Find $\frac{d y}{d x}$ if $y = x^{x}$

x^{x} (l n x + 1)

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x^{x} (l n x - 1)

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x . x^{x - 1}

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x^{x - 1} (l n x + 1)

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Solution

The correct option is A $x^{x} (l n x + 1)$
$y = x^{x}$
Take log both sides,
$log y = x log x$
Now differentiate both side w.r.t $x$
$\Rightarrow \frac{1}{y} \frac{d y}{d x} = 1 + log x$
$\Rightarrow \frac{d y}{d x} = x^{x} (1 + log x)$

Suggest Corrections

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