Proof of the derivative of $a^x$

Question

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.

Answer 1

Using the chain rule and assuming you already know that $(\exp)^\prime=\exp,$ you have:

\begin{align} \frac{\mathrm d}{\mathrm dx}a^x=\frac{\mathrm d}{\mathrm dx}e^{x\ln a}=\ln a\cdot e^{x\ln a}=\ln a\cdot a^x. \end{align}

Answer 2

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 $y'=a^x\cdot ln(a)$.