Deriving the Derivative of e^x

Introduction

A derivative describes the instantaneous rate of change of a function. For the natural exponential, the result is especially simple: at every input, its derivative equals its value there. But this fact needs a proof from the definition of derivative; it is not enough to recognize a familiar pattern.

The proof turns on a limit that measures the change in \(e^h\) relative to a small increment \(h\). We will establish that limit from the compound-growth construction of the exponential, then use it in the difference quotient at an arbitrary point.

The difference quotient

For a function \(f\), its derivative at \(a\), when the limit exists, is defined by

f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.

Here \(h\) is a nonzero increment that approaches 0. The numerator is the change in the function's value, and dividing by \(h\) gives the average rate of change over that increment. The limit, if it exists, is the instantaneous rate of change at \(a\).

For \(f(x)=e^x\), the quotient becomes

\frac{e^{a+h}-e^a}{h}.

The exponential law \(e^{a+h}=e^a e^h\) lets us factor this expression:

\frac{e^{a+h}-e^a}{h} = e^a\frac{e^h-1}{h}.

For fixed \(a\), the factor \(e^a\) does not depend on \(h\). Therefore the central question is whether the limit

\lim_{h\to 0}\frac{e^h-1}{h}

exists, and what its value is.

Key limit: We will prove that \(\displaystyle \lim_{h\to 0}\frac{e^h-1}{h}=1\). The proof must handle increments approaching 0 from both the positive and negative sides.

Bounds from the compound-growth construction

The natural exponential can be constructed using the compound-growth limit: for each fixed real \(x\),

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

where \(n\) is taken large enough that the base is positive. In particular, \(e=e^1\). We use this construction to bound \(e^x\) when \(x\) is positive and small.

Suppose \(0<x<1\). For each positive integer \(n\), the binomial theorem gives

\left(1+\frac{x}{n}\right)^n = \sum_{k=0}^{n}\binom{n}{k}\frac{x^k}{n^k}.

The term for \(k=0\) is 1, and the term for \(k=1\) is \(x\). Every remaining term is nonnegative, so

\left(1+\frac{x}{n}\right)^n\ge 1+x.

For an upper bound, when \(0\le k\le n\),

\binom{n}{k}\frac{1}{n^k} = \frac{1}{k!}\frac{n(n-1)\cdots(n-k+1)}{n^k} \le \frac{1}{k!}\le 1.

Consequently,

\left(1+\frac{x}{n}\right)^n \le \sum_{k=0}^{n}x^k \le \sum_{k=0}^{\infty}x^k =\frac{1}{1-x}.

Taking the limit as \(n\to\infty\) preserves these inequalities. Thus, for \(0<x<1\),

1+x\le e^x\le \frac{1}{1-x}.

These bounds are useful because both sides are close to 1 when \(x\) is close to 0. We now use them to prove the key limit separately for positive and negative increments.

Proof of the key limit

Positive increments.
Let \(0<h<1\). Subtract 1 from the bounds \(1+h\le e^h\le 1/(1-h)\), then divide by the positive number \(h\). This gives

Negative increments.
Write \(h=-t\), where \(0<t<1\). Use the positive-increment bounds for \(e^t\), and take reciprocals, using \(e^{-t}=1/e^t\).

Combine the two sides.
In both cases, the quotient is trapped between expressions that approach 1. The squeeze theorem then gives the two-sided limit.

For \(0<h<1\), the positive-increment bounds give

1+h\le e^h\le \frac{1}{1-h}.

Subtracting 1 and dividing by \(h>0\) yields

1\le \frac{e^h-1}{h}\le \frac{1}{1-h}.

As \(h\to 0^+\), the lower bound approaches 1 and the upper bound also approaches 1. The squeeze theorem therefore gives

\lim_{h\to 0^+}\frac{e^h-1}{h}=1.

For negative increments, put \(h=-t\), where \(0<t<1\). The positive bounds for \(e^t\) are

1+t\le e^t\le \frac{1}{1-t}.

All quantities here are positive, so taking reciprocals reverses the inequalities:

1-t\le \frac{1}{e^t}\le \frac{1}{1+t}.

Since \(e^{-t}=1/e^t\), subtracting from 1 and dividing by \(t>0\) gives

\frac{1}{1+t} \le \frac{1-e^{-t}}{t} \le 1.

As \(t\to 0^+\), both bounds approach 1. Also,

\frac{1-e^{-t}}{t} = \frac{e^h-1}{h} \qquad\text{when }h=-t.

It follows that the quotient approaches 1 from the negative side as well. Since its one-sided limits agree, the two-sided limit exists and equals 1:

\lim_{h\to 0}\frac{e^h-1}{h}=1.

Why the bounds matter: The compound-growth construction supplies estimates on \(e^h\) that hold for small positive \(h\). The reciprocal relation \(e^{-h}=1/e^h\) transfers those estimates to negative increments. Together they justify the limit from both sides, not merely for selected values of \(h\).

Deriving the derivative at any point

We can now return to the difference quotient at an arbitrary real number \(a\). Using the exponential law and factoring out the constant \(e^a\), we obtain

\begin{aligned} \frac{e^{a+h}-e^a}{h} &=\frac{e^a e^h-e^a}{h}\\ &=e^a\frac{e^h-1}{h}. \end{aligned}

As \(h\to 0\), the second factor tends to 1 by the limit just proved. The constant multiple rule for limits then gives

\begin{aligned} \frac{d}{dx}e^x\bigg|_{x=a} &=\lim_{h\to 0}\frac{e^{a+h}-e^a}{h}\\ &=e^a\lim_{h\to 0}\frac{e^h-1}{h}\\ &=e^a\cdot 1\\ &=e^a. \end{aligned}

Theorem: The natural exponential is differentiable at every real number \(a\), and its derivative there is \(e^a\). Equivalently, \(\displaystyle \frac{d}{dx}e^x=e^x\).

The proof establishes the derivative directly from its definition. It does not assume a general rule for differentiating exponentials with more complicated exponents.

Worked examples

Worked Example: Derivative at \(x=0\)

Find the derivative of \(e^x\) at 0 from the difference quotient. Begin by substituting \(a=0\) into the definition:

\begin{aligned} \frac{d}{dx}e^x\bigg|_{x=0} &=\lim_{h\to 0}\frac{e^{0+h}-e^0}{h}\\ &=\lim_{h\to 0}\frac{e^h-1}{h}\\ &=1. \end{aligned}

The last equality is the key limit proved above. The general result gives \(e^0=1\), which agrees with this calculation. As a check, the difference quotient at 0 is exactly the expression whose limit was established, so its value is indeed 1.

Worked Example: Slope at \(x=1\)

Find the slope of the graph \(y=e^x\) at \(x=1\) by evaluating the difference quotient:

\begin{aligned} \lim_{h\to 0}\frac{e^{1+h}-e^1}{h} &=\lim_{h\to 0}\frac{e\cdot e^h-e}{h}\\ &=\lim_{h\to 0}e\frac{e^h-1}{h}\\ &=e\cdot 1\\ &=e. \end{aligned}

Thus the slope at the point \((1,e)\) is \(e\). Checking against the theorem, the derivative at \(x=1\) is \(e^1=e\), the same value obtained from the quotient.

Worked Example: Slope at \(x=-2\)

Evaluate the difference quotient at \(a=-2\):

\begin{aligned} \lim_{h\to 0}\frac{e^{-2+h}-e^{-2}}{h} &=\lim_{h\to 0}\frac{e^{-2}e^h-e^{-2}}{h}\\ &=\lim_{h\to 0}e^{-2}\frac{e^h-1}{h}\\ &=e^{-2}\cdot 1\\ &=e^{-2}. \end{aligned}

The slope is \(e^{-2}=1/e^2\). Substituting \(a=-2\) into the theorem gives \(e^a=e^{-2}\), confirming the quotient calculation.

A common mistake: losing the factor \(e^a\)

A tempting but incorrect simplification is to treat \(e^{a+h}-e^a\) as though it were just \(e^h-1\). For instance, at \(a=1\), this mistake would give

\frac{e^{1+h}-e}{h} \overset{\text{wrong}}{=} \frac{e^h-1}{h} \longrightarrow 1.

The equality is wrong because \(e^{1+h}=e\cdot e^h\), so the correct factorization is

\frac{e^{1+h}-e}{h} = e\frac{e^h-1}{h} \longrightarrow e\cdot 1=e.

The factor \(e^a\) is constant with respect to \(h\), but it is not generally equal to 1 and cannot be dropped. Keeping it visible in the factorization prevents this error.

Why this result is distinctive

The natural exponential is distinguished by the fact that its instantaneous rate of change at each input equals its value at that input. For example, at \(x=0\), both the value and the slope are 1; at \(x=1\), both are \(e\). This property follows here from the compound-growth limit and the derivative definition, rather than from a memorized differentiation rule.

The argument also illustrates a useful proof strategy: rewrite a difference quotient so that its dependence on the increment is isolated in one standard limit. In this case the exponential law separates the fixed input \(a\) from the changing increment \(h\). When differentiating functions with more complicated exponents, additional reasoning will be needed; the present result concerns \(e^x\) itself.

Check Your Understanding

Use the definition of derivative and the key limit to answer these questions.

What difference quotient defines the derivative of \(e^x\) at a real number \(a\)?
Why is it necessary to establish the limit of \((e^h-1)/h\) from both sides?
What bounds for \(e^h\) hold when \(0<h<1\), and how do they bound the quotient?
How does the identity \(e^{-t}=1/e^t\) help handle negative increments?
At which step in the derivative proof does the factor \(e^a\) arise, and why must it remain?

What You'll Learn

Introduction

The difference quotient

Bounds from the compound-growth construction

Proof of the key limit

Deriving the derivative at any point