d dx ( x + x) = - Vidyadip

MCQ

${d \over {dx}}\{ \log (\sec x + \tan x)\} = $

A
$\cos x$
✓
$\sec x$
C
$\tan x$
D
$\cot x$

✓

Answer

Correct option: B.

$\sec x$

b
(b) $\frac{d}{{dx}}\{ \log (\sec x + \tan x)\} = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}} = \sec x$.

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