Differentiate the function with respect to $x$: $(x \cos x)^{x} + (x \sin x)^{\frac{1}{x}}$

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