Let y=xxTake “log” on both sides logy=xlogx diff w.r.to x on both sides 1ydydx=x(1x)+logx(1) dydx=xx(1+logx)

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$\frac{d}{d x} (x^{x}) =$

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$x . x^{x - 1}$

$x^{x}$

$x^{x} \log x$

$x^{x} (1 + \log_{e} x)$

answer is B.

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Detailed Solution

Let $y = x^{x}$
Take “log” on both sides
$\log y = x \log x d i f f w . r . t o x o n b o t h s i d e s \frac{1}{y} \frac{d y}{d x} = x (\frac{1}{x}) + \log x (1) \frac{d y}{d x} = x^{x} (1 + \log x)$

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