10^ - x , x [ d dx ( 10 ^ x x ) ] is equal to - Vidyadip

MCQ

${10^{ - x\,\tan x}}\left[ {{d \over {dx}}({{10}^{x\tan x}})} \right]$ is equal to

A
$\tan x\, + x\,\,{\sec ^2}x$
✓
$\ln \,10\,(\tan x + x{\sec ^2}x)$
C
$\ln \,10\,\left( {\tan x + {x \over {{{\cos }^2}x}} + \tan x\sec x} \right)$
D
$x\tan x\,{\rm{ln}}\,\,10$

✓

Answer

Correct option: B.

$\ln \,10\,(\tan x + x{\sec ^2}x)$

b
(b) ${10^{ - x\tan x}}\frac{d}{{dx}}({10^{x\tan x}})$

$= {10^{ - x\tan x}}{.10^{x\tan x}}.\log 10(\tan x + x{\sec ^2}x)$

$= \log 10(\tan x + x{\sec ^2}x)$.

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