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

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

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

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

Explanation:

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

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

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

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