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} }}$
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
$\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{{dy}}{{dx}} + \frac{1}{x}\sin 2y = {x^3}\,{\cos ^2}\,y$ represented by family of curves which is is givey by