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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$ for $ \delta>0$, it follows that there exists a positive real number we'll call $ e$ such that for $ a=e$ we get

$\displaystyle \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

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

and use the chain rule,3.3 where $ \ln(a)\isdef \log_e(a)$ denotes the log-base-$ e$ of $ a$.3.4 Formally, the chain rule tells us how to differentiate a function of a function as follows:

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

Evaluated at a particular point $ x_0$, we obtain

$\displaystyle \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.,

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


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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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