Deriving the Derivative of the Natural Exponential

Why the Derivative of \(e^x\) Is Special

The natural exponential function has a remarkable property: its derivative at each input equals its value there. To establish that property from the definition of a derivative, we need more than a pattern or a memorized rule. We need to evaluate a limit that describes how \(e^x\) changes when its input increases by a small amount.

The calculation will reduce the derivative to one central limit involving \(e^h\), where \(h\) is the small change in the input. We will establish that limit using the defining limit for the natural exponential and then use it to evaluate derivatives at specified inputs.

We use the standard exponential laws \(e^{x+h}=e^x e^h\), \(e^0=1\), and \(e^{-h}=1/e^h\). These laws express how the exponential function responds to adding, zero, and negating inputs. The defining limit is one way to construct this function; the exponential laws are properties of the resulting function.

Start with the Difference Quotient

For a function \(f\), its derivative at \(x\), when the limit exists, is defined by $$ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. $$ Here \(h\ne 0\) in the quotient, and the limit describes the behavior as \(h\) approaches zero. Apply this definition to \(f(x)=e^x\):

The factor \(e^x\) does not depend on \(h\), so it can be taken outside the limit. Thus the derivative calculation depends on whether \(\lim_{h\to 0}(e^h-1)/h\) exists and what its value is. We will show that this limit equals \(1\).

Establishing the Key Limit

A useful starting point is the inequality \(e^t\ge 1+t\), valid for every real \(t\). We can obtain it from the defining limit. Bernoulli’s inequality says that \((1+u)^n\ge 1+nu\) for every positive integer \(n\) and \(u\ge -1\). It follows by induction: the statement holds for \(n=1\), and multiplying the inequality for \(n\) by \(1+u\ge 0\) gives \((1+u)^{n+1}\ge (1+nu)(1+u)=1+(n+1)u+nu^2\ge 1+(n+1)u\).

For a fixed real number \(t\), take \(u=t/n\). For sufficiently large \(n\), \(t/n\ge -1\), so \((1+t/n)^n\ge 1+t\). Taking the limit as \(n\to\infty\), as in the definition of \(e^t\), gives \(e^t\ge 1+t\).

First suppose \(0<h<1\). Applying the inequality with \(t=h\) gives \(e^h\ge 1+h\), so \((e^h-1)/h\ge 1\). Applying it with \(t=-h\) gives \(e^{-h}\ge 1-h>0\). Since \(e^{-h}=1/e^h\), this implies \(e^h\le 1/(1-h)\). Therefore \((e^h-1)/h\le ((1/(1-h))-1)/h=1/(1-h)\).

Now suppose \(h<0\) and write \(s=-h\), so \(s>0\). Then \((e^h-1)/h=(1-e^{-s})/s\). The inequality \(e^{-s}\ge 1-s\) gives \((1-e^{-s})/s\le 1\). Also \(e^s\ge 1+s\), so \(e^{-s}\le 1/(1+s)\). It follows that \((1-e^{-s})/s\ge ((1-1/(1+s))/s)=1/(1+s)\).

These bounds show that the quotient approaches \(1\) from either side. The Squeeze Theorem applies because the lower and upper bounds both approach \(1\) as the increment approaches zero.

The Derivative Formula

We can now finish the derivation. Since \(e^x\) is constant with respect to \(h\), and the key limit has value \(1\), the derivative definition gives the following result.

This is a complete consequence of the difference quotient: for each fixed \(x\), we have \(\lim_{h\to 0}(e^{x+h}-e^x)/h=e^x\cdot 1=e^x\). The same key limit works at every input; only the constant factor \(e^x\) changes.

Worked Evaluations

Why the Result Matters

The derivative measures instantaneous rate of change. For the natural exponential, that rate at any input is exactly the function’s current value. For example, at \(x=2\), the function’s value and its instantaneous rate of change are both \(e^2\). At \(x=-1\), both are \(e^{-1}\).

The proof also illustrates a useful structure in derivative calculations. A difference quotient may simplify into a factor depending on the input and a second factor depending only on the small increment. The derivative follows once the limit of the increment factor is established. Here the exponential law creates that separation, and the defining limit for \(e\) underlies the squeeze argument.

A Common Mistake

A tempting but incorrect step is to cancel the \(h\) in the numerator and denominator of the quotient \((e^h-1)/h\). The expression \(e^h-1\) is not equal to \(h\), so this cancellation is not valid.