# Proof of the derivative of $a^x$ [duplicate]

I've tried for a while myself from first principles and applying various rules, but always end up going in circles. I've gotten as far as

$$y = a^x$$ $$\frac{dy}{dx} = a^x \left( \lim_{x \rightarrow 0} \frac{a^h-1}{h} \right)$$

but I have no idea how I should go about cancelling the $h$ in the denominator. Any help is appreciated.

• Do you know this relation ? $$a^x=\exp(x\ln(a))$$ – Fabien Jun 18 '14 at 11:12
• Here it's derived: math.com/tables/derivatives/more/b%5Ex.htm (Using an logarithm/exponential-Identity) – Mario Krenn Jun 18 '14 at 11:13
• @NicoDean Interesting, do you have a proof for that identity? – user157789 Jun 18 '14 at 11:15
• @user157789 Some people would take it as the definition of $a^x$. – Jack M Jun 18 '14 at 11:33
Using the chain rule and assuming you already know that $(\exp)^\prime=\exp,$ you have:
Let $y=a^x$. Take the natural logarithm of both sides and rewrite the right-hand side using rules of logarithms to obtain $ln(y)=x\cdot ln(a)$. Differentiate both sides implicitly with respect to $x$. So ${1\over y}y'=ln(a)$. Now multiply both sides of the equation by $y=a^x$ and we have $y'=a^x\cdot ln(a)$.