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

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