MCQ
If $y = {x^2} + {x^{\log x}},$ then ${{dy} \over {dx}} = $
- A${{{x^2} + \log x.{x^{\log x}}} \over x}$
- B${x^2} + \log x.{x^{\log x}}$
- ✓${{2({x^2} + \log x.{x^{\log x}})} \over x}$
- DNone of these
✓
Answer
Correct option: C.
${{2({x^2} + \log x.{x^{\log x}})} \over x}$
c
(c) $y = {x^2} + {x^{\log x}}$
(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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