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

Answer: $-2$

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

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

Explanation:

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

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

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

\begin{aligned} &\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}

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