Differentiate the function with respect to $x$: $(\log x)^{x}+x^{\log x}$

Let $y=(\log x)^{x}+x^{\log x}$.
Let $u=(\log x)^{x}$ and $v=x^{\log x}$.
Then $y=u+v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ...........$(1)$
For $u=(\log x)^{x}$,taking log on both sides:
$\log u = x \log(\log x)$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log(\log x) + \frac{1}{\log x}$.
$\frac{du}{dx} = (\log x)^{x} \left[ \log(\log x) + \frac{1}{\log x} \right] = (\log x)^{x-1} [1 + \log x \cdot \log(\log x)]$ ...........$(2)$
For $v=x^{\log x}$,taking log on both sides:
$\log v = \log x \cdot \log x = (\log x)^{2}$.
Differentiating with respect to $x$:
$\frac{1}{v} \frac{dv}{dx} = 2 \log x \cdot \frac{1}{x}$.
$\frac{dv}{dx} = x^{\log x} \cdot \frac{2 \log x}{x} = 2 x^{\log x-1} \log x$ ...........$(3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$\frac{dy}{dx} = (\log x)^{x-1} [1 + \log x \cdot \log(\log x)] + 2 x^{\log x-1} \log x$.