Find d y d x if y = x ^ x +(7 x -1)^ x - Vidyadip

Question

Find $\frac{d y}{d x}$ if $y = x ^{ x }+(7 x -1)^{ x }$

✓

Answer

$
y=x^x+(7 x-1)^x
$
Let $u=x^x$ and $v=(7 x-1)^x$
Then $y=u+v$
$
\therefore \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}
$
Take $u=x^x$
$
\therefore \log u=\log x^x=x \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 \log x) \\
& =x \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}
$
Also, $v=(7 x-1)^x$
$
\begin{aligned}
& \therefore \log v=\log (7 x-1)^x=x \log (7 x-1) \\
& \begin{aligned}
\frac{1}{v} \cdot \frac{d v}{d x} & =\frac{d}{d x}[x \log (7 x-1)] \\
& =x \frac{d}{d x}[\log (7 x-1)]+[\log (7 x-1)] \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{7 x-1} \cdot \frac{d}{d x}(7 x-1)+[\log (7 x-1)] \times 1
\end{aligned}
\end{aligned}
$
$
=\frac{x}{7 x-1} \times(7 \times 1-0)+\log (7 x-1)
$
$
\begin{aligned}
\therefore \frac{d v}{d x} & =v\left[\frac{7 x}{7 x-1}+\log (7 x-1)\right] \\
& =(7 x-1)^x\left[\frac{7 x}{7 x-1}+\log (7 x-1)\right]
\end{aligned}
$
From (1), (2) and (3), we get
$
\frac{d y}{d x}=x^x(1+\log x)+(7 x-1)^x\left[\frac{7 x}{7 x-1}+\log (7 x-1)\right] .
$

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