If y = ^ - 1 1 - x^2 , then dy/dx = - Vidyadip

MCQ

If $y = {\sin ^{ - 1}}\sqrt {1 - {x^2}} $, then $dy/dx = $

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

✓

Answer

Correct option: C.

$ - {1 \over {\sqrt {1 - {x^2}} }}$

c
(c) $y = {\sin ^{ - 1}}(\sqrt {1 - {x^2}} )$

Let $\sqrt {1 - {x^2}} = \sin \theta \Rightarrow 1 - {x^2} = {\sin ^2}\theta $

==> ${x^2} = 1 - {\sin ^2}\theta = {\cos ^2}\theta $

$\therefore x = \cos \theta $ or $\theta = {\cos ^{ - 1}}x$

$ \Rightarrow y = {\cos ^{ - 1}}x$

Differentiating w.r.t. $x $ of $y,$ we get

$\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}$.

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