#### Please solve RD Sharma class 12 chapter 10 Differentiation exercise 10.3 question 5 maths textbook solution.

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

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

$\tan ^{-1}\left\{\frac{x}{\sqrt{a^{2}-x^{2}}}\right\},-a

Solution:

$Let \ \ y=\tan ^{-1}\left\{\frac{x}{\sqrt{a^{2}-x^{2}}}\right\}\\\\$

$\operatorname{let} \mathrm{x}=\operatorname{asin} \theta\\\\$

$Now,\\\\$

$\mathrm{y}=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{\sqrt{a^{2}-a^{2} \\\\\sin ^{2} \theta}}\right\}$

$\begin{array}{l}Using \ \ \sin ^{2} \theta+\cos ^{2} \theta=1\\\\ y=\tan ^{-1}\left\{\frac{\operatorname{asin} \theta}{a \sqrt{1-\sin ^{2} \theta}}\right\}\\\\ y=\tan ^{-1}\frac{a\sin \theta}{a\cos \theta} .\\\\ \mathrm{y}=\tan ^{-1}\{\tan \theta\}\\\\ Using \ \ \frac{\sin \theta}{\cos \theta}=\tan \theta \end{array}$
Considering the limits,

$\begin{array}{l} -a

Differentiating with Respects to x, we get

$\begin{array}{l} \frac{d y}{d x}=\frac{d}{\partial x}\left(\sin ^{-1}\left(\frac{x}{a}\right)\right) \\\\ \therefore \frac{\partial}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \\\\ \frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}} \times \frac{1}{a} \\\\ \end{array}$

$\begin{array}{l} \frac{d y}{d x}=\frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{a}{\sqrt{a^{2}-x^{2}}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{1}{\sqrt{a^{2}-x^{2}}} \end{array}$