\displaystyle \textbf{Question 1: }~\text{Differentiate }x^2\text{ with respect to }x^3.
\displaystyle \text{Answer:}

\displaystyle \text{Let } u=x^{2} \text{ and } v=x^{3}

\displaystyle \Rightarrow \frac{du}{dx}=2x \text{ and } \frac{dv}{dx}=3x^{2}

\displaystyle \therefore \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2x}{3x^{2}} = \frac{2}{3x}

\displaystyle \textbf{Question 2: }~\text{Differentiate }\log(1+x^2)\text{ with respect to }\tan^{-1}x.
\displaystyle \text{Answer:}

\displaystyle \text{Let } u=\log(1+x^{2}) \text{ and } v=\tan^{-1}x

\displaystyle \Rightarrow \frac{du}{dx} = \frac{1}{1+x^{2}}\frac{d}{dx}(1+x^{2}) = \frac{2x}{1+x^{2}} \text{ and } \frac{dv}{dx} = \frac{1}{1+x^{2}}

\displaystyle \therefore \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2x}{1+x^{2}}\times\frac{1+x^{2}}{1} = 2x

\displaystyle \textbf{Question 3: }~\text{Differentiate }(\log x)^x\text{ with respect to }\log x.
\displaystyle \text{Answer:}

\displaystyle \text{Let } u=(\log x)^{x}

\displaystyle \text{Taking log on both sides,}

\displaystyle \log u=\log\big((\log x)^{x}\big)

\displaystyle \Rightarrow \log u=x\log(\log x)

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\{\log(\log x)\} + \log(\log x)\frac{d}{dx}(x)

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = x\left(\frac{1}{\log x}\right)\frac{d}{dx}(\log x) + \log(\log x)\cdot 1

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} = x\left(\frac{1}{\log x}\right)\left(\frac{1}{x}\right) + \log(\log x)

\displaystyle \Rightarrow \frac{du}{dx} = u\left[\frac{1}{\log x}+\log(\log x)\right] \ \ldots (i)

\displaystyle \text{Again, let } v=\log x

\displaystyle \Rightarrow \frac{dv}{dx}=\frac{1}{x} \ \ldots (ii)

\displaystyle \text{Dividing equation (i) by (ii), we get}

\displaystyle \frac{du}{dv} = \frac{(\log x)^{x}\left[\frac{1}{\log x}+\log(\log x)\right]} {\frac{1}{x}}

\displaystyle \Rightarrow \frac{du}{dv} = x(\log x)^{x}\left[\frac{1}{\log x}+\log(\log x)\right]

\displaystyle \Rightarrow \frac{du}{dv} = x(\log x)^{x-1}\left[1+\log x\,\log(\log x)\right]

\displaystyle \textbf{Question 4: }~\text{Differentiate }\sin^{-1}\!\sqrt{1-x^2}\text{ with respect to }\cos^{-1}x,\text{ if }(i)\;x\in(0,1)\quad (ii)\;x\in(-1,0).
\displaystyle \text{Answer: (i) }

\displaystyle \text{Let } u=\sin^{-1}\!\sqrt{1-x^{2}}

\displaystyle \text{Put } x=\cos\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\sqrt{1-\cos^{2}\theta} = \sin^{-1}(\sin\theta)\quad\ldots (i)

\displaystyle \text{And let } v=\cos^{-1}x \quad\ldots (ii)

\displaystyle \text{Now, } x\in(0,1) \Rightarrow \cos\theta\in(0,1) \Rightarrow \theta\in\left(0,\frac{\pi}{2}\right)

\displaystyle \text{So, from (i), } u=\theta \quad\big[\text{since } \sin^{-1}(\sin\theta)=\theta \text{ for } \theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\big]

\displaystyle \Rightarrow u=\cos^{-1}x \quad\big[\text{since } \cos\theta=x\big]

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

\displaystyle \frac{du}{dx} = -\frac{1}{\sqrt{1-x^{2}}} \quad\ldots (iii)

\displaystyle \text{From (ii), } v=\cos^{-1}x

\displaystyle \Rightarrow \frac{dv}{dx} = -\frac{1}{\sqrt{1-x^{2}}} \quad\ldots (iv)

\displaystyle \text{Dividing (iii) by (iv),}

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

\displaystyle \text{Answer: (ii) }

\displaystyle \text{Let } u=\sin^{-1}\!\sqrt{1-x^{2}}

\displaystyle \text{Put } x=\cos\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\sqrt{1-\cos^{2}\theta}

\displaystyle \Rightarrow u=\sin^{-1}(\sin\theta)\quad ...(i)

\displaystyle \text{And, } v=\cos^{-1}x\quad ...(ii)

\displaystyle \text{Now, } x\in(-1,0)

\displaystyle \Rightarrow \cos\theta\in(-1,0)

\displaystyle \Rightarrow \theta\in\left(\frac{\pi}{2},\pi\right)

\displaystyle \text{So, from equation }(i),

\displaystyle u=\pi-\theta \quad\left[\text{Since } \sin^{-1}(\sin\theta)=\pi-\theta \text{ if } \theta\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\right]

\displaystyle \Rightarrow u=\pi-\cos^{-1}x \quad\left[\text{Since } x=\cos\theta\right]

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

\displaystyle \frac{du}{dx} =0-\left(-\frac{1}{\sqrt{1-x^{2}}}\right) =\frac{1}{\sqrt{1-x^{2}}} \quad ...(iii)

\displaystyle \text{From equation }(ii),

\displaystyle v=\cos^{-1}x

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

\displaystyle \frac{dv}{dx} =-\frac{1}{\sqrt{1-x^{2}}} \quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \textbf{Question 5: }~\text{Differentiate }\sin^{-1}\!\bigl(4x\sqrt{1-4x^2}\bigr)\text{ with respect to }\sqrt{1-4x^2},\text{ if }(i)\;x\in\left(-\frac{1}{2\sqrt2},\frac{1}{2\sqrt2}\right)\;(ii)\;x\in\left(\frac{1}{2\sqrt2},\frac12\right)\;(iii)\;x\in\left(-\frac12,-\frac{1}{2\sqrt2}\right).
\displaystyle \text{Answer: (i) }

\displaystyle \text{Let } u=\sin^{-1}\!\left(4x\sqrt{1-4x^{2}}\right)

\displaystyle \text{Put } 2x=\cos\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\left(2\cos\theta\sqrt{1-\cos^{2}\theta}\right)

\displaystyle \Rightarrow u=\sin^{-1}(2\cos\theta\sin\theta)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad ...(i)

\displaystyle \text{Let } v=\sqrt{1-4x^{2}}\quad ...(ii)

\displaystyle \text{Here, } x\in\left(-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}\right)

\displaystyle \Rightarrow 2x\in\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)

\displaystyle \Rightarrow \theta\in\left(\frac{\pi}{4},\frac{3\pi}{4}\right)

\displaystyle \text{So, from equation }(i),

\displaystyle u=\pi-2\theta \quad\left[\text{Since } \sin^{-1}(\sin\alpha)=\pi-\alpha \text{ if } \alpha\in\left(\frac{\pi}{2},\pi\right)\right]

\displaystyle \Rightarrow u=\pi-2\cos^{-1}(2x) \quad\left[\text{Since } 2x=\cos\theta\right]

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

\displaystyle \frac{du}{dx} =0-2\left(-\frac{1}{\sqrt{1-(2x)^{2}}}\right)\frac{d}{dx}(2x)

\displaystyle \Rightarrow \frac{du}{dx} =\frac{2}{\sqrt{1-4x^{2}}}\cdot 2 =\frac{4}{\sqrt{1-4x^{2}}} \quad ...(iii)

\displaystyle \text{From equation }(ii),

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

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

\displaystyle \frac{dv}{dx} =\frac{1}{2\sqrt{1-4x^{2}}}\frac{d}{dx}(1-4x^{2}) =\frac{-8x}{2\sqrt{1-4x^{2}}}

\displaystyle \Rightarrow \frac{dv}{dx} =\frac{-4x}{\sqrt{1-4x^{2}}} \quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \text{Answer: (ii) }

\displaystyle \text{Let } u=\sin^{-1}\!\left(4x\sqrt{1-4x^{2}}\right)

\displaystyle \text{Put } 2x=\cos\theta

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

\displaystyle \Rightarrow u=\sin^{-1}(2\cos\theta\sin\theta)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad ...(i)

\displaystyle \text{Let } v=\sqrt{1-4x^{2}}\quad ...(ii)

\displaystyle \text{Here, } x\in\left(\frac{1}{2\sqrt{2}},\frac{1}{2}\right)

\displaystyle \Rightarrow 2x\in\left(\frac{1}{\sqrt{2}},1\right)

\displaystyle \Rightarrow \cos\theta\in\left(\frac{1}{\sqrt{2}},1\right)

\displaystyle \Rightarrow \theta\in\left(0,\frac{\pi}{4}\right)

\displaystyle \text{So, from equation }(i),

\displaystyle u=2\theta \quad\left[\text{Since } \sin^{-1}(\sin\theta)=\theta \text{ if } \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow u=2\cos^{-1}(2x) \quad\left[\text{Since } 2x=\cos\theta\right]

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

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

\displaystyle \Rightarrow \frac{du}{dx} =\frac{-2}{\sqrt{1-4x^{2}}}\cdot 2 =\frac{-4}{\sqrt{1-4x^{2}}} \quad ...(iii)

\displaystyle \text{Differentiating equation }(ii)\text{ with respect to }x,

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

\displaystyle \Rightarrow \frac{dv}{dx} =\frac{1}{2\sqrt{1-4x^{2}}}(-8x) =\frac{-4x}{\sqrt{1-4x^{2}}} \quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \text{Answer: (iii) }

\displaystyle \text{Let } u=\sin^{-1}\!\left(4x\sqrt{1-4x^{2}}\right)

\displaystyle \text{Put } 2x=\cos\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\left(2\cos\theta\sqrt{1-\cos^{2}\theta}\right)

\displaystyle \Rightarrow u=\sin^{-1}(2\cos\theta\sin\theta)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad ...(i)

\displaystyle \text{Let } v=\sqrt{1-4x^{2}}\quad ...(ii)

\displaystyle \text{Here, } x\in\left(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\right)

\displaystyle \Rightarrow 2x\in\left(-1,-\frac{1}{\sqrt{2}}\right)

\displaystyle \Rightarrow \theta\in\left(\frac{3\pi}{4},\pi\right)

\displaystyle \text{So, from equation }(i),

\displaystyle u=\pi-2\theta \quad\left[\text{Since } \sin^{-1}(\sin\alpha)=\pi-\alpha \text{ if } \alpha\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\right]

\displaystyle \Rightarrow u=\pi-2\cos^{-1}(2x) \quad\left[\text{Since } 2x=\cos\theta\right]

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

\displaystyle \frac{du}{dx} =0-2\left(-\frac{1}{\sqrt{1-(2x)^{2}}}\right)\frac{d}{dx}(2x)

\displaystyle \Rightarrow \frac{du}{dx} =\frac{2}{\sqrt{1-4x^{2}}}\cdot 2 =\frac{4}{\sqrt{1-4x^{2}}} \quad ...(iii)

\displaystyle \text{From equation }(ii),

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

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

\displaystyle \frac{dv}{dx} =\frac{1}{2\sqrt{1-4x^{2}}}\frac{d}{dx}(1-4x^{2}) =\frac{-8x}{2\sqrt{1-4x^{2}}}

\displaystyle \Rightarrow \frac{dv}{dx} =\frac{-4x}{\sqrt{1-4x^{2}}} \quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \textbf{Question 6: }~\text{Differentiate }\tan^{-1}\!\left(\frac{\sqrt{1+x^2}-1}{x}\right)\text{ with respect to }\sin^{-1}\!\left(\frac{2x}{1+x^2}\right),\text{ if }-1<x<1,\;x\neq0.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } x=\tan\theta

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{\sqrt{1+\tan^{2}\theta}-1}{\tan\theta}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{\sec\theta-1}{\tan\theta}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{1-\cos\theta}{\sin\theta}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{2\sin^{2}\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\tan\frac{\theta}{2}\right)\quad ...(i)

\displaystyle \text{And, } v=\sin^{-1}\!\left(\frac{2x}{1+x^{2}}\right)

\displaystyle \Rightarrow v=\sin^{-1}\!\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right)

\displaystyle \Rightarrow v=\sin^{-1}(\sin 2\theta)\quad ...(ii)

\displaystyle \text{Here, } -1<x<1

\displaystyle \Rightarrow -1<\tan\theta<1

\displaystyle \Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}\quad ...(A)

\displaystyle \text{So, from equation }(i),

\displaystyle u=\frac{\theta}{2} \quad\left[\text{Since } \tan^{-1}(\tan\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow u=\frac{1}{2}\tan^{-1}x \quad\left[\text{Since } x=\tan\theta\right]

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

\displaystyle \frac{du}{dx} =\frac{1}{2}\cdot\frac{1}{1+x^{2}} =\frac{1}{2(1+x^{2})}\quad ...(iii)

\displaystyle \text{Now, from equation }(ii)\text{ and }(A),

\displaystyle v=2\theta \quad\left[\text{Since } \sin^{-1}(\sin\beta)=\beta,\ \beta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\right]

\displaystyle \Rightarrow v=2\tan^{-1}x \quad\left[\text{Since } x=\tan\theta\right]

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

\displaystyle \frac{dv}{dx} =2\cdot\frac{1}{1+x^{2}} =\frac{2}{1+x^{2}}\quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \textbf{Question 7: }~\text{Differentiate }\sin^{-1}\!\bigl(2x\sqrt{1-x^2}\bigr)\text{ with respect to }\sec^{-1}\!\left(\frac{1}{\sqrt{1-x^2}}\right),\text{ if }(i)\;x\in\left(0,\frac{1}{\sqrt2}\right)\;(ii)\;x\in\left(\frac{1}{\sqrt2},1\right).
\displaystyle \text{Answer:(i)}

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

\displaystyle \text{Put } x=\sin\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\left(2\sin\theta\sqrt{1-\sin^{2}\theta}\right)

\displaystyle \Rightarrow u=\sin^{-1}(2\sin\theta\cos\theta)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad ...(i)

\displaystyle \text{And, let } v=\sec^{-1}\!\left(\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \Rightarrow v=\sec^{-1}\!\left(\frac{1}{\sqrt{1-\sin^{2}\theta}}\right)

\displaystyle \Rightarrow v=\sec^{-1}\!\left(\frac{1}{\cos\theta}\right)

\displaystyle \Rightarrow v=\sec^{-1}(\sec\theta)

\displaystyle \Rightarrow v=\cos^{-1}\!\left(\frac{1}{\sec\theta}\right) \quad\left[\text{Since } \sec^{-1}z=\cos^{-1}\!\left(\frac{1}{z}\right)\right]

\displaystyle \Rightarrow v=\cos^{-1}(\cos\theta)\quad ...(ii)

\displaystyle \text{Here, } x\in\left(0,\frac{1}{\sqrt{2}}\right)

\displaystyle \Rightarrow \sin\theta\in\left(0,\frac{1}{\sqrt{2}}\right)

\displaystyle \Rightarrow \theta\in\left(0,\frac{\pi}{4}\right)

\displaystyle \text{So, from equation }(i),

\displaystyle u=2\theta \quad\left[\text{Since } \sin^{-1}(\sin\alpha)=\alpha \text{ if } \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow u=2\sin^{-1}x \quad\left[\text{Since } x=\sin\theta\right]

\displaystyle \text{Also, from equation }(ii),

\displaystyle v=\theta \quad\left[\text{Since } \cos^{-1}(\cos\theta)=\theta \text{ for } \theta\in(0,\pi)\right]

\displaystyle \Rightarrow v=\sin^{-1}x \quad\left[\text{Since } x=\sin\theta\right]

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

\displaystyle \frac{du}{dx} =2\cdot\frac{1}{\sqrt{1-x^{2}}} =\frac{2}{\sqrt{1-x^{2}}}\quad ...(iii)

\displaystyle \frac{dv}{dx} =\frac{1}{\sqrt{1-x^{2}}}\quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \text{Answer: (ii)}

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

\displaystyle \text{Put } x=\sin\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\left(2\sin\theta\sqrt{1-\sin^{2}\theta}\right)

\displaystyle \Rightarrow u=\sin^{-1}(2\sin\theta\cos\theta)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad ...(i)

\displaystyle \text{And, let } v=\sec^{-1}\!\left(\frac{1}{\sqrt{1-x^{2}}}\right)

\displaystyle \Rightarrow v=\sec^{-1}\!\left(\frac{1}{\sqrt{1-\sin^{2}\theta}}\right)

\displaystyle \Rightarrow v=\sec^{-1}\!\left(\frac{1}{\cos\theta}\right)

\displaystyle \Rightarrow v=\sec^{-1}(\sec\theta)

\displaystyle \Rightarrow v=\cos^{-1}\!\left(\frac{1}{\sec\theta}\right) \quad\left[\text{Since } \sec^{-1}x=\cos^{-1}\!\left(\frac{1}{x}\right)\right]

\displaystyle \Rightarrow v=\cos^{-1}(\cos\theta)\quad ...(ii)

\displaystyle \text{Here, } x\in\left(\frac{1}{\sqrt{2}},1\right)

\displaystyle \Rightarrow \sin\theta\in\left(\frac{1}{\sqrt{2}},1\right)

\displaystyle \Rightarrow \theta\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)

\displaystyle \text{So, from equation }(i),

\displaystyle u=2\theta \quad\left[\text{Since } \sin^{-1}(\sin\alpha)=\alpha \text{ if } \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow u=2\sin^{-1}x \quad\left[\text{Since } x=\sin\theta\right]

\displaystyle \text{And, from equation }(ii),

\displaystyle v=\theta \quad\left[\text{Since } \cos^{-1}(\cos\theta)=\theta \text{ if } \theta\in[0,\pi]\right]

\displaystyle \Rightarrow v=\sin^{-1}x \quad\left[\text{Since } x=\sin\theta\right]

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

\displaystyle \frac{du}{dx} =2\cdot\frac{1}{\sqrt{1-x^{2}}} =\frac{2}{\sqrt{1-x^{2}}}\quad ...(iii)

\displaystyle \frac{dv}{dx} =\frac{1}{\sqrt{1-x^{2}}}\quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \textbf{Question 8: }~\text{Differentiate }(\cos x)^{\sin x}\text{ with respect to }(\sin x)^{\cos x}.
\displaystyle \text{Answer:}

\displaystyle \text{Let, } u=(\cos x)^{\sin x}

\displaystyle \text{Taking log on both sides,}

\displaystyle \log u=\log(\cos x)^{\sin x}

\displaystyle \Rightarrow \log u=\sin x\,\log(\cos x)

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{1}{u}\frac{du}{dx} =\sin x\,\frac{d}{dx}\{\log(\cos x)\}+\log(\cos x)\,\frac{d}{dx}(\sin x) \ \ [\text{using product rule}]

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} =\sin x\left(\frac{1}{\cos x}\right)\frac{d}{dx}(\cos x)+\log(\cos x)\,(\cos x)

\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx} =\sin x\left(\frac{1}{\cos x}\right)(-\sin x)+\cos x\,\log(\cos x)

\displaystyle \Rightarrow \frac{du}{dx} =u\left[-\frac{\sin^{2}x}{\cos x}+\cos x\,\log(\cos x)\right]

\displaystyle \Rightarrow \frac{du}{dx} =(\cos x)^{\sin x}\left[\cos x\,\log(\cos x)-\sin x\,\tan x\right]\quad ...(i)

\displaystyle \text{Let, } v=(\sin x)^{\cos x}

\displaystyle \text{Taking log on both sides,}

\displaystyle \log v=\log(\sin x)^{\cos x}

\displaystyle \Rightarrow \log v=\cos x\,\log(\sin x)

\displaystyle \text{Differentiating with respect to }x\text{ using chain rule,}

\displaystyle \frac{1}{v}\frac{dv}{dx} =\cos x\,\frac{d}{dx}\{\log(\sin x)\}+\log(\sin x)\,\frac{d}{dx}(\cos x) \ \ [\text{using product rule}]

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} =\cos x\left(\frac{1}{\sin x}\right)\frac{d}{dx}(\sin x)+\log(\sin x)(-\sin x)

\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx} =\cos x\left(\frac{1}{\sin x}\right)(\cos x)-\sin x\,\log(\sin x)

\displaystyle \Rightarrow \frac{dv}{dx} =v\left[\frac{\cos^{2}x}{\sin x}-\sin x\,\log(\sin x)\right]

\displaystyle \Rightarrow \frac{dv}{dx} =(\sin x)^{\cos x}\left[\cot x\,\cos x-\sin x\,\log(\sin x)\right]\quad ...(ii)

\displaystyle \text{Dividing equation }(i)\text{ by }(ii),

\displaystyle \frac{du}{dv} =\frac{(\cos x)^{\sin x}\left[\cos x\,\log(\cos x)-\sin x\,\tan x\right]} {(\sin x)^{\cos x}\left[\cot x\,\cos x-\sin x\,\log(\sin x)\right]}

\displaystyle \textbf{Question 9: }~\text{Differentiate }\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\text{ with respect to }\cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right),\text{ if } \\ 0<x<1.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } x=\tan\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad ...(i)

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

\displaystyle \Rightarrow v=\cos^{-1}\!\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)

\displaystyle \Rightarrow v=\cos^{-1}(\cos 2\theta)\quad ...(ii)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow 0<\tan\theta<1

\displaystyle \Rightarrow 0<\theta<\frac{\pi}{4}

\displaystyle \text{So, from equation }(i),

\displaystyle u=2\theta \quad\left[\text{Since } \sin^{-1}(\sin\alpha)=\alpha,\ \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow u=2\tan^{-1}x \quad\left[\text{Since } x=\tan\theta\right]

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

\displaystyle \frac{du}{dx} =\frac{2}{1+x^{2}}\quad ...(iii)

\displaystyle \text{From equation }(ii),

\displaystyle v=2\theta \quad\left[\text{Since } \cos^{-1}(\cos\beta)=\beta,\ \beta\in[0,\pi]\right]

\displaystyle \Rightarrow v=2\tan^{-1}x \quad\left[\text{Since } x=\tan\theta\right]

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

\displaystyle \frac{dv}{dx} =\frac{2}{1+x^{2}}\quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

\displaystyle \frac{du}{dv} =\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}} =1

\displaystyle \textbf{Question 10: }~\text{Differentiate }\tan^{-1}\!\left(\frac{1+ax}{1-ax}\right)\text{ with respect to }\sqrt{1+a^2x^2}.
\displaystyle \text{Answer:}

\displaystyle \text{Let, } u=\tan^{-1}\!\left(\frac{1+ax}{1-ax}\right)

\displaystyle \text{Put } ax=\tan\theta

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{1+\tan\theta}{1-\tan\theta}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left[\tan\left(\frac{\pi}{4}+\theta\right)\right]

\displaystyle \Rightarrow u=\frac{\pi}{4}+\theta

\displaystyle \Rightarrow u=\frac{\pi}{4}+\tan^{-1}(ax)\quad [\text{Since } \tan\theta=ax]

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

\displaystyle \frac{du}{dx}=0+\frac{1}{1+(ax)^{2}}\frac{d}{dx}(ax)

\displaystyle \Rightarrow \frac{du}{dx}=\frac{a}{1+a^{2}x^{2}}\quad ...(i)

\displaystyle \text{Now, let } v=\sqrt{1+a^{2}x^{2}}

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

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

\displaystyle \Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{1+a^{2}x^{2}}}(2a^{2}x)

\displaystyle \Rightarrow \frac{dv}{dx}=\frac{a^{2}x}{\sqrt{1+a^{2}x^{2}}}\quad ...(ii)

\displaystyle \text{Dividing equation }(i)\text{ by }(ii),

\displaystyle \frac{du}{dv} =\frac{\frac{a}{1+a^{2}x^{2}}}{\frac{a^{2}x}{\sqrt{1+a^{2}x^{2}}}} =\frac{a}{1+a^{2}x^{2}}\cdot\frac{\sqrt{1+a^{2}x^{2}}}{a^{2}x}

\displaystyle \Rightarrow \frac{du}{dv} =\frac{1}{ax\sqrt{1+a^{2}x^{2}}}

\displaystyle \textbf{Question 11: }~\text{Differentiate }\sin^{-1}\!\bigl(2x\sqrt{1-x^2}\bigr)\text{ with respect to }\tan^{-1}\!\left(\frac{x}{\sqrt{1-x^2}}\right),\text{ if }-\frac{1}{\sqrt2}<x<\frac{1}{\sqrt2}.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } x=\sin\theta

\displaystyle \Rightarrow u=\sin^{-1}\!\left(2\sin\theta\sqrt{1-\sin^{2}\theta}\right)

\displaystyle \Rightarrow u=\sin^{-1}(2\sin\theta\cos\theta)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad ...(i)

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

\displaystyle \Rightarrow v=\tan^{-1}\!\left(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\right)

\displaystyle \Rightarrow v=\tan^{-1}\!\left(\frac{\sin\theta}{\cos\theta}\right)

\displaystyle \Rightarrow v=\tan^{-1}(\tan\theta)\quad ...(ii)

\displaystyle \text{Here, } -\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}

\displaystyle \Rightarrow -\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}

\displaystyle \Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}

\displaystyle \text{So, from equation }(i),

\displaystyle u=2\theta \quad\left[\text{Since } \sin^{-1}(\sin\alpha)=\alpha \text{ if } \alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow u=2\sin^{-1}x \quad\left[\text{Since } x=\sin\theta\right]

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

\displaystyle \frac{du}{dx} =2\cdot\frac{1}{\sqrt{1-x^{2}}} =\frac{2}{\sqrt{1-x^{2}}}\quad ...(iii)

\displaystyle \text{From equation }(ii),

\displaystyle v=\theta \quad\left[\text{Since } \tan^{-1}(\tan\theta)=\theta \text{ if } \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow v=\sin^{-1}x \quad\left[\text{Since } x=\sin\theta\right]

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

\displaystyle \frac{dv}{dx} =\frac{1}{\sqrt{1-x^{2}}}\quad ...(iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

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

\displaystyle \textbf{Question 12: }~\text{Differentiate }\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)\text{ with respect to }\cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right),\text{ if }0<x<1.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } x = \tan\theta

\displaystyle \Rightarrow u = \tan^{-1}\!\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)

\displaystyle \Rightarrow u = \tan^{-1}(\tan 2\theta)\ \ldots (i)

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

\displaystyle \Rightarrow v = \cos^{-1}\!\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)

\displaystyle \Rightarrow v = \cos^{-1}(\cos 2\theta)\ \ldots (ii)

\displaystyle \text{Here, } 0<x<1

\displaystyle \Rightarrow 0<\tan\theta<1

\displaystyle \Rightarrow 0<\theta<\frac{\pi}{4}

\displaystyle \text{So, from equation }(i),

\displaystyle u = 2\theta \qquad \left[\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right]

\displaystyle \Rightarrow u = 2\tan^{-1}x

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

\displaystyle \frac{du}{dx}=\frac{2}{1+x^2}\ \ldots (iii)

\displaystyle \text{From equation }(ii),

\displaystyle v = 2\theta \qquad \left[\text{Since } \cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi]\right]

\displaystyle \Rightarrow v = 2\tan^{-1}x

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

\displaystyle \frac{dv}{dx}=\frac{2}{1+x^2}\ \ldots (iv)

\displaystyle \text{Dividing equation }(iii)\text{ by }(iv),

\displaystyle \frac{du}{dv}=\frac{2}{1+x^2}\times\frac{1+x^2}{2}

\displaystyle \therefore\ \frac{du}{dv}=1

\displaystyle \textbf{Question 13: }~\text{Differentiate }\tan^{-1}\!\left(\frac{x-1}{x+1}\right)\text{ with respect to }\sin^{-1}(3x-4x^3),\text{ if }-\frac12<x<\frac12.
\displaystyle \text{Answer:}

\displaystyle \text{Let } u=\tan^{-1}\!\left(\frac{x-1}{x+1}\right)

\displaystyle \text{Put } x=\tan\theta

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{\tan\theta-1}{\tan\theta+1}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{\tan\theta-\tan\frac{\pi}{4}}{1+\tan\theta\tan\frac{\pi}{4}}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left[\tan\!\left(\theta-\frac{\pi}{4}\right)\right]\qquad\text{(i)}

\displaystyle \text{Here, }-\frac12<x<\frac12

\displaystyle \Rightarrow -\frac12<\tan\theta<\frac12

\displaystyle \Rightarrow -\tan^{-1}\!\left(\frac12\right)<\theta<\tan^{-1}\!\left(\frac12\right)

\displaystyle \text{So, from (i), } u=\theta-\frac{\pi}{4}\qquad\text{since }\tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

\displaystyle \Rightarrow u=\tan^{-1}x-\frac{\pi}{4}

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

\displaystyle \frac{du}{dx}=\frac{1}{1+x^{2}}\qquad\text{(ii)}

\displaystyle \text{And, let } v=\sin^{-1}(3x-4x^{3})

\displaystyle \text{Put } x=\sin\theta

\displaystyle \Rightarrow v=\sin^{-1}(3\sin\theta-4\sin^{3}\theta)

\displaystyle \Rightarrow v=\sin^{-1}(\sin3\theta)\qquad\text{(iii)}

\displaystyle \text{Now, }-\frac12<x<\frac12

\displaystyle \Rightarrow -\frac12<\sin\theta<\frac12

\displaystyle \Rightarrow -\frac{\pi}{6}<\theta<\frac{\pi}{6}

\displaystyle \text{So, from (iii), } v=3\theta\qquad\text{since }\sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

\displaystyle \Rightarrow v=3\sin^{-1}x

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

\displaystyle \frac{dv}{dx}=\frac{3}{\sqrt{1-x^{2}}}\qquad\text{(iv)}

\displaystyle \text{Dividing (ii) by (iv),}

\displaystyle \frac{du}{dv}=\frac{\frac{1}{1+x^{2}}}{\frac{3}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{3(1+x^{2})}

\displaystyle \textbf{Question 14: }~\text{Differentiate }\tan^{-1}\!\left(\frac{\cos x}{1+\sin x}\right)\text{ with respect to }\sec^{-1}x.
\displaystyle \text{Answer:}

\displaystyle \text{Let } u=\tan^{-1}\!\left(\frac{\cos x}{1+\sin x}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left[\tan\!\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]

\displaystyle \Rightarrow u=\frac{\pi}{4}-\frac{x}{2}

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

\displaystyle \frac{du}{dx}=0-\frac{1}{2}=-\frac{1}{2}\qquad\text{(i)}

\displaystyle \text{Let } v=\sec^{-1}x

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

\displaystyle \frac{dv}{dx}=\frac{1}{x\sqrt{x^{2}-1}}\qquad\text{(ii)}

\displaystyle \text{Dividing (i) by (ii),}

\displaystyle \frac{du}{dv}=\frac{-\tfrac12}{\tfrac{1}{x\sqrt{x^{2}-1}}}=-\frac{x\sqrt{x^{2}-1}}{2}

\displaystyle \textbf{Question 15: }~\text{Differentiate }\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\text{ with respect to }\tan^{-1}\!\left(\frac{2x}{1-x^2}\right),\text{ if }-1<x<1.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } x=\tan\theta \Rightarrow \theta=\tan^{-1}x

\displaystyle \Rightarrow u=\sin^{-1}\!\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right)

\displaystyle \Rightarrow u=\sin^{-1}(\sin 2\theta)\quad\text{...(i)}

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

\displaystyle \Rightarrow v=\tan^{-1}\!\left(\frac{2\tan\theta}{1-\tan^{2}\theta}\right)

\displaystyle \Rightarrow v=\tan^{-1}(\tan 2\theta)\quad\text{...(ii)}

\displaystyle \text{Here, } -1<x<1

\displaystyle \Rightarrow -1<\tan\theta<1

\displaystyle \Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}

\displaystyle \text{So, from equation (i),}

\displaystyle u=2\theta \qquad [\text{Since } \sin^{-1}(\sin\theta)=\theta,\ \theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})]

\displaystyle \Rightarrow u=2\tan^{-1}x

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

\displaystyle \frac{du}{dx}=\frac{2}{1+x^{2}}\quad\text{...(iii)}

\displaystyle \text{From equation (ii),}

\displaystyle v=2\theta \qquad [\text{Since } \tan^{-1}(\tan\theta)=\theta,\ \theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})]

\displaystyle \Rightarrow v=2\tan^{-1}x

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

\displaystyle \frac{dv}{dx}=\frac{2}{1+x^{2}}\quad\text{...(iv)}

\displaystyle \text{Dividing equation (iii) by (iv),}

\displaystyle \frac{du/dx}{dv/dx}=\frac{2}{1+x^{2}}\times\frac{1+x^{2}}{2}

\displaystyle \therefore \frac{du}{dv}=1

\displaystyle \textbf{Question 16: }~\text{Differentiate }\cos^{-1}(4x^3-3x)\text{ with respect to }\tan^{-1}\!\left(\frac{\sqrt{1-x^2}}{x}\right),\text{ if }\frac12<x<1.
\displaystyle \text{Answer:}

\displaystyle \text{Let } u=\cos^{-1}(4x^3-3x)

\displaystyle \text{Put } x=\cos\theta

\displaystyle \Rightarrow \theta=\cos^{-1}x

\displaystyle \text{Now, } u=\cos^{-1}(4\cos^3\theta-3\cos\theta)

\displaystyle \Rightarrow u=\cos^{-1}(\cos 3\theta)\quad\ldots(i)

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

\displaystyle \Rightarrow v=\tan^{-1}\!\left(\frac{\sqrt{1-\cos^2\theta}}{\cos\theta}\right)

\displaystyle \Rightarrow v=\tan^{-1}\!\left(\frac{\sin\theta}{\cos\theta}\right)

\displaystyle \Rightarrow v=\tan^{-1}(\tan\theta)\quad\ldots(ii)

\displaystyle \text{Here, } \frac12<x<1

\displaystyle \Rightarrow \frac12<\cos\theta<1

\displaystyle \Rightarrow 0<\theta<\frac{\pi}{3}

\displaystyle \text{So, from equation (i),}

\displaystyle u=3\theta\quad\text{[Since, }\cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi]]

\displaystyle \Rightarrow u=3\cos^{-1}x

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

\displaystyle \frac{du}{dx}=\frac{-3}{\sqrt{1-x^2}}\quad\ldots(iii)

\displaystyle \text{From equation (ii),}

\displaystyle v=\theta\quad\text{[Since, }\tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)]

\displaystyle \Rightarrow v=\cos^{-1}x

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

\displaystyle \frac{dv}{dx}=\frac{-1}{\sqrt{1-x^2}}\quad\ldots(iv)

\displaystyle \text{Dividing equation (iii) by (iv),}

\displaystyle \frac{du/dx}{dv/dx}=\left(\frac{-3}{\sqrt{1-x^2}}\right)\left(\frac{\sqrt{1-x^2}}{-1}\right)

\displaystyle \Rightarrow \frac{du}{dv}=3

\displaystyle \textbf{Question 17: }~\text{Differentiate }\tan^{-1}\!\left(\frac{x}{\sqrt{1-x^2}}\right)\text{ with respect to }\sin^{-1}\!\bigl(2x\sqrt{1-x^2}\bigr),\text{ if }-\frac{1}{\sqrt2}<x<\frac{1}{\sqrt2}.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } x=\sin\theta \;\Rightarrow\; \theta=\sin^{-1}x

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}\right)

\displaystyle \Rightarrow u=\tan^{-1}\!\left(\frac{\sin\theta}{\cos\theta}\right)

\displaystyle \Rightarrow u=\tan^{-1}(\tan\theta)\qquad\text{(i)}

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

\displaystyle \Rightarrow v=\sin^{-1}\!\left(2\sin\theta\sqrt{1-\sin^{2}\theta}\right)

\displaystyle \Rightarrow v=\sin^{-1}(2\sin\theta\cos\theta)

\displaystyle \Rightarrow v=\sin^{-1}(\sin2\theta)\qquad\text{(ii)}

\displaystyle \text{Here, } -\frac{1}{\sqrt2}<x<\frac{1}{\sqrt2}

\displaystyle \Rightarrow -\frac{1}{\sqrt2}<\sin\theta<\frac{1}{\sqrt2}

\displaystyle \Rightarrow -\frac{\pi}{4}<\theta<\frac{\pi}{4}

\displaystyle \text{So, from (i), } u=\theta \quad\big[\text{since } \tan^{-1}(\tan\theta)=\theta,\; \theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\big]

\displaystyle \Rightarrow u=\sin^{-1}x

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

\displaystyle \frac{du}{dx}=\frac{1}{\sqrt{1-x^{2}}}\qquad\text{(iii)}

\displaystyle \text{From (ii), } v=2\theta\quad\big[\text{since } \sin^{-1}(\sin\theta)=\theta,\; \theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\big]

\displaystyle \Rightarrow v=2\sin^{-1}x

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

\displaystyle \frac{dv}{dx}=\frac{2}{\sqrt{1-x^{2}}}\qquad\text{(iv)}

\displaystyle \text{Dividing (iii) by (iv),}

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

\displaystyle \textbf{Question 18: }~\text{Differentiate }\sin^{-1}\!\sqrt{1-x^2}\text{ with respect to }\cot^{-1}\!\left(\frac{x}{\sqrt{1-x^2}}\right),\text{ if }0<x<1.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } x=\cos\theta

\displaystyle \Rightarrow \theta=\cos^{-1}x

\displaystyle \text{We get } u=\sin^{-1}(\sin\theta)\qquad\text{(i)}

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

\displaystyle \Rightarrow v=\cot^{-1}\!\left(\frac{\cos\theta}{\sqrt{1-\cos^{2}\theta}}\right)

\displaystyle \Rightarrow v=\cot^{-1}\!\left(\frac{\cos\theta}{\sin\theta}\right)

\displaystyle \Rightarrow v=\cot^{-1}(\cot\theta)\qquad\text{(ii)}

\displaystyle \text{Here, }0<x<1

\displaystyle \Rightarrow 0<\cos\theta<1

\displaystyle \Rightarrow 0<\theta<\frac{\pi}{2}

\displaystyle \text{So, from (i),}

\displaystyle u=\theta \qquad\left[\text{since }\sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\right]

\displaystyle \Rightarrow u=\cos^{-1}x

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

\displaystyle \frac{du}{dx}=-\frac{1}{\sqrt{1-x^{2}}}\qquad\text{(iii)}

\displaystyle \text{From (ii),}

\displaystyle v=\theta \qquad\left[\text{since }\cot^{-1}(\cot\theta)=\theta,\ \theta\in(0,\pi)\right]

\displaystyle \Rightarrow v=\cos^{-1}x

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

\displaystyle \frac{dv}{dx}=-\frac{1}{\sqrt{1-x^{2}}}\qquad\text{(iv)}

\displaystyle \text{Dividing (iii) by (iv),}

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

\displaystyle \textbf{Question 19: }~\text{Differentiate }\sin^{-1}\!\bigl(2ax\sqrt{1-a^2x^2}\bigr)\text{ with respect to }\sqrt{1-a^2x^2},\text{ if }-\frac{1}{\sqrt2}<ax<\frac{1}{\sqrt2}.
\displaystyle \text{Answer:}

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

\displaystyle \text{Put } ax = \sin\theta \Rightarrow \theta = \sin^{-1}(ax)

\displaystyle \Rightarrow u = \sin^{-1}\!\left(2\sin\theta\sqrt{1-\sin^{2}\theta}\right)

\displaystyle \Rightarrow u = \sin^{-1}(2\sin\theta\cos\theta)

\displaystyle \Rightarrow u = \sin^{-1}(\sin 2\theta) \qquad \text{(i)}

\displaystyle \text{And let } v = \sqrt{1-a^{2}x^{2}}

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

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

\displaystyle \Rightarrow \frac{dv}{dx} = \frac{-2a^{2}x}{2\sqrt{1-a^{2}x^{2}}}

\displaystyle \Rightarrow \frac{dv}{dx} = -\frac{a^{2}x}{\sqrt{1-a^{2}x^{2}}} \qquad \text{(ii)}

\displaystyle \text{Here, } -\frac{1}{\sqrt{2}} < ax < \frac{1}{\sqrt{2}}

\displaystyle \Rightarrow -\frac{1}{\sqrt{2}} < \sin\theta < \frac{1}{\sqrt{2}}

\displaystyle \Rightarrow -\frac{\pi}{4} < \theta < \frac{\pi}{4}

\displaystyle \text{So from (i), } u = 2\theta \quad \text{since } \sin^{-1}(\sin y)=y,\; y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

\displaystyle \Rightarrow u = 2\sin^{-1}(ax)

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

\displaystyle \frac{du}{dx} = 2\cdot\frac{1}{\sqrt{1-(ax)^{2}}}\cdot\frac{d}{dx}(ax)

\displaystyle \Rightarrow \frac{du}{dx} = \frac{2a}{\sqrt{1-a^{2}x^{2}}} \qquad \text{(iii)}

\displaystyle \text{Dividing (iii) by (ii),}

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

\displaystyle \Rightarrow \frac{du}{dv} = -\frac{2}{ax}

\displaystyle \textbf{Question 20: }~\text{Differentiate }\tan^{-1}\!\left(\frac{1-x}{1+x}\right)\text{ with respect to }\sqrt{1-x^2},\text{ if }-1<x<1.
\displaystyle \text{Answer:}

\displaystyle \text{Let } u = \tan^{-1}\!\left(\frac{1-x}{1+x}\right)

\displaystyle \text{Put } x = \tan\theta

\displaystyle \Rightarrow \theta = \tan^{-1}x

\displaystyle \Rightarrow u = \tan^{-1}\!\left(\frac{1-\tan\theta}{1+\tan\theta}\right)

\displaystyle \Rightarrow u = \tan^{-1}\!\left[\tan\!\left(\frac{\pi}{4}-\theta\right)\right] \qquad \text{(i)}

\displaystyle \text{Here, } -1 < x < 1

\displaystyle \Rightarrow -1 < \tan\theta < 1

\displaystyle \Rightarrow -\frac{\pi}{4} < \theta < \frac{\pi}{4}

\displaystyle \Rightarrow 0 < \frac{\pi}{4}-\theta < \frac{\pi}{2}

\displaystyle \text{So from (i), } u = \frac{\pi}{4}-\theta \quad \text{since } \tan^{-1}(\tan y)=y,\; y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

\displaystyle \Rightarrow u = \frac{\pi}{4}-\tan^{-1}x

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

\displaystyle \frac{du}{dx} = 0-\frac{1}{1+x^{2}}

\displaystyle \Rightarrow \frac{du}{dx} = -\frac{1}{1+x^{2}} \qquad \text{(ii)}

\displaystyle \text{And let } v = \sqrt{1-x^{2}}

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

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

\displaystyle \Rightarrow \frac{dv}{dx} = \frac{1}{2\sqrt{1-x^{2}}}(-2x)

\displaystyle \Rightarrow \frac{dv}{dx} = -\frac{x}{\sqrt{1-x^{2}}} \qquad \text{(iii)}

\displaystyle \text{Dividing (ii) by (iii),}

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

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

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