If y = x^ x ,then dy dx = - Vidyadip

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
${{2 + \log x} \over {2\sqrt x }}$
D
None of these

✓

Answer

Correct option: A.

${x^{\sqrt x }}{{2 + \log x} \over {2\sqrt x }}$

a
(a) $y = {x^{\sqrt x }} \Rightarrow {\log _e}y = \sqrt x \log x$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \sqrt x \frac{1}{x} + \frac{1}{{2\sqrt x }}\log x$ or

$\frac{{dy}}{{dx}} = {x^{\sqrt x }}\left[ {\frac{{2 + {{\log }_e}x}}{{2\sqrt x }}} \right]$

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