Find d y d x if, : y = ( x)^x + x^ x - Vidyadip

Question

Find $\frac{d y}{d x}$ if, :
$y = (\log x)^x + x^{\log x}$

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Answer

$y = (\log x)^x + x^{\log x}$
Let $u=(\log x)^x$ and $v=x^{\log x}$
Then $y=u+v$
$\therefore \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
Take $u=(\log x)^x \quad$
$\therefore \log u=\log (\log x)^x=x \log (\log x)$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{u} \cdot \frac{d u}{d x} =\frac{d}{d x}[x \log (\log x)]$
$=x \cdot \frac{d}{d x}[\log (\log x)]+[\log (\log x)] \cdot \frac{d}{d x}(x)$
$=x \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)+[\log (\log x)] \times 1$
$=x \times \frac{1}{\log x} \times \frac{1}{x}+\log (\log x)$
\therefore \frac{d u}{d x} $=u\left[\frac{1}{\log x}+\log (\log x)\right]$
$=(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]$
Also, $v=x^{\log x}$
$\therefore \log v=\log x^{\log x}=\log x \cdot \log x=(\log x)^2$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{v} \cdot \frac{d v}{d x} =\frac{d}{d x}(\log x)^2$
$=2(\log x) \cdot \frac{d}{d x}(\log x)$
$=2 \log x \times \frac{1}{x}$
$\therefore \frac{d v}{d x} =v\left[\frac{2 \log x}{x}\right]$
$=x^{\log x\left[\frac{2 \log x}{x}\right]}$
From (1), (2) and (3), we get
$\frac{d y}{d x}=(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]+x^{\log x}\left[\frac{2 \log x}{x}\right]$

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