d dx ^ - 1 1 + x^2 2 = - Vidyadip

MCQ

${d \over {dx}}{\cos ^{ - 1}}\sqrt {{{1 + {x^2}} \over 2}} = $

A
${{ - 1} \over {2\sqrt {1 - {x^4}} }}$
B
${1 \over {2\sqrt {1 - {x^4}} }}$
✓
${{ - x} \over {\sqrt {1 - {x^4}} }}$
D
${x \over {\sqrt {1 - {x^4}} }}$

✓

Answer

Correct option: C.

${{ - x} \over {\sqrt {1 - {x^4}} }}$

c
(c) Putting ${x^2} = \cos 2\theta $, we have

$\frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\sqrt {\frac{{1 + {x^2}}}{2}} } \right] = \frac{d}{{dx}}[{\cos ^{ - 1}}\cos \theta ]$

$ = \frac{d}{{dx}}[\theta ] = \frac{d}{{dx}}\left[ {\frac{1}{2}{{\cos }^{ - 1}}{x^2}} \right] = \frac{{ - x}}{{\sqrt {1 - {x^4}} }}$.

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