Find d y d x if, : y=e^ x^x - Vidyadip

Question

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

✓

Answer

$
\begin{aligned}
& y=e^{x^x} \\
& \therefore \log y=\log e^{x^x}=x^x \log e \\
& \therefore \log y=x^x \quad \ldots[\because \log e=1]
\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}\left(x^x\right) \\
& \therefore \frac{d y}{d x}=y \cdot \frac{d}{d x}\left(x^x\right)=e^{x^x} \cdot \frac{d}{d x}\left(x^x\right)
\end{aligned}
$
Let $u=x^x $
Then $\log u=\log x^x=x \log x$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \begin{aligned}
\frac{1}{u} \cdot \frac{d u}{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
\end{aligned} \\
& \therefore \frac{d u}{d x}=u(1+\log x) \\
& \therefore \frac{d}{d x}\left(x^x\right)=x^x(1+\log x) \\
& \therefore \text { from }(1), \\
& \frac{d y}{d x}=e^{x^x} \cdot x^x(1+\log x) .
\end{aligned}
$

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