Question

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

• 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

$\frac{1}{\sqrt{1-x^2}}$

Explanation for the correct option.

Find the value of $\frac{dy}{dx}$.

Let $x=\sin \theta$, then $\theta ={\sin }^{-1}x$

In the equation $y={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$, substitute $\sin \theta$ for $x$.

$\begin{array}{rcl}y& =& {\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{1-{\left(\sin \theta \right)}^2}}\right)\\ & =& {\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{{\cos }^2\theta }}\right)\left[\because {\sin }^{2}A+{\cos }^{2}A=1\right]\\ & =& {\tan }^{-1}\left(\frac{\sin \theta }{\cos \theta }\right)\\ & =& {\tan }^{-1}\left(\tan \theta \right)\\ & =& \theta \end{array}$

But $\theta ={\sin }^{-1}x$, so the value of $y={\sin }^{-1}x$.

Now differentiate it with both sides with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left({\sin }^{-1}x\right)\\ & =& \frac{1}{\sqrt{1-x^2}}\end{array}$

Hence, the correct option is C.