MCQ
${d \over {dx}}\,\,\left[ {{{\tan }^{ - 1}}\left( {{{\sqrt x (3 - x)} \over {1 - 3x}}} \right)} \right] =$
- A${1 \over {2(1 + x)\,\sqrt x }}$
- B${3 \over {(1 + x)\,\sqrt x }}$
- C${2 \over {(1 + x)\,\sqrt x }}$
- ✓$\frac{3}{{2(1 + x)\sqrt x }}$
Put $\sqrt x = \tan \theta \Rightarrow \theta = {\tan ^{ - 1}}\sqrt x $
$\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{{(\tan \theta (3 - {{\tan }^2}\theta )}}{{1 - 3{{\tan }^2}\theta }}} \right)$
$= \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{{(3\tan \theta - {{\tan }^3}\theta )}}{{1 - 3{{\tan }^2}\theta }}} \right)$
$= \frac{d}{{dx}}({\tan ^{ - 1}}(\tan 3\theta ) = \frac{d}{{dx}}(3\theta )$
$= \frac{d}{{dx}}(3.{\tan ^{ - 1}}\sqrt x ) = \frac{3}{{2\sqrt x (1 + x)}}$.
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