Question

If $y=sec\left(ta{n}^{-1}x\right)$, then $\frac{dy}{dx}$ is.

• A. $\frac{x}{\sqrt{1+x^2}}$
• B. $\frac{-x}{\sqrt{1+x^2}}$
• C. $\frac{x}{\sqrt{1-x^2}}$
• D. None of these

Answer 1

Answer: (A)

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

Given, $y=sec\left({\tan }^{-1}x\right)$
On differentiating w.r.t. $x,$ we get
$\frac{dy}{dx}=sec\left({\tan }^{-1}x\right)\cdot \tan \left({\tan }^{-1}x\right)\frac{1}{1+x^2}$
$=\frac{x}{1+x^2}\sqrt{1+x^2}$
$[\because {\tan }^{-1}x=se{c}^{-1}(\sqrt{1+x^2}\left)\right]$
$=\frac{x}{\sqrt{1+x^2}}$