# Get Answers to all your Questions

#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 64 maths

Hint: you must know the rules of derivative of inverse trigonometric functions.

Given:  $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}$

Prove   $\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{y}{x}$

Solution:

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

Differentiate with respect to x,

$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\right)$                [   quotient rule    ]

$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\sqrt{1-x^{2}} \frac{d}{d x}\left(x \sin ^{-1} x\right)-\left(x \sin ^{-1} x\right) \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)}{\left(\sqrt{1-x^{2}}\right)^{2}}\right] . \cdot \frac{d}{d x} u \cdot v=$$\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

$\frac{d y}{d x}=\left[\frac{\sqrt{1-x^{2}}\left\{x \frac{d}{d x}\left(\sin ^{-1} x\right)+\left(\sin ^{-1} x\right) \frac{d}{d x}(x)\right\}-\left(x \sin ^{-1} x\right)\left(\frac{1}{2 \sqrt{1-x^{2}}}\right) \frac{d\left(1-x^{2}\right)}{d x}}{\left(\sqrt{1-x^{2}}\right)^{2}}\right]$

$\frac{d y}{d x}=\left[\frac{\sqrt{1-x^{2}}\left\{\frac{x}{\sqrt{1-x^{2}}}+\sin ^{-1} x\right\}-\frac{x \sin ^{-1} x(-2 x)}{2\left(\sqrt{1-x^{2}}\right)}}{\left(\sqrt{1-x^{2}}\right)^{2}}\right]$

$\frac{d y}{d x}=\left[\frac{x+\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}}{\left(1-x^{2}\right)}\right]$

$\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)}{1}+\frac{x^{2} \sin ^{-1} x}{\sqrt{1-x^{2}}}$

$\left(1-x^{2}\right) \frac{d y}{d x}=x+\left(\frac{\sin ^{-1} x-x^{2}\left(\sin ^{-1} x\right)+x^{2}\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}}\right)$

$\left(1-x^{2}\right) \frac{d y}{d x}=x+\left(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\right)$

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

$\left(1-x^{2}\right) \frac{d y}{d x}=x+\frac{y}{x}$

∴ Proved