MCQ
If $y = {\tan ^{ - 1}}\left( {{x \over {\sqrt {1 - {x^2}} }}} \right)$, then ${{dy} \over {dx}} = $
- A$ - {1 \over {\sqrt {1 - {x^2}} }}$
- B${x \over {\sqrt {1 - {x^2}} }}$
- ✓${1 \over {\sqrt {1 - {x^2}} }}$
- D${{\sqrt {1 - {x^2}} } \over x}$
Put $x = \sin \theta $,
$\therefore $ $dx = \cos \theta d\theta $, $\frac{{d\theta }}{{dx}} = \frac{1}{{\cos \theta }}$
$\therefore y = {\tan ^{ - 1}}\left( {\frac{{\sin \theta }}{{\cos \theta }}} \right)$$ \Rightarrow $$y = \theta $
$\therefore $ $dy = d\theta $
Now $\frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}.\frac{{d\theta }}{{dx}}$ = $1.\frac{1}{{\cos \theta }} = \sec \theta = \frac{1}{{\sqrt {1 - {x^2}} }}$.
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