Find d y d x if, : y=x^ x^ 2 x - Vidyadip

Question

Find $\frac{d y}{d x}$ if, :
$y=x^{x^{2 x}}$

✓

Answer

$
\begin{aligned}
& y=x^{x^{2 x}} \\
& \therefore \log y=\log x^{x^{2 x}}=x^{2 x} \cdot \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}\left(x^{2 x} \cdot \log x\right) \\
& =x^{2 x} \cdot \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right) \\
& =x^{2 x} \times \frac{1}{x}+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right) \\
\therefore \frac{d y}{d x} & =y\left[\frac{x^{2 x}}{x}+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right)\right] \\
& =x^{x^{2 x}}\left[\frac{x^{2 x}}{x}+(\log x) \cdot \frac{d}{d x}\left(x^{2 x}\right)\right]
\end{aligned}
$
Let $u=x^{2 x}$
Then $\log u=\log x^{2 x}=2 x \log x$ Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{u} \cdot \frac{d u}{d x} & =2 \frac{d}{d x}(x \log x) \\
& =2\left[x \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x)\right] \\
& =2\left[x \times \frac{1}{x}+(\log x) \times 1\right]
\end{aligned}
$
$
\begin{aligned}
& \therefore \frac{d u}{d x}=2 u(1+\log x) \\
& \therefore \frac{d}{d x}\left(x^{2 x}\right)=2 x^{2 x}(1+\log x)
\end{aligned}
$
$\therefore$ from (1),
$
\begin{aligned}
\frac{d y}{d x} & =x^{2^{2 x}}\left[\frac{x^{2 x}}{x}+(\log x) \times 2 x^{2 x}(1+\log x)\right] \\
& =x^{2^{2 x}} \cdot x^{2 x} \cdot \log x\left[2(1+\log x)+\frac{1}{x \log x}\right]
\end{aligned}
$

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