Differentiate

Solution:

Let $u=(\log x)^{x}$ and $v=\log x$

We need to differentiate $u$ with respect to $v$ that is find $\frac{\text { du }}{\text { dv }}$.

We have $u=(\log x)^{x}$

Taking log on both sides, we get

$\log u=\log (\log x)^{x}$

$\Rightarrow \log u=x \times \log (\log x)\left[\because \log a^{m}=m \times \log a\right]$

On differentiating both the sides with respect to $x$, we get

$\frac{\mathrm{d}}{\mathrm{du}}(\log \mathrm{u}) \times \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \times \log (\log \mathrm{x})]$

Recall that (uv) $^{\prime}=v u^{\prime}+u v^{\prime}$ (product rule)

$\Rightarrow \frac{\mathrm{d}}{\mathrm{du}}(\log \mathrm{u}) \times \frac{\mathrm{du}}{\mathrm{dx}}=\log (\log \mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}[\log (\log \mathrm{x})]$

We know $\frac{d}{d x}(\log x)=\frac{1}{x}$ and $\frac{d}{d x}(x)=1$

$\Rightarrow \frac{1}{u} \times \frac{d u}{d x}=\log (\log x) \times 1+x\left[\frac{1}{\log x} \frac{d}{d x}(\log x)\right]$

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=\log (\log x)+\frac{x}{\log x} \frac{d}{d x}(\log x)$

But, $u=(\log x)^{x}$ and $\frac{d}{d x}(\log x)=\frac{1}{x}$

$\Rightarrow \frac{1}{(\log x)^{x}} \frac{d u}{d x}=\log (\log x)+\frac{x}{\log x} \times \frac{1}{x}$

$\Rightarrow \frac{1}{(\log x)^{x}} \frac{d u}{d x}=\log (\log x)+\frac{1}{\log x}$

$\therefore \frac{\mathrm{du}}{\mathrm{dx}}=(\log \mathrm{x})^{\mathrm{x}}\left[\log (\log \mathrm{x})+\frac{1}{\log \mathrm{x}}\right]$

Now, on differentiating $v$ with respect to $x$, we get

$\frac{d v}{d x}=\frac{d}{d x}(\log x)$

$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{\mathrm{x}}$

We have $\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dv}}}{\frac{\mathrm{dx}}{\mathrm{dx}}}$

$\Rightarrow \frac{d u}{d v}=\frac{(\log x)^{x}\left[\log (\log x)+\frac{1}{\log x}\right]}{\frac{1}{x}}$

$\Rightarrow \frac{d u}{d v}=x(\log x)^{x}\left[\log (\log x)+\frac{1}{\log x}\right]$

$\Rightarrow \frac{d u}{d v}=x(\log x)^{x}\left[\frac{\log (\log x) \log x+1}{\log x}\right]$

$\Rightarrow \frac{d u}{d v}=\frac{x(\log x)^{x}}{\log x}[\log (\log x) \log x+1]$

$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\mathrm{x}(\log \mathrm{x})^{\mathrm{x}-1}[1+\log \mathrm{x} \log (\log \mathrm{x})]$

Thus, $\frac{d u}{d v}=x(\log x)^{x-1}[1+\log x \log (\log x)]$