Derivative of the Exponential Function

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It's a known piece of math folklore that e was "discovered" by Jacob Bernoulli in the 17th century, when he was pondering compound interest, and defined thus [1]:

\[e=\displaystyle \lim_{n \to \infty}\left ( 1+\frac{1}{n}\right )^n\]

e is extremely important in mathematics for several reasons; one of them is its useful behavior under derivation and integration; specifically, that:

\[\frac{\mathrm{d} }{\mathrm{d} x}e^x=e^x\]

In this post I want to present a couple of simple proofs of this fundamental fact.

Proof using the limit definition

As a prerequisite for this proof, let's reorder the original definition of e slightly. If we perform a change of variable replacing n by \frac{1}{m}, we get:

\[e=\displaystyle \lim_{m \to 0}\left ( 1+m \right )^\frac{1}{m} \tag{1}\]

This equation will become useful a bit later.

Let's start our proof by spelling out the definition of a derivative:

\[\frac{\mathrm{d} }{\mathrm{d} x}e^x=\displaystyle \lim_{h \to 0}\left ( \frac{e^{x+h}-e^x}{h} \right )\]

A bit of algebra and observing that e^x does not depend on h gives us:

\[\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x}e^x= \displaystyle \lim_{h \to 0}\left ( \frac{e^{x+h}-e^x}{h} \right )&= \displaystyle \lim_{h \to 0}\left ( \frac{e^x(e^{h}-1)}{h} \right )\\ &=\displaystyle e^x\lim_{h \to 0}\left ( \frac{e^{h}-1}{h} \right ) \end{align*}\]

At this point we're stuck; clearly as h approaches 0, both the numerator and denominator approach 0 as well. The way out - as is often the case in such scenarios - is a sneaky change of variable. Recall equation (1) - how could we use it here?

The change of variable we'll use is m=e^h-1, which implies that h=ln(m+1). Note that as h approaches zero, so does m. Rewriting our last expression, we get:

\[\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x}e^x= \displaystyle e^x\lim_{m \to 0}\left ( \frac{m}{ln(m+1)} \right ) &=\displaystyle e^x\lim_{m \to 0}\left ( \frac{1}{\frac{1}{m}ln(m+1)} \right )\\ &=\displaystyle e^x\lim_{m \to 0}\left ( \frac{1}{ln(m+1)^\frac{1}{m}} \right ) \end{align*}\]

Equation (1) tells us that as m approaches zero, (m+1)^\frac{1}{m} approaches e. Substituting that into the denominator we get:

\[\frac{\mathrm{d} }{\mathrm{d} x}e^x= \displaystyle e^x\lim_{m \to 0}\left ( \frac{1}{ln(e)} \right )=e^x \quad \blacksquare\]

Proof using power series expansion

It's always fun to prove the same thing in multiple ways; while I'm sure there are many other techniques to find the derivative of e^x, one I particularly like for its simplicity uses its power series expansion.

Similarly to the way e itself was defined empirically, one can show that:

\[e^x=\displaystyle \lim_{n \to \infty}\left ( 1+\frac{x}{n}\right )^n\]

(For a proof of this equation, see the Appendix)

Let's use the Binomial theorem to open up the parentheses inside the limit:

\[e^x=\lim_{n \to \infty}\left ( 1+\frac{x}{n} \right )^n = \lim_{n \to \infty}\sum_{k=0}^n {n \choose k}1^{n-k}\left (\frac{x}{n}\right )^k\]

We'll unroll the sum a bit, so it's easier to manipulate algebraically. We can use the standard formula for "choose n out of k" and get:

\[e^x=\lim_{n \to \infty}\left ( 1+n\frac{x}{n}+\frac{n(n-1)}{2!}\frac{x^2}{n^2}+\frac{n(n-1)(n-2)}{3!}\frac{x^3}{n^3}+\dotsb \right )\]

Inside the limit, we can simplify all the n-c terms with a constant c to just n, since compared to infinity c is negligible. This means that all these terms can be simplified as n(n-1)\approx n^2, n(n-1)(n-2)\approx n^3 and so on. All these powers of n cancel out in the numerator and denominator, and we get:

\[e^x=\lim_{n \to \infty}\left ( 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb \right )\]

And since the contents of the limit don't actually depend on n any more, this leaves us with a well-known formula for approximating e^x [2]:

\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb\]

We can finally use this power series expansion to calculate the derivative of e^x quite trivially. Since it's a sum of terms, the derivative is the sum of the derivatives of the terms:

\[\begin{align*} (e^x)^\prime&=1^\prime+x^\prime+\left (\frac{x^2}{2!}\right )^\prime+\left (\frac{x^3}{3!}\right )^\prime+\left (\frac{x^4}{4!}\right )^\prime+\dotsb\\ &=0+1+\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}+\dotsb\\ &=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb \end{align*}\]

Look at that, we've got e^x back, \blacksquare

Appendix

Let's see why:

\[e^x=\displaystyle \lim_{n \to \infty}\left ( 1+\frac{x}{n}\right )^n\]

We'll start with the limit and will arrive at e^x. Using a change of variable m=\frac{n}{x}:

\[\lim_{n \to \infty}\left (1 + \frac{x}{n}\right )^{n}=\lim_{n \to \infty}\left (1 + \frac{1}{m}\right )^{mx} =\lim_{n \to \infty}\left [\left (1 + \frac{1}{m}\right )^{m}\right ]^{x}\]

Given our change of variable, since n approaches infinity, so does m. Therefore, we get:

\[\lim_{m \to \infty}\left [\left (1 + \frac{1}{m}\right )^{m}\right ]^{x}\]

Nothing in the limit depends on x, so that exponent can be seen as applying to the whole limit. And the limit is the definition of e; therefore, we get e^x, \blacksquare

[1]What I love about this definition is that it's entirely empirical. Try to substitute successively larger numbers for n in the equation, and you'll see that the result approaches the value e more and more closely. The limit of this process for an infinite n was called e. Bernoulli did all of this by hand, which is rather tedious. His best estimate was that e is "larger than 2 and a half but smaller than 3".
[2]Another way to get this formula is from the Maclaurin series expansion of e^x, but we couldn't use that here since Maclaurin series require derivatives, while we're trying to figure out what the derivative of e^x is.

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