d dx ^ - 1 x = - Vidyadip

MCQ

${d \over {dx}}{\cos ^{ - 1}}\sqrt {\cos x} = $

✓
${1 \over 2}\sqrt {1 + \sec x} $
B
$\sqrt {1 + \sec x} $
C
$ - {1 \over 2}\sqrt {1 + \sec x} $
D
$ - \sqrt {1 + \sec x} $

✓

Answer

Correct option: A.

${1 \over 2}\sqrt {1 + \sec x} $

a
(a) $\frac{d}{{dx}}{\cos ^{ - 1}}\sqrt {\cos x} = \frac{{\sin x}}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }}$

$ = \frac{{\sqrt {1 - {{\cos }^2}x} }}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }} = \frac{1}{2}\sqrt {\frac{{1 + \cos x}}{{\cos x}}} $.

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