Derivatives of f(x)=a^x

Let's apply the definition of differentiation and see what happens:

$$\begin{eqnarray*} f^\prime(x_0) &\isdef & \lim_{\delta\to0} \frac{f(x_0+\delta)-... ...{\delta} = a^{x_0}\lim_{\delta\to0} \frac{a^\delta-1}{\delta}. \end{eqnarray*}$$

Since the limit of $(a^\delta-1)/\delta$ as $\delta\to 0$ is less than 1 for $a=2$ and greater than $1$ for $a=3$ (as one can show via direct calculations), and since $(a^\delta-1)/\delta$ is a continuous function of $a$, it follows that there exists a positive real number we'll call $e$ such that for $a=e$ we get

$$\lim_{\delta\to 0} \frac{e^\delta-1}{\delta} \isdef 1 .$$

For $a=e$, we thus have $\left(a^x\right)^\prime = (e^x)^\prime = e^x$.

So far we have proved that the derivative of $e^x$ is $e^x$. What about $a^x$ for other values of $a$? The trick is to write it as

$$a^x = e^{\ln\left(a^x\right)}=e^{x\ln(a)}$$

and use the chain rule, where $\ln(a)\isdef \log_e(a)$ denotes the log-base-$e$ of $a$.^3.2Formally, the chain rule tells us how to differentiate a function of a function as follows:

$$\frac{d}{dx} f(g(x)) = f^\prime(g(x)) g^\prime(x)$$

Evaluated at a particular point, we obtain

$$\frac{d}{dx} f(g(x))\vert _{x=x_0} = f^\prime(g(x_0)) g^\prime(x_0).$$

In this case, $g(x)=x\ln(a)$ so that $g^\prime(x) = \ln(a)$, and $f(y)=e^y$ which is its own derivative. The end result is then $\left(a^x\right)^\prime = \left(e^{x\ln a}\right)^\prime = e^{x\ln(a)}\ln(a) = a^x \ln(a)$, i.e.,

$$\zbox {\frac{d}{dx} a^x = a^x \ln(a).}$$