#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 11

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

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

Explanation:

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

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

\begin{aligned} &=\sin ^{-1}(2 \cos \theta \sin \theta) \\\\ &=\sin ^{-1}(\sin 2 \theta) \end{aligned}

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

Now,

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

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

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

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

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