =\frac{\sec^{2}x}{1-\tan^{2}x}

\displaystyle =\frac{1+\tan^{2}x}{1-\tan^{2}x}

\displaystyle =\frac{1}{\dfrac{1-\tan^{2}x}{1+\tan^{2}x}}

\displaystyle =\frac{1}{\cos 2x} \quad[\because\ \cos 2x=\frac{1-\tan^{2}x}{1+\tan^{2}x}]

\displaystyle =\sec 2x

\displaystyle \text{So, } \frac{dy}{dx}=\sec 2x

\displaystyle \textbf{Question 63. }\ \text{If } y=\sqrt{x}+\frac{1}{\sqrt{x}},\ \text{prove that }2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}}

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\sqrt{x}+\frac{1}{\sqrt{x}}

\displaystyle \text{Differentiate with respect to }x,

\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{d}{dx}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)

\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{d}{dx}(\sqrt{x})+\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)

\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}

\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{x-1}{2x\sqrt{x}}

\displaystyle \Rightarrow\ 2x\,\frac{dy}{dx}=\frac{x-1}{\sqrt{x}}

\displaystyle \Rightarrow\ 2x\,\frac{dy}{dx}=\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}}

\displaystyle \Rightarrow\ 2x\,\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}}

\displaystyle \textbf{Question 64. }\ \text{If } y=\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}},\ \text{prove that }(1-x^{2})\frac{dy}{dx}=x+\frac{y}{x}

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}\right)

\displaystyle =\frac{\sqrt{1-x^{2}}\;\frac{d}{dx}(x\sin^{-1}x)-(x\sin^{-1}x)\;\frac{d}{dx}(\sqrt{1-x^{2}})} {(\sqrt{1-x^{2}})^{2}}

\displaystyle =\frac{\sqrt{1-x^{2}}\left[x\frac{d}{dx}(\sin^{-1}x)+\sin^{-1}x\frac{d}{dx}(x)\right] -(x\sin^{-1}x)\left(\frac{-x}{\sqrt{1-x^{2}}}\right)} {1-x^{2}}

\displaystyle =\frac{\sqrt{1-x^{2}}\left(\frac{x}{\sqrt{1-x^{2}}}+\sin^{-1}x\right) +\frac{x^{2}\sin^{-1}x}{\sqrt{1-x^{2}}}} {1-x^{2}}

\displaystyle =\frac{x+\sqrt{1-x^{2}}\sin^{-1}x+\frac{x^{2}\sin^{-1}x}{\sqrt{1-x^{2}}}} {1-x^{2}}

\displaystyle (1-x^{2})\frac{dy}{dx} =x+\frac{(1-x^{2})\sin^{-1}x+x^{2}\sin^{-1}x}{\sqrt{1-x^{2}}}

\displaystyle (1-x^{2})\frac{dy}{dx} =x+\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}

\displaystyle (1-x^{2})\frac{dy}{dx} =x+\frac{y}{x}\qquad\left[\because y=\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}\right]

\displaystyle \textbf{Question 65. }\ \text{If } y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}},\ \text{prove that }\frac{dy}{dx}=1-y^{2}.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\left(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right)

\displaystyle =\frac{(e^{x}+e^{-x})\frac{d}{dx}(e^{x}-e^{-x})-(e^{x}-e^{-x})\frac{d}{dx}(e^{x}+e^{-x})} {(e^{x}+e^{-x})^{2}}

\displaystyle =\frac{(e^{x}+e^{-x})(e^{x}+e^{-x})-(e^{x}-e^{-x})(e^{x}-e^{-x})} {(e^{x}+e^{-x})^{2}}

\displaystyle =\frac{(e^{x}+e^{-x})^{2}-(e^{x}-e^{-x})^{2}} {(e^{x}+e^{-x})^{2}}

\displaystyle =1-\left(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right)^{2}

\displaystyle =1-y^{2}

\displaystyle \text{So, } \frac{dy}{dx}=1-y^{2}

\displaystyle \textbf{Question 66. }\ \text{If } y=(x-1)\log(x-1)-(x+1)\log(x+1),\ \text{prove that }\frac{dy}{dx}=\log\left(\frac{x-1}{1+x}\right).

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=(x-1)\log(x-1)-(x+1)\log(x+1)

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\big[(x-1)\log(x-1)-(x+1)\log(x+1)\big]

\displaystyle =\Big[(x-1)\frac{d}{dx}\log(x-1)+\log(x-1)\frac{d}{dx}(x-1)\Big] -\Big[(x+1)\frac{d}{dx}\log(x+1)+\log(x+1)\frac{d}{dx}(x+1)\Big]

\displaystyle =\Big[(x-1)\frac{1}{x-1}(1)+\log(x-1)(1)\Big] -\Big[(x+1)\frac{1}{x+1}(1)+\log(x+1)(1)\Big]

\displaystyle =\big[1+\log(x-1)\big]-\big[1+\log(x+1)\big]

\displaystyle =\log(x-1)-\log(x+1)

\displaystyle =\log\!\left(\frac{x-1}{x+1}\right)

\displaystyle \text{So, }\frac{dy}{dx}=\log\!\left(\frac{x-1}{x+1}\right)

\displaystyle \textbf{Question 67. }\ \text{If } y=e^{x}\cos x,\ \text{prove that }\frac{dy}{dx}=\sqrt{2}\,e^{x}\cos\left(x+\frac{\pi}{4}\right).

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=e^{x}\cos x

\displaystyle \text{Differentiating with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}(e^{x}\cos x)

\displaystyle =e^{x}\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(e^{x}) \quad[\text{Using product rule}]

\displaystyle =e^{x}(-\sin x)+\cos x\cdot e^{x}

\displaystyle =e^{x}(\cos x-\sin x)

\displaystyle =\sqrt{2}\,e^{x}\left(\frac{\cos x}{\sqrt{2}}-\frac{\sin x}{\sqrt{2}}\right)

\displaystyle =\sqrt{2}\,e^{x}\left(\cos\frac{\pi}{4}\cos x-\sin\frac{\pi}{4}\sin x\right)

\displaystyle =\sqrt{2}\,e^{x}\cos\left(x+\frac{\pi}{4}\right) \quad[\because\ \cos(A+B)=\cos A\cos B-\sin A\sin B]

\displaystyle \text{So, } \frac{dy}{dx}=\sqrt{2}\,e^{x}\cos\left(x+\frac{\pi}{4}\right)

\displaystyle \textbf{Question 68. }\ \text{If } y=\frac{1}{2}\log\left(\frac{1-\cos 2x}{1+\cos 2x}\right),\ \text{prove that }\frac{dy}{dx}=2\mathrm{cosec} 2x.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\frac12\log\left(\frac{1-\cos 2x}{1+\cos 2x}\right)

\displaystyle \Rightarrow\ y=\frac12\log\left(\frac{2\sin^{2}x}{2\cos^{2}x}\right)

\displaystyle \Rightarrow\ y=\frac12\log(\tan^{2}x)

\displaystyle \Rightarrow\ y=\frac12\cdot 2\log(\tan x)

\displaystyle \Rightarrow\ y=\log(\tan x)

\displaystyle \text{Differentiate with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\log\tan x)

\displaystyle =\frac{1}{\tan x}\cdot \frac{d}{dx}(\tan x)

\displaystyle =\frac{\sec^{2}x}{\tan x}

\displaystyle =\frac{1}{\sin x\cos x}

\displaystyle =\frac{2}{2\sin x\cos x}

\displaystyle =\frac{2}{\sin 2x}

\displaystyle \text{So, } \frac{dy}{dx}=2\,\mathrm{cosec}\,2x

\displaystyle \textbf{Question 69. }\ \text{If } y=x\sin^{-1}x+\sqrt{1-x^{2}},\ \text{prove that }\frac{dy}{dx}=\sin^{-1}x.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=x\sin^{-1}x+\sqrt{1-x^{2}}

\displaystyle \text{Differentiate with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}\big[x\sin^{-1}x+\sqrt{1-x^{2}}\big]

\displaystyle =\frac{d}{dx}(x\sin^{-1}x)+\frac{d}{dx}(\sqrt{1-x^{2}})

\displaystyle =\Big[x\frac{d}{dx}(\sin^{-1}x)+\sin^{-1}x\frac{d}{dx}(x)\Big] +\frac{1}{2\sqrt{1-x^{2}}}\frac{d}{dx}(1-x^{2})

\displaystyle =\left[\frac{x}{\sqrt{1-x^{2}}}+\sin^{-1}x\right] -\frac{2x}{2\sqrt{1-x^{2}}}

\displaystyle =\frac{x}{\sqrt{1-x^{2}}}+\sin^{-1}x-\frac{x}{\sqrt{1-x^{2}}}

\displaystyle =\sin^{-1}x

\displaystyle \text{So, }\frac{dy}{dx}=\sin^{-1}x

\displaystyle \textbf{Question 70. }\ \text{If } y=\sqrt{x^{2}+a^{2}},\ \text{prove that }y\frac{dy}{dx}-x=0.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\sqrt{x^{2}+a^{2}}

\displaystyle \text{Squaring both sides, we get}

\displaystyle y^{2}=x^{2}+a^{2}

\displaystyle \Rightarrow\ 2y\frac{dy}{dx}=\frac{d}{dx}(x^{2}+a^{2})

\displaystyle \Rightarrow\ 2y\frac{dy}{dx}=2x

\displaystyle \Rightarrow\ y\frac{dy}{dx}=x

\displaystyle \Rightarrow\ y\frac{dy}{dx}-x=0

\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 71. }\ \text{If } y=e^{x}+e^{-x},\ \text{prove that }\frac{dy}{dx}=\sqrt{y^{2}-4}.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=e^{x}+e^{-x}

\displaystyle \text{Differentiate it with respect to }x,

\displaystyle \frac{dy}{dx} =\frac{d}{dx}(e^{x}+e^{-x})

\displaystyle =\frac{d}{dx}(e^{x})+\frac{d}{dx}(e^{-x})

\displaystyle =e^{x}+e^{-x}\frac{d}{dx}(-x) \quad[\text{Using chain rule}]

\displaystyle =e^{x}-e^{-x}

\displaystyle =\sqrt{(e^{x}-e^{-x})^{2}}

\displaystyle =\sqrt{(e^{x}+e^{-x})^{2}-4e^{x}e^{-x}} \quad\left[\because (a-b)^{2}=(a+b)^{2}-4ab\right]

\displaystyle =\sqrt{y^{2}-4} \quad\left[\because e^{x}+e^{-x}=y\right]

\displaystyle \textbf{Question 72. }\ \text{If } y=\sqrt{a^{2}-x^{2}},\ \text{prove that }y\frac{dy}{dx}+x=0.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } y=\sqrt{a^{2}-x^{2}}

\displaystyle \text{Squaring both sides, we get}

\displaystyle y^{2}=a^{2}-x^{2}

\displaystyle \text{Differentiating both sides w.r.t. }x,

\displaystyle 2y\frac{dy}{dx}=\frac{d}{dx}(a^{2}-x^{2})

\displaystyle 2y\frac{dy}{dx}=0-2x

\displaystyle y\frac{dy}{dx}=-x

\displaystyle y\frac{dy}{dx}+x=0

\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 73. }\ \text{If } xy=4,\ \text{prove that }x\left(\frac{dy}{dx}+y^{2}\right)=3y.

\displaystyle \text{Answer:}

\displaystyle \text{We have, } xy=4

\displaystyle \Rightarrow\ y=\frac{4}{x}

\displaystyle \text{Differentiate it with respect to }x,

\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{4}{x}\right)

\displaystyle \Rightarrow\ \frac{dy}{dx}=4\frac{d}{dx}(x^{-1})

\displaystyle \Rightarrow\ \frac{dy}{dx}=4(-1)x^{-2}

\displaystyle \Rightarrow\ \frac{dy}{dx}=-\frac{4}{x^{2}}

\displaystyle \Rightarrow\ \frac{dy}{dx}=-\frac{4}{\left(\frac{4}{y}\right)^{2}} \qquad[\because x=\frac{4}{y}]

\displaystyle \Rightarrow\ \frac{dy}{dx}=-\frac{4y^{2}}{16}

\displaystyle \Rightarrow\ \frac{dy}{dx}=-\frac{y^{2}}{4}

\displaystyle \Rightarrow\ 4\frac{dy}{dx}=-y^{2}

\displaystyle \Rightarrow\ 4\frac{dy}{dx}+y^{2}=0

\displaystyle \Rightarrow\ 4\left(\frac{dy}{dx}+y^{2}\right)=3y^{2}

\displaystyle \Rightarrow\ \frac{4}{x}\left(\frac{dy}{dx}+y^{2}\right)=\frac{3y^{2}}{x}

\displaystyle \Rightarrow\ y\left(\frac{dy}{dx}+y^{2}\right)=\frac{3y^{2}}{x}

\displaystyle \Rightarrow\ x\left(\frac{dy}{dx}+y^{2}\right)=\frac{3y^{2}}{y}

\displaystyle \Rightarrow\ x\left(\frac{dy}{dx}+y^{2}\right)=3y

\displaystyle \textbf{Question 74. }\ \text{Prove that }\frac{d}{dx}\left[\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}\right)\right]=\sqrt{a^{2}-x^{2}}.

\displaystyle \text{Answer:}

\displaystyle \frac{d}{dx}\left\{\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\right\} =\sqrt{a^{2}-x^{2}}

\displaystyle \text{LHS }= \frac{d}{dx}\left\{\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\right\}

\displaystyle =\frac{d}{dx}\left(\frac{x}{2}\sqrt{a^{2}-x^{2}}\right) +\frac{d}{dx}\left(\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\right)

\displaystyle =\frac12\left[x\frac{d}{dx}\!\left(\sqrt{a^{2}-x^{2}}\right) +\sqrt{a^{2}-x^{2}}\frac{d}{dx}(x)\right] +\frac{a^{2}}{2}\cdot\frac{1}{\sqrt{1-(x/a)^{2}}}\cdot\frac{d}{dx}\!\left(\frac{x}{a}\right)

\displaystyle =\frac12\left[x\cdot\frac{1}{2\sqrt{a^{2}-x^{2}}}\frac{d}{dx}(a^{2}-x^{2}) +\sqrt{a^{2}-x^{2}}\right] +\frac{a^{2}}{2}\cdot\frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}}\cdot\frac{1}{a}

\displaystyle =\frac12\left[\frac{x(-2x)}{2\sqrt{a^{2}-x^{2}}} +\sqrt{a^{2}-x^{2}}\right] +\frac{a^{2}}{2}\cdot\frac{a}{\sqrt{a^{2}-x^{2}}}\cdot\frac{1}{a}

\displaystyle =\frac12\left[\frac{-2x^{2}+2(a^{2}-x^{2})}{2\sqrt{a^{2}-x^{2}}}\right] +\frac{a^{2}}{2\sqrt{a^{2}-x^{2}}}

\displaystyle =\frac{a^{2}-2x^{2}}{2\sqrt{a^{2}-x^{2}}} +\frac{a^{2}}{2\sqrt{a^{2}-x^{2}}}

\displaystyle =\frac{2a^{2}-2x^{2}}{2\sqrt{a^{2}-x^{2}}}

\displaystyle =\frac{2(a^{2}-x^{2})}{2\sqrt{a^{2}-x^{2}}}

\displaystyle =\frac{a^{2}-x^{2}}{\sqrt{a^{2}-x^{2}}}

\displaystyle =\sqrt{a^{2}-x^{2}} \;=\; \text{RHS}

\displaystyle \text{Hence proved.}

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