Question

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

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

Answer 1

Answer: (A)

$y=\sec \left({\tan }^{-1}x\right)$
$=\sec \left({\sec }^{-1}\sqrt{1+x^2}\right)$
$=\sqrt{1+x^2}$
On differentiating both sides w.r.t. $X$, we get
$\frac{dy}{dx}=\frac{1}{2\sqrt{1+x^2}}(2x)$
$\Rightarrow \frac{dy}{dx}=\frac{x}{\sqrt{1+x^2}}$