MCQ
જો $y = {\sin ^{ - 1}}\sqrt {1 - {x^2}} $, તો $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} }}$
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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