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