MCQ
${d \over {dx}}{\tan ^{ - 1}}{x \over {\sqrt {{a^2} - {x^2}} }} = $
- A${a \over {{a^2} + {x^2}}}$
- B${{ - a} \over {{a^2} + {x^2}}}$
- C${1 \over {a\sqrt {{a^2} - {x^2}} }}$
- ✓${1 \over {\sqrt {{a^2} - {x^2}} }}$
Putting $x = a\sin \theta ,$ we get
$ = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\frac{{a\sin \theta }}{{a\cos \theta }}} \right] = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\tan \theta } \right] = \frac{d}{{dx}}[\theta ]$
Substituting value of $\theta $, so
$= \frac{d}{dx}\left[ {{\sin }^{-1}}\tan \left( \frac{x}{a} \right) \right]=\frac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}.$
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