Find the derivative of $a^x$ using the difference quotient, but without using the chain rule 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.
 A: We have to show that $$\frac{d}{dx}a^x=a^x\log(a).$$
Proof: Recall that for any function $f(x)$, by definition, $$f'(x) =\lim _{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right), \forall f.$$ Let $f(x)=a^x$ s.t. $$\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}} \implies \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x+h}-a^{\displaystyle x}}{h}}. $$ $$\implies \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x} \times a^{\displaystyle h}-a^{\displaystyle x}}{h}}=\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \bigg(a^{\displaystyle x} \times \dfrac{a^{\displaystyle h}-1}{h}\bigg)}.$$ We can now evaluate $f'(x)$ by the constant multiple rule of calculus s.t. $$a^{\displaystyle x} \times\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle h}-1}{h}} \implies a^x\ln(a)$$ $$\therefore \forall f=a^x,f'(x)=a^x\ln(a)$$ Q.E.D
A: 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
