= Solution
For $x\ne0$, $f(x)=|x|^x=e^{x\log|x|}$, so $f'(x)=|x|^x(\log|x|+1)$. Although $x\log|x|\to0$ makes $f$ continuous at zero, $(f(x)-1)/x\sim\log|x|\to-\infty$, so it is not differentiable there.
Since the <cosine function> is even, $g(x)=\cos|x|=\cos x$, so it is differentiable everywhere with $g'(x)=-\sin x$. Finally, $h(x)=x|x|$ is differentiable everywhere, including at zero, and $h'(x)=2|x|$. Therefore
$$
\boxed{f'(x)=|x|^x(\log|x|+1)\ (x\ne0),\qquad g'(x)=-\sin x,\qquad h'(x)=2|x|.}
$$
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