The derivative

The limit of a rate of change, and why eˣ is its own derivative.

about 9 min

Start from a problem

Problem

An object's position is $x(t)$ . Its average velocity from $t$ to $t+\Delta t$ is $\frac{x(t+\Delta t)-x(t)}{\Delta t}.$ But what we really want to ask is: at the instant $t$ , how fast is it going?

An instant has no length; at $\Delta t=0$ the quotient is $\tfrac00$ . Yet "instantaneous speed" clearly means something: the speedometer shows it. So we are forced to consider the limit of the average velocity as $\Delta t\to0$ . If that limit exists, it deserves the name instantaneous velocity: $x'(t)=\lim_{\Delta t\to0}\frac{x(t+\Delta t)-x(t)}{\Delta t}.$ The derivative is not a notation handed down from above; it is the only reasonable answer to the question "what is the rate of change right now?"

Observe

Play first, read later. Below are $f(x)=a^x$ and its difference quotient $\frac{f(x+h)-f(x)}{h}.$ Drag the base $a$ and make the step $h$ small. The difference-quotient curve always looks like $a^x$ itself, just taller or shorter. Is there a base $a$ for which the two curves coincide exactly?

Interactive experimentWhy eˣ is its own derivative

base a2.000

step h0.500

\frac{a^{x+h}-a^{x}}{h}=a^{x}\cdot\frac{a^{h}-1}{h}

difference quotient ÷ function 0.8284

Conjecture

The difference-quotient curve is a constant multiple of $a^x$ ; the constant $\dfrac{a^h-1}{h}$ depends only on $a$ and $h$ . As $h\to0$ it tends to some number $c(a)$ depending only on $a$ . $c(a)$ increases with $a$ : about $0.69$ at $a=2$ , about $1.10$ at $a=3$ . So between $2$ and $3$ there should be a unique $a$ with $c(a)=1$ . We call it $e$ .

First a geometric look: shrink the distance $h$ between two points and the secant turns into the tangent.

Interactive experimentSecant becomes tangent

Function

point x₀1.00

increment h1.200

\frac{f(x_0+h)-f(x_0)}{h}

secant slope -0.0275tangent slope f′(x₀) 0.5403gap 0.5678

Definition

Definition 2.1Derivative

Let $f$ be defined on a neighbourhood of $x_0$ . If the limit $f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$ exists, $f$ is differentiable at $x_0$ and $f'(x_0)$ is its derivative there.

Geometrically the difference quotient is the slope of a secant and the derivative the slope of the tangent. Physically, average velocity and instantaneous velocity. The definition contains one limit and nothing else.

Definition 2.2The number e

$e$ is the unique positive real number with $\displaystyle\lim_{h\to0}\frac{e^h-1}{h}=1$ .

This looks engineered to make $(e^x)'=e^x$ true, and that is exactly what it is. We now show such a number exists, is unique, and identify it.

Derivation

Proposition 2.1Derivative of a^x

For $a>0$ , $\frac{d}{dx}a^x=a^x\cdot\lim_{h\to0}\frac{a^h-1}{h}=a^x\ln a.$

Proof

The first equality is the factorisation seen in the experiment: $a^{x+h}-a^x=a^x(a^h-1)$ .

For the second we need $\lim_{h\to0}\frac{a^h-1}h=\ln a$ . Put $u=a^h-1$ , so $h=\log_a(1+u)=\dfrac{\ln(1+u)}{\ln a}$ and $u\to0$ as $h\to0$ . Then $\frac{a^h-1}{h}=\frac{u\ln a}{\ln(1+u)}=\ln a\cdot\frac1{\frac{\ln(1+u)}u}\to\ln a,$ using $\lim_{u\to0}\frac{\ln(1+u)}u=1$ , which is equivalent to $\lim_{u\to0}(1+u)^{1/u}=e$ , the other common definition $e=\lim_{n\to\infty}(1+\tfrac1n)^n$ .

So $c(a)=\ln a$ , and $c(a)=1\iff a=e$ . The "coincidence" you saw in the experiment is $\ln e=1$ .

Basic rules and their proofs

Theorem 2.2Rules of differentiation

Let $f,g$ be differentiable at $x$ . Then

$(f+g)'=f'+g'$ ;
$(fg)'=f'g+fg'$ ;
if $g(x)\ne0$ , $\left(\dfrac fg\right)'=\dfrac{f'g-fg'}{g^2}$ ;
(chain rule) if $g$ is differentiable at $x$ and $f$ at $g(x)$ , then $(f\circ g)'(x)=f'(g(x))\,g'(x)$ .

Proof

We prove the product rule; the method is the same add-and-subtract trick as for products of limits. $\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=\frac{f(x+h)-f(x)}{h}\,g(x+h)+f(x)\,\frac{g(x+h)-g(x)}{h}.$ As $h\to0$ the first term tends to $f'(x)g(x)$ (differentiability of $g$ implies continuity, so $g(x+h)\to g(x)$ ) and the second to $f(x)g'(x)$ .

Theorem 2.3Differentiable implies continuous

If $f$ is differentiable at $x_0$ , then $f$ is continuous at $x_0$ .

Proof

$f(x_0+h)-f(x_0)=\dfrac{f(x_0+h)-f(x_0)}h\cdot h\to f'(x_0)\cdot0=0$ .

The converse fails: $|x|$ is continuous at $0$ but not differentiable; the one-sided difference quotients tend to $-1$ and $1$ .

Common mistake

$f'(x_0)$ is a number; $f'(x)$ is a function. In $f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}h$ the point $x_0$ is fixed and $h$ is what moves. Writing " $f'(x)=2x$ " means: for every $x$ , this limit was computed, and it happened to equal $2x$ .

Applications

ApplicationWhy e appears everywhere

$(e^x)'=e^x$ says: the rate of change is proportional to the current value, with constant 1. Any process whose growth rate is proportional to its size (compound interest, populations, radioactive decay, a discharging capacitor) is solved by $Ce^{kt}$ . This is the first equation, $y'=ky$ , in What is an ODE.

ApplicationTrigonometric functions

From $\lim_{x\to0}\frac{\sin x}x=1$ and $\lim_{x\to0}\frac{\cos x-1}x=0$ , $\frac{\sin(x+h)-\sin x}h=\sin x\,\frac{\cos h-1}h+\cos x\,\frac{\sin h}h\to\cos x.$

Remark

Observe → conjecture → define → derive → prove → apply. We did not start from " $e\approx2.718$ ". We first saw that a special base exists, then defined it, then proved it is the base of $\ln$ . A definition is the tidying-up of a conclusion, not something handed down from above.

Exercises

Let

f(x)=x^3-2x

. Find

f'(2)

Compute

\displaystyle\lim_{h\to0}\frac{2^h-1}{h}

to three decimals. This is the derivative of

2^x

x=0