Find d y d x if, : y=x^x+a^x

Question

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

✓

Answer

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

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