Differentiate the function with respect to $x$: $x^{x}+x^{a}+a^{x}+a^{a}$,for some fixed $a > 0$ and $x > 0$.

(N/A) Let $y = x^{x} + x^{a} + a^{x} + a^{a}$.
Let $u = x^{x}$,$v = x^{a}$,$w = a^{x}$,and $s = a^{a}$.
Then $y = u + v + w + s$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} + \frac{ds}{dx} \dots (1)$.
For $u = x^{x}$,taking log on both sides: $\log u = x \log x$. Differentiating w.r.t $x$: $\frac{1}{u} \frac{du}{dx} = \log x + x(\frac{1}{x}) = \log x + 1$. Thus,$\frac{du}{dx} = x^{x}(1 + \log x) \dots (2)$.
For $v = x^{a}$,using the power rule: $\frac{dv}{dx} = a x^{a-1} \dots (3)$.
For $w = a^{x}$,using the exponential derivative rule: $\frac{dw}{dx} = a^{x} \log a \dots (4)$.
For $s = a^{a}$,since $a$ is a constant,$s$ is a constant,so $\frac{ds}{dx} = 0 \dots (5)$.
Substituting $(2), (3), (4),$ and $(5)$ into $(1)$,we get $\frac{dy}{dx} = x^{x}(1 + \log x) + a x^{a-1} + a^{x} \log a$.