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

Answer

$
\begin{aligned}
& y=x^x \\
& \therefore \log y=\log x^x=x \log x
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}(x \log x) \\
& =x \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{x}+(\log x) \times 1 \\
& \therefore \frac{d y}{d x}=y(1+\log x) \\
& =x^x(1+\log x) . \\
\end{aligned}
$

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