#### Need solution for R.D. Sharma class 12 chapter 10  Differentiation exercise 10.3 question 7 math textbook solution.

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

$\frac{\mathrm{d}}{\mathrm{dx}}( Constant )=0 ; \frac{\partial}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-\mathbf{1}}$
Given:

$\sin ^{-1}\left(2 x^{2}-1\right), 0

Solution:

$\begin {array}{l} Let,\\ y=\sin ^{-1}\left(2 x^{2}-1\right)\\\\ Let \\ \mathrm{x}=\cos \theta\\\\ Now,\\ y=\sin ^{-1}\left\{2 \cos ^{2} \theta-1\right\}\\\\ Using\\\\ 2 \cos ^{2} \theta-1=\cos 2 \theta\\\\ y=\sin ^{-1}(\cos 2 \theta)\\\\ y=\sin ^{-1}\left\{\sin \left(\frac{n}{2}-2 \theta\right)\right\} \end{}$

As we know,

$\sin \left(\frac{n}{2}-\theta\right)=\cos \theta$
Considering the limits, $0<\mathrm{x}<1$

$\begin {array}{l} 0<\cos \theta<1\\\\ 0<\theta<\frac{n}{2} \ \ \ \ \ \ \left\{\cos \frac{\pi}{2}=0, \cos 0=1\right\}\\\\ 0>-2 \theta>-\pi\\\\ \frac{n}{2}>\frac{n}{2}-2 \theta>-\frac{\pi}{2}\\\\ \text { Now, } \mathrm{y}=\sin ^{-1}\left\{\sin \left(\frac{\mathrm{n}-20}{2}\right)\right\}\\\\ \mathrm{y}=\frac{\mathrm{n}}{2}-2 \theta \ \ \ \ \ \ \ \ \ \left\{\sin ^{-1}(\sin \theta)=\theta\right\} \text { if } \theta \in\left[-\frac{n}{2}, \frac{n}{2}\right]\\\\ \therefore \mathrm{x}=\cos \theta, \theta=\cos ^{-1} \mathrm{x}\\\\ y=\frac{n}{2}-2 \cos ^{-1} x \end{}$

Differentiating with respect ts? x, weget

$\frac{d y}{d x}=\frac{d}{\partial x}\left(\frac{n}{2}-2 \cos ^{-1} x\right)$
As we know

$\begin {array}{l} \frac{d}{d x}(constant)=0, \frac{\partial}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}}\\\\ \frac{\mathrm{dy}}{\mathrm{dx}}=0-2\left(-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)\\\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\sqrt{1-\mathrm{x}^{2}}} \end{}$