#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 15

Hint:  $\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

Given:   $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \text { w.r.t } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

$-1

Explanation:

\begin{aligned} &\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \\\\ &v=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\\\ &\text { Let } x=\tan \theta \end{aligned}

\begin{aligned} &u=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right), v=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right) \\\\ &u=\sin ^{-1}(\sin 2 \theta), v=\tan ^{-1}(\tan 2 \theta) \end{aligned}

\begin{aligned} &-1

\begin{aligned} &-\frac{\pi}{4}<\theta<\frac{\pi}{4} \\\\ &-\frac{\pi}{2}<2 \theta<\frac{\pi}{2} \end{aligned}

\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta)=2 \theta \\\\ &v=\tan ^{-1}(\tan 2 \theta)=2 \theta \\\\ &u=2 \tan ^{-1} x \end{aligned}

\begin{aligned} &\frac{d u}{d x}=\frac{2}{1+x^{2}} \\\\ &v=2 \tan ^{-1} x \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{2}{1+x^{2}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}=1 \end{aligned}