Video Lecture

In this video we will give an algebraic derivation of the derivative for functions of the form $a^x$ for $a > 0$; with special attention to the function $e^x$.

For a look at the special role exponential functions, and $e^x$ specifically, play in calculus, there is an optional supplemental video you can access here:

https://youtu.be/4PJTEV8lYDk

(Supplemental Videos are included via external link so you don’t have to watch them to earn credit.)

Text and Additional Details

We will begin by trying to determine the derivative of $e^x$ and, as always, let’s start by looking at the difference quotient.

$$\lim \limits _{\Delta x\rightarrow 0} \frac {e^{x+\Delta x} - e^x}{\Delta x}$$

Unlike most of our previous derivations for derivative rules, this one isn’t going to be solved by adding zero or multiplying by one “cleverly”. Indeed, the underlying problem here is actually that there is a hidden limit that we haven’t noticed!

$e^x$ has a very special property in calculus, which we will discover shortly, but $e$ itself can be defined in many ways. You may have encountered it as simply an irrational constant approximately equal to 2.7182818... You may have seen it defined in it’s more proper form however, which is the value of a limit (although they may not have used limit notation explicitly at the time). Indeed, $e$ was originally defined by Jacob Bernoulli in the late 1600s as the following:

$$e = \lim \limits _{n\rightarrow \infty } \left (1 + \frac {1}{n}\right )^n$$

So, this means that in our difference quotient, the $e$ that is used as a base is actually a limit itself, which explains why the algebra might get a bit messy trying to simplify this expression!

To proceed, instead of trying to manipulate the difference quotient directly, we will do a little manipulation to the definition of $e$. In particular, let’s substitute $\Delta x = \frac {1}{n}$. Notice that as $n\rightarrow \infty$, $\Delta x = \frac {1}{n} \rightarrow 0$. With this in mind we can rewrite the definition of $e$ as follows:

$$e = \lim \limits _{n\rightarrow \infty } \left (1 + \frac {1}{n}\right )^n = \lim \limits _{\Delta x\rightarrow 0} \left (1 + \Delta x\right )^{\frac {1}{\Delta x}}$$

Believe it or not, with this somewhat simple (but admittedly non-obvious) substitution, we are ready to proceed!

At this point you might be thinking, “Wait... did we just do all that work to get back where we started?” And the answer is, Yes! Yes we did (cue maniacal laughter)! Indeed, we just showed that the derivative of $e^x$ is itself!

And as a first use, we will use this to develop a rule to take a derivative of $a^x$ for $a > 0$.

Rather than looking at the difference quotient, we are going to manipulate the expression directly by using the fact that $e^{\ln (u)} = u$ and $\ln (a^u) = u\ln (a)$.

$$a^x = e^{\ln (a^x)} = e^{x\ln (a)}$$ Thus, we can use the chain rule on the above expression when taking a derivative:

So, using the $e^x$ rule we can deduce a general rule for $a^x$ for any $a > 0$. Notice here that, if $a=e$ this rule still works because we get $\ln (e)=1$, so if we apply the $a^x$ rule with base $e$ we get: $$\frac {d}{dx}\left [e^x\right ] = \ln (e)e^x = 1e^x = e^x$$

Moreover, if $a=1$ the rule also still works, it is just somewhat silly since $1^x$ is not a particularly interesting function (at least in the real numbers), so using our rule we get:

$$\frac {d}{dx}\left [1^x\right ] = \ln (1)e^x = 0\cdot e^x = 0$$

So, as we saw, there is something special going on with $e$ that makes the derivation of the derivative rule for $e^x$ a pretty different process. Nonetheless, this rule immediately generalizes to $a^x$ for any $a>0$. We conclude by formally recording our rule: