MCQ
જો $y = \log \tan \sqrt x $ તો ${{dy} \over {dx}}$ = . . . ..
- A${1 \over {2\sqrt x }}$
- B${{{{\sec }^2}\sqrt x } \over {\sqrt x \tan x}}$
- C$2{\sec ^2}\sqrt x $
- ✓${{{{\sec }^2}\sqrt x } \over {2\sqrt x \tan \sqrt x }}$
$\frac{{dy}}{{dx}} = \frac{1}{{\tan \sqrt x }}.{\sec ^2}\sqrt x .\frac{1}{{2\sqrt x }} = \frac{{{{\sec }^2}\sqrt x }}{{2\sqrt x \tan \sqrt x }}$.
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