Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 13 maths textbook solution

Answer:   $\frac{\sqrt{1-x^{2}}}{3\left(1+x^{2}\right)}$

Hint:   $\text { Let } u=\tan ^{-1}\left(\frac{x-1}{x+1}\right), v=\sin ^{-1}\left(3 x-4 x^{3}\right)$

Given:  $\tan ^{-1}\left(\frac{x-1}{x+1}\right) \text { w.r.t } \sin ^{-1}\left(3 x-4 x^{3}\right)$

$\frac{-1}{2}

Explanation:  $\text { Let } u=\tan ^{-1}\left(\frac{x-1}{x+1}\right)$

\begin{aligned} &v=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ &u=\tan ^{-1}\left(\frac{x-1}{x+1}\right) \\\\ &=\tan ^{-1} x-\tan ^{-1} 1 \\\\ &=\tan ^{-1} x-\frac{\pi}{4} \end{aligned}

\begin{aligned} &\frac{d u}{d x}=\frac{1}{1+x^{2}} \\\\ &v=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ &\text { Let } x=\sin \theta \end{aligned}

\begin{aligned} &\frac{-1}{2}

\begin{aligned} &v=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right) \\\\ &=\sin ^{-1}(\sin 3 \theta) \\\\ &=3 \theta \\\\ &v=3 \sin ^{-1} x \end{aligned}

\begin{aligned} &\frac{d v}{d x}=\frac{3}{\sqrt{1-x^{2}}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{1}{1+x^{2}}}{\frac{3}{\sqrt{1-x^{2}}}}=\frac{\sqrt{1-x^{2}}}{3\left(1+x^{2}\right)} \end{aligned}