Question

If $y={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right),$ $-1<x<1$ then $\frac{dy}{dx}$ is equal to-

• A. $-\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{x}{\sqrt{1-x^2}}$
• C. $\frac{1}{\sqrt{1-x^2}}$
• D. $\frac{\sqrt{1-x^2}}{x}$

Answer 1

The correct option is C $\frac{1}{\sqrt{1-x^2}}$

$y={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{\frac{d}{dx}\left(\frac{x}{\sqrt{1-x^2}}\right)}{1+\frac{x^2}{{1-x}^2}}$
$=\frac{x\frac{d}{dx}\left(\frac{1}{\sqrt{1+x^2}}\right)+1\left(\frac{1}{\sqrt{1-x^2}}\right)}{1+\frac{x^2}{1-x^2}}$
$=\frac{\frac{1}{\sqrt{1-x^2}}+2x\frac{x}{2{(1-x^2)}^{\frac{3}{2}}}}{1+\frac{x^2}{{1-x}^2}}=\frac{1}{\sqrt{1-x^2}}$