(a) Differentiate tan^-1 (√(1-x^2))/x w. r. t. cos^-1(2 x √(1-x^2)), x ∈(1/(√ 2), 1)

Put $\displaystyle x=\cos \theta \Rightarrow \theta=\cos ^{-1} x$ Let $\displaystyle u=\tan ^{-1} \frac{\sqrt{1-x^{2}}}{x}=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta)=\theta=\cos ^{-1} x$ \[\Rightarrow \frac{d u}{d x}=-\frac{1}{\sqrt{1-x^{2}}} \] Let $\displaystyle v=\cos ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=\cos ^{-1}(\sin 2 \theta)=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)=\frac{\pi}{2}-2 \cos ^{-1} x$ \[\Rightarrow \frac{d v}{d x}=\frac{2}{\sqrt{1-x^{2}}} \] $\displaystyle \therefore \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=-\frac{1}{2}$ Let $\displaystyle \mathrm{y}=u+v \Rightarrow \frac{d \mathrm{y}}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$, where $\displaystyle u=x^{\tan x}, v=\frac{\sqrt{x^{2}+1}}{2}$ $\displaystyle u=x^{\boldsymbol{\operatorname { t a n }} x} \Rightarrow \boldsymbol{\operatorname { l o g }} u=\boldsymbol{\operatorname { t a n }} x \boldsymbol{\operatorname { l o g }} x$, differentiating with respect to ' $\displaystyle x$ ', we get $\displaystyle \Rightarrow \frac{\mathbf{1}}{\boldsymbol{u}} \frac{\boldsymbol{d} \boldsymbol{u}}{\boldsymbol{d} \boldsymbol{x}}=\frac{\boldsymbol{\operatorname { t a n }} \boldsymbol{x}}{\boldsymbol{x}}+\boldsymbol{\operatorname { s e c }}^{\mathbf{2}} \boldsymbol{x} \boldsymbol{\operatorname { l o g }} \boldsymbol{x} \Rightarrow \frac{d u}{d x}=u\left(\frac{\tan x}{x}+\sec ^{2} x \log x\right)=x^{\tan x}\left(\frac{\tan x}{x}+\sec ^{2} x \log x\right)$ \[v=\frac{\sqrt{x^{2}+1}}{2} \Rightarrow \frac{d v}{d x}=\frac{2 x}{4 \sqrt{x^{2}+1}}=\frac{x}{2 \sqrt{x^{2}+1}} \] $\displaystyle \Rightarrow \frac{d \mathrm{y}}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\sec ^{2} x \log x\right)+\frac{x}{2 \sqrt{x^{2}+1}}$