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Let $A \subseteq \mathbb{R}$ and $f:A \to (0, \infty), \: g:A \to \mathbb{R}$ be two differentiable functions. Prove that $f^g:A \to \mathbb{R}$ is also differentiable, using limits.

Let $x_0 \in A$ be arbitrary. In other words, we have to prove that there is a finite number $$l=\lim_{x\to x_0}\frac{f(x)^{g(x)}-f(x_0)^{g(x_0)}}{x-x_0}$$ I tried to get a common factor ouside the limit: $$l=f(x_0)^{g(x_0)}\lim_{x \to x_0}\frac{\frac{f(x)^{g(x)}}{f(x_0)^{g(x_0)}}-1}{x-x_0}$$ but I'm not quite sure how to continue.

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3 Answers 3

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You are proceeding in right direction. The next step is to write $f^{g} /f_{0}^{g_{0}}$ in terms of exponential and logarithmic functions as $$\exp\log\frac{f(x) ^{g(x)}} {f(x_{0})^{g(x_{0})}} =\exp(g(x)\log f(x) - g(x_{0})\log f(x_{0}))$$ and note that the argument of $\exp$ function tends to $0$ as $x\to x_{0}$. Denoting this complicated argument by $t$ we can see that the derivative is given by $$\lim_{x\to x_{0}}f(x_{0})^{g(x_{0})}\cdot\frac{\exp(t)-1}{t}\cdot\frac{t}{x-x_{0}}$$ and this is same as $$f(x_{0})^{g(x_{0})}\lim_{x\to x_{0}}\frac{g(x)\log f(x) - g(x_{0})\log f(x_{0})}{x-x_{0}}$$ I hope you can take it from here.

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  • $\begingroup$ Yes! Thank you very much, this is brilliant! $\endgroup$
    – Shroud
    Commented Aug 23, 2017 at 8:44
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The composition of differentiable functions is differentiable (chain rule), as is the product of differentiable functions (product rule). Therefore the function $\ln f(x)$ is differentiable, hence $g(x)\ln f(x)$ is differentiable. Thus

$$f(x)^{g(x)} = \exp (g(x)\ln f(x))$$

is differentiable.

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Unfortunately, I'm also not sure how to go about this using limit. Let $F=f^g$.

\begin{align*} \log F(x) &= g(x) \log f(x)\\ \frac{F'(x)}{F(x)} &= g'(x)\log f(x) + \frac{g(x)f'(x)}{f(x)}\\ F'(x) &= f(x)^{g(x)} \left\{g'(x)\log f(x) + \frac{g(x)f'(x)}{f(x)}\right\} \end{align*}

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  • $\begingroup$ I believe that, without limits, we can write $f^g=e^{g\ln f}$. Since the composition of two differentiable functions is also differentiable (can be proven with limits) we see that $\ln f$ is differentiable, then $g\ln f$ is differentiable, thus $e^{g\ln f}=f^g$ is finally differentiable. $\endgroup$
    – Shroud
    Commented Aug 23, 2017 at 6:42

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