#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 17 maths textbook solution

Answer: $\frac{1}{2}$

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

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

$-\frac{1}{\sqrt{2}}

Explanation:

$u=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)$

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

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

$\begin{array}{ll} u=\tan ^{-1}(\tan \theta)=\theta & \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \\\\ v=\sin ^{-1}(\sin 2 \theta)=2 \theta & 2 \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{array}$

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

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

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

$\frac{d u}{d v}=\frac{1}{2}$