I have to prove that $\frac{d}{dx}a^x=a^x\cdot \log(a)$ for $a>0$ and $x>0$, using the difference quotient $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ but without using the chain rule.
I know that $\log(a^x)=x\cdot \log(a)$, and that for $y>0$ we have $y=\exp(\log(y))$, which means that $a^x=\exp(\log(a)\cdot x)$.
I have absolutely no idea how to solve this, as L'Hôpital's rule has yet to be covered and cannot be used here.
Best Answer
you need to know that $$\lim_{x \rightarrow 0} \frac{e^x-1}x =1$$
which is proved here: proof of derivative of an exponential function