MCQ
જો $y = {x^x}$, તો ${{dy} \over {dx}} = $
- ✓${x^x}\log ex$
- B${x^x}\left( {1 + {1 \over x}} \right)$
- C$(1 + \log x)$
- D${x^x}\log x$
Taking $\log $ on both sides, ==> $\log y = x\log x$
Differentiating with respect to $x,$ we get
==> $\frac{1}{y}\frac{{dy}}{{dx}} = 1 + \log x$;
$\therefore \frac{{dy}}{{dx}} = {x^x}(1 + \log x) = {x^x}\log ex$.
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