Q. $10^{- x\tan x}\left\lbrack \frac{d}{dx}(10^{x\tan x}) \right\rbrack$ is equal to
A) $\tan{x + x\sec^{2}x}$
B) In 10 $(\tan{x + x\sec^{2}{x)}}$
C) In 10 $\left( \tan{x + \frac{x}{\cos^{2}x} + \tan{x\sec x}} \right)$
D) $x\tan xIn\ 10$
Explanation
(b)
$10^{- x\tan x}\frac{d}{dx}(10^{x\tan x})$
$= 10^{- x\tan\ x}10^{x\tan\ x}\log{10(\tan{x + x\sec^{2}{x)}}}$
$= \log{10(\tan{x + x\sec^{2}{x)}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.