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Let $f: \mathbb{R}^n \to (0, \infty)$ and $g: \mathbb{R}^n \to \mathbb{R}$ be totally differentiable functions. Prove that $$f(x)^{g(x)}$$ is also totally differentiable.

I wanted to use the chain rule. I let $h:= f^g$. I wanted to write $h$ as a composition, but I'm not sure of what. I think I will need the function $k: \mathbb{R} \to (0, \infty) : x \mapsto x^x$. But how can I write $h$ as a suitable composition of differentiable functions?

What is the domain and codomain of the function $f^g$?

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$f^{g(x)}=exp(g(x)log(f(x))$.

I assume $f>0$, $log(f(x))$ and $g(x)log(f(x))$ are differentiable since the composition and the product of differentiable functions is differentiable. $exp(g(x)log(x))$ is differentiable since the composition of differentiable functions is differentiable.

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Hint: note that $$a^b = e^{b \log a}$$

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    Ah, ok I found it with that relation. But what is the domain and codomain of $f^g$? Is it a function from $(0, \infty)$ to $\mathbb{R}$? – 2017-01-28
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    $f(x)^g(x)$ is defined precisely when the following two conditions hold: $f(x)$ and $g(x)$ are defined, and $f(x)$ and $g(x)$ are not both zero. That will tell you the domain. The smallest codomain is not easy to predict in general, I think. – 2017-01-28