Differentiate the function with respect to x.

Solution:

Let $y=(\log x)^{x}+x^{\log x}$

Also, let $u=(\log x)^{x}$ and $v=x^{\log x}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$   ...(1)

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

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

$\Rightarrow \log u=x \log (\log x)$

Differentiating both sides with respect to x, we obtain

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

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

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

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

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

$\Rightarrow \frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]$    ...(2)

$v=x^{\log x}$

$\Rightarrow \log v=\log \left(x^{\log x}\right)$

$\Rightarrow \log v=\log x \log x=(\log x)^{2}$

Differentiating both sides with respect to x, we obtain

$\frac{1}{v} \cdot \frac{d v}{d x}=\frac{d}{d x}\left[(\log x)^{2}\right]$

$\Rightarrow \frac{1}{v} \cdot \frac{d v}{d x}=2(\log x) \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d v}{d x}=2 v(\log x) \cdot \frac{1}{x}$

$\Rightarrow \frac{d v}{d x}=2 x^{\log x} \frac{\log x}{x}$

$\Rightarrow \frac{d v}{d x}=2 x^{\log x-1} \cdot \log x$   ...(3)

Therefore, from (1), (2), and (3), we obtain

$\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 x^{\log x-1} \cdot \log x$