MCQ
If $y = {x^{\sqrt x }},$then ${{dy} \over {dx}} =$
- ✓${x^{\sqrt x }}{{2 + \log x} \over {2\sqrt x }}$
- B${x^{\sqrt x }}{{2 + \log x} \over {\sqrt x }}$
- C${{2 + \log x} \over {2\sqrt x }}$
- DNone of these
==> $\frac{1}{y}\frac{{dy}}{{dx}} = \sqrt x \frac{1}{x} + \frac{1}{{2\sqrt x }}\log x$ or
$\frac{{dy}}{{dx}} = {x^{\sqrt x }}\left[ {\frac{{2 + {{\log }_e}x}}{{2\sqrt x }}} \right]$
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