#### Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 7 sub question (i)

Hint:    $\text { Let } u=\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right]$

$v=\sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right]$

Given: $\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right] \text { w.r.t } \sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right]$

$x \in\left(0, \frac{1}{\sqrt{2}}\right)$

Explanation:

$\text { Let } x=\sin \theta$

\begin{aligned} &u=\sin ^{-1}\left[2 x \sqrt{1-x^{2}}\right] \\ &v=\sec ^{-1}\left[\frac{1}{\sqrt{1-x^{2}}}\right] \end{aligned}

$u=\sin ^{-1}\left[2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right]$

\begin{aligned} &=\sin ^{-1}[2 \sin \theta \cos \theta] \\\\ &=\sin ^{-1}[\sin 2 \theta] \end{aligned}

\begin{aligned} &x \in\left(0, \frac{1}{\sqrt{2}}\right) \\\\ &\sin \theta \in\left(0, \frac{1}{\sqrt{2}}\right) \\\\ &\theta \in\left(0, \frac{\pi}{4}\right) \quad 2 \theta \in\left(0, \frac{\pi}{2}\right) \end{aligned}

\begin{aligned} &u=\sin ^{-1}[\sin 2 \theta]=2 \theta \quad \text { when } 2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &u=2 \sin ^{-1} x \end{aligned}

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

\begin{aligned} &v=\sec ^{-1}\left[\frac{1}{\sqrt{1-\sin ^{2} \theta}}\right] \\\\ &\sec ^{-1}\left(\frac{1}{\cos \theta}\right)=\sec ^{-1}(\sec \theta) \\\\ &=\theta \quad \text { when } \theta \in\left(0, \frac{\pi}{4}\right) \end{aligned}

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

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