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.

Here it's derived: http://www.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
Actually, you find the answere already at this page: http://math.stackexchange.com/questions/587275/finding-the-derivative-of-2x-from-first-terms?rq=1 — Mario Krenn, Jun 18 '14 at 11:17
@user157789 Some people would take it as the definition of $a^x$. — Jack M, Jun 18 '14 at 11:33

score 6 · Answer 1 · edited Jun 18 '14 at 11:25

6

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}

edited Jun 18 '14 at 11:25

mike

5,604

answered Jun 18 '14 at 11:17

Hakim

10,213

2

What if $a$ is negative? – user599310 Nov 21 '20 at 17:34

score 0 · Answer 2 · answered Jun 18 '14 at 11:28

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

Proof of the derivative of $a^x$

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