Question:
If $y=x^{x}$, find $\frac{d y}{d x}$ at $x=e$.
Solution:
$y=x^{x}$
Taking logarithm on both sides,
$\log y=x \log x$
Differentiating w.r.t. $x$ on both sides,
$\frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{x}+1 \cdot \log x$
$=1+\log x$
$\Rightarrow \frac{d y}{d x}=y(1+\log x)$
$=x^{x}(1+\log x)$
So, at $x=e$,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{e}}(1+\log \mathrm{e})$
$=e^{e}(1+1)$
$=2 e^{e}($ Ans $)$
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