For constant \(a, \frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)\) is

\(\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)\)
\(\begin{aligned}
&=\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(x^{a}\right)+\frac{d}{d x}\left(a^{x}\right)+\frac{d}{d x}\left(a^{a}\right) \\
&=\frac{d}{d x}(y)+a x^{a-1}+a^{x} \log a+0...(i)
\end{aligned}\)
\(\left[\operatorname{let} x^{x}=y\right]\)
\(y=x^{x}\)
Taking log on both sides,
\(\log y=x \log x\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned}
\frac{1}{y} \frac{d y}{d x} &=x \times \frac{1}{x}+\log x \\
\Rightarrow \quad \quad \frac{d y}{d x} &=x^{x}(1+\log x)
\end{aligned}\)
Substitute the value of \(\frac{d y}{d x}\) in Eq. (i),
\(\begin{aligned}
&\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right) \\
&=x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a
\end{aligned}\)