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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$(A)$ There exist $r , s \in R$, where $r < s$, such that $f$ is one-one on the open interval $( r , s )$
$(B)$ There exists $x 0 \in(-4,0)$ such that $\left| f ^{\prime}\left( x _0\right)\right| \leq 1$
$(C)$ $\lim _{x \rightarrow \infty} f(x)=1$
$(D)$ There exists a $\in(-4,4)$ such that $f(a)+f^{\prime \prime}(a)=0$ and $f^{\prime}(a) \neq 0$