#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.8 question 4 sub question (ii) maths

Answer: $-1$

Hint: $\text { Let } u=\sin ^{-1} \sqrt{1-x^{2}}, v=\cos ^{-1} x$

$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

Given:    $\sin ^{-1} \sqrt{1-x^{2}} \text { w.r.t } \cos ^{-1} x, x \in(-1,0)$

Explanation:

$\text { Let } u=\sin ^{-1} \sqrt{1-x^{2}}, v=\cos ^{-1} x$

\begin{aligned} &u=\sin ^{-1} \sqrt{1-x^{2}} \\\\ &\text { Let } x=\cos \theta \\\\ &u=\sin ^{-1} \sqrt{1-\cos ^{2} \theta} \end{aligned}

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

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

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

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