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

MCQ

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

✓
${x^x}\log ex$
B
${x^x}\left( {1 + {1 \over x}} \right)$
C
$(1 + \log x)$
D
${x^x}\log x$

✓

Answer

Correct option: A.

${x^x}\log ex$

a
(a) $y = {x^x}$

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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