If y = x^2 + x^ x , then dy dx = - Vidyadip

MCQ

If $y = {x^2} + {x^{\log x}},$ then ${{dy} \over {dx}} = $

A
${{{x^2} + \log x.{x^{\log x}}} \over x}$
B
${x^2} + \log x.{x^{\log x}}$
✓
${{2({x^2} + \log x.{x^{\log x}})} \over x}$
D
None of these

✓

Answer

Correct option: C.

${{2({x^2} + \log x.{x^{\log x}})} \over x}$

c
(c) $y = {x^2} + {x^{\log x}}$

==> $\frac{{dy}}{{dx}} = 2x + {x^{\log x}}\left( {2{{\log }_e}x.\frac{1}{x}} \right)$

$ = \frac{{2({x^2} + {x^{\log x}}{{\log }_e}x)}}{x}$.

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