Explain solution for RD Sharma class class 12 chapter 10 Differentiation exercise 10.3 question 4 math textbook solution.

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

Hint:

$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} ; \frac{d}{d x} (constant) = 0$
Given:

$\sin ^{-1}\left\{\sqrt{1-\mathrm{x}^{2}}\right\}, 0<\mathrm{x}<1$

Solution:

$\begin {array}{l} Let \ \ x=\cos \theta\\\\ \theta=\cos ^{-1} x\\\\ y=\sin ^{-1}\left\{\sqrt{ \left.1-\cos ^{2} \theta\right\}}\right.\\\\ \therefore \cos ^{2} \theta+\sin ^{2} \theta=1\\\\ y=\sin ^{-1}\left\{\sqrt{\sin ^{2} \theta}\right\}\\\\ \mathrm{y}=\sin ^{-1}\{\sin \theta\} \end{}$

Considering the limits,

$\begin {array}{l} 0<\cos \theta<1 \\\\ 0<\theta<\frac{\pi}{2} \ \ \ \ \ \ \ \ \ \ \left\{\cos \frac{1}{2}=1\right\}\\\\ Now, y=\sin ^{-1}(\sin \theta)\\\\ \mathrm{y}=\theta\\\\ \left\{\sin ^{-1}(\sin \theta)=\theta\right\}\\\\ if \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \\\\ y=\cos ^{-1} x \end{}$
differentiating with respects to x, we get

$\begin {array}{l} \frac{d y}{d x}=\frac{\mathrm{d} (\cos^{-1}x)}{\mathrm{d} x}\\\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right) \Rightarrow \frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\\\\ \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{-1}{\sqrt{1-\mathrm{x}^{2}}} \end{}$