Trouble understanding variable exponent derivatives

Question

If $a \in \mathbb{R}$ then we know (by plugging into wolfram) that $ \frac{d} {dx} a^x=\log(a) a^x$. Of course, this is taken from the fact that $\lim_{h\to 0} \frac{a^{x+h}-a^x}{h}=a^x \lim_{h\to 0} \frac{a^{h}-1}{h}$. So, my question is, does anyone have a proof or way I can understand why the last limit is the same as $\log(a) $?

Answer 1

Do you already have the chain rule available? If so, $a^x=e^{x \log a}$, so $\frac d{dx}a^x=\frac d{dx}e^{x \log a}=(\log a) e^{x \log a}=(\log a)a^x$