Q. If $y = \sec{(\tan^{- 1}{x),}}$then $\frac{dy}{dx}$ is equal to
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
Explanation
(a)
Given, $y = \sec{(\tan^{- 1}{x)}}$
$\Rightarrow \ y = \sec{(\sec^{- 1}{\sqrt{1 + x^{2}}) = \sqrt{1 + x^{2}}}}$
On differentiating w.r.t. $x,$we get
$\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^{2}}}(2x) = \frac{x}{\sqrt{1 + x^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.