Definition of the Derivative

Calculus begins with a question about change: how fast is a quantity changing right now? Before we can answer it, we need the simpler idea of change over an interval.

Definition (average rate of change). The average rate of change of $f$ on the interval from $x = a$ to $x = b$ is

$$\frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}.$$

Geometrically, this is the slope of the secant line: the line through the two points $(a, f(a))$ and $(b, f(b))$ on the graph. Physically, if $f(t)$ is the position of a car at time $t$, it is the car's average velocity over that stretch of time: total distance covered divided by time taken.

The difference quotient

It is often convenient to describe the second point by how far it is from the first. Write $b = a + h$, where $h \ne 0$ is the step size. Then $b - a = h$, and the average rate of change becomes the difference quotient

$$\frac{f(a + h) - f(a)}{h}.$$

The step $h$ can be positive (the second point is to the right) or negative (to the left). Everything in this lesson grows out of this one expression.

Example

For $f(x) = x^2$ on $[1, 2]$, the average rate of change is

$$\frac{f(2) - f(1)}{2 - 1} = \frac{4 - 1}{1} = 3.$$

Using the difference quotient with $a = 1$:

$$\frac{(1 + h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = 2 + h \qquad (h \ne 0).$$

With $h = 1$ this gives $3$, as before. With $h = 0.5$ it gives $2.5$; with $h = 0.1$, $2.1$; with $h = 0.01$, $2.01$.

As the second point slides toward $(1, 1)$, the secant lines tilt and their slopes $3, 2.5, 2.1, 2.01, \dots$ close in on a single number, $2$. The line they settle onto touches the parabola at $(1, 1)$: it is the tangent line. Making "close in on" precise is exactly what limits are for.

Common mistake. $f(a + h)$ means "evaluate $f$ at the input $a + h$." It is not $f(a) + h$. For $f(x) = x^2$, $f(1 + h) = (1 + h)^2 = 1 + 2h + h^2$, not $1 + h$.

The difference quotient measures change over an interval of width $h$. To measure change at a single instant, we let the interval shrink.

Definition (derivative at a point). The derivative of $f$ at $a$ is

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

provided the limit exists. When it does, $f$ is differentiable at $a$.

We can't simply substitute $h = 0$: that gives $\frac{f(a) - f(a)}{0} = \frac{0}{0}$, which is meaningless. As in section 04.03, the plan is to simplify the quotient algebraically until the troublesome factor of $h$ cancels, and then let $h \to 0$.

Example 1

Find $f'(3)$ for $f(x) = x^2$.

$$f'(3) = \lim_{h \to 0}\frac{(3 + h)^2 - 9}{h} = \lim_{h \to 0}\frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0}\frac{6h + h^2}{h} = \lim_{h \to 0}(6 + h) = 6.$$

The cancellation $\frac{6h + h^2}{h} = 6 + h$ is valid because $h \ne 0$ inside the limit.

Example 2

Find $g'(1)$ for $g(x) = x^3 - 2x$. First, $g(1) = -1$ and

$$g(1 + h) = (1 + h)^3 - 2(1 + h) = 1 + 3h + 3h^2 + h^3 - 2 - 2h = -1 + h + 3h^2 + h^3.$$

So

$$g'(1) = \lim_{h \to 0}\frac{(-1 + h + 3h^2 + h^3) - (-1)}{h} = \lim_{h \to 0}\frac{h + 3h^2 + h^3}{h} = \lim_{h \to 0}(1 + 3h + h^2) = 1.$$

An equivalent form

Setting $x = a + h$, so that $h = x - a$ and $h \to 0$ exactly when $x \to a$, gives a second, equivalent formula:

$$f'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}.$$

Use whichever makes the algebra easier. For $f(x) = x^2$ at $a = 3$, the second form factors neatly: $\dfrac{x^2 - 9}{x - 3} = x + 3 \to 6$.

The pattern to remember. Every derivative from the definition follows the same four steps:

1. Write out $f(a + h)$ carefully.
2. Subtract $f(a)$ and simplify the numerator.
3. Divide by $h$; a factor of $h$ should cancel.
4. Let $h \to 0$.

Computing $f'(3)$, then $f'(1)$, then $f'(5)$, each from scratch, repeats the same algebra. It is more efficient to leave the point general.

Definition (the derivative function). The derivative of $f$ is the function $f'$ defined by

$$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$

at every $x$ where the limit exists. Its domain is the set of points where $f$ is differentiable, which can be smaller than the domain of $f$ (section 05.02 looks at where this fails).

Example 1: $f(x) = x^2$

$$f'(x) = \lim_{h \to 0}\frac{(x + h)^2 - x^2}{h} = \lim_{h \to 0}\frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0}\frac{2xh + h^2}{h} = \lim_{h \to 0}(2x + h) = 2x.$$

So $f'(3) = 6$ and $f'(1) = 2$, agreeing with the examples above, and now every slope is available at once. Notice that $x$ is held fixed while $h$ varies: inside the limit, $x$ behaves like a constant.

Example 2: $f(x) = \dfrac{1}{x}$

The numerator is a difference of fractions, so combine them over a common denominator:

$$\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}.$$

Dividing by $h$ cancels the $h$:

$$f'(x) = \lim_{h \to 0}\frac{-1}{x(x + h)} = -\frac{1}{x^2} \qquad (x \ne 0).$$

The slope is negative everywhere, which matches the graph of $y = \frac{1}{x}$: on each side of $0$ it falls from left to right.

Example 3: $f(x) = \sqrt{x}$

Square roots call for the conjugate. Multiply the top and bottom by $\sqrt{x + h} + \sqrt{x}$:

$$\frac{\sqrt{x + h} - \sqrt{x}}{h}\cdot\frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} = \frac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} = \frac{1}{\sqrt{x + h} + \sqrt{x}}.$$

Letting $h \to 0$:

$$f'(x) = \frac{1}{2\sqrt{x}} \qquad (x > 0).$$

Here $f$ is defined at $x = 0$ but $f'$ is not: the graph of $\sqrt{x}$ starts with a vertical tangent at the origin.

Common mistakes.

• Cancelling $h$ before simplifying the numerator. In $\frac{(x + h)^2 - x^2}{h}$ you can't "cancel the $h$" in $(x + h)^2$; expand first.
• Letting $h \to 0$ too early. If you still have $\frac{0}{0}$, there is more algebra to do.
• Forgetting the domain: the formula for $f'$ may need restrictions such as $x \ne 0$ or $x > 0$.

We have been calling $f'(a)$ "the slope of the tangent line." Now we can make that the definition.

Definition (tangent line). If $f$ is differentiable at $a$, the tangent line to $y = f(x)$ at $(a, f(a))$ is the line through that point with slope $f'(a)$. In point-slope form:

$$y = f(a) + f'(a)(x - a).$$

So finding a tangent line takes three numbers: the point's $x$-coordinate $a$, its $y$-coordinate $f(a)$, and the slope $f'(a)$.

Example 1

Find the tangent line to $y = x^2$ at $x = 3$. The point is $(3, 9)$ and the slope is $f'(3) = 6$, so

$$y = 9 + 6(x - 3) = 6x - 9.$$

Example 2

Find the tangent line to $y = \dfrac{1}{x}$ at $x = 2$. The point is $\left(2, \frac{1}{2}\right)$ and, from the derivative of $\frac{1}{x}$ above, $f'(2) = -\frac{1}{2^2} = -\frac{1}{4}$. So

$$y = \frac{1}{2} - \frac{1}{4}(x - 2) = -\frac{1}{4}x + 1.$$

Why it matters: the best straight-line approximation

Zoom in far enough on a differentiable function and its graph looks straight; the line it looks like is the tangent line. Near $x = a$,

$$f(x) \approx f(a) + f'(a)(x - a).$$

For example, the tangent line to $y = \sqrt{x}$ at $x = 4$ is $y = 2 + \frac{1}{4}(x - 4)$, which estimates $\sqrt{4.1} \approx 2.025$. The true value is $2.02485\ldots$. This idea, linear approximation, returns in chapter 06.

Common mistake. The slope of the tangent line is the number $f'(a)$, not the function $f'(x)$. "The tangent line to $y = x^2$ at $x = 3$ is $y = 9 + 2x(x - 3)$" is wrong: that isn't even a line. Evaluate $f'$ at $a$ first.

Rates of change

The derivative is the instantaneous rate of change: the limit of average rates over shorter and shorter intervals. Its meaning depends on what $f$ measures.

• If $s(t)$ is position at time $t$, then $s'(t)$ is the velocity at time $t$.
• If $C(q)$ is the cost of producing $q$ items, $C'(q)$ is the marginal cost: approximately the cost of producing one more item.
• If $T(h)$ is the temperature at altitude $h$, $T'(h)$ says how fast temperature changes as you climb.

Units. Because $f'(a)$ is a limit of $\frac{\Delta y}{\Delta x}$, its units are "units of output per unit of input." If $s$ is in meters and $t$ in seconds, $s'(t)$ is in meters per second. If $C$ is in dollars and $q$ in items, $C'(q)$ is in dollars per item. So $C'(500) = 80$ means: at a production level of 500 items, cost is rising at about $80 per additional item.

Notation

Several notations are in common use, all meaning the same thing:

$$f'(x), \qquad y', \qquad \frac{dy}{dx}, \qquad \frac{df}{dx}, \qquad \frac{d}{dx}f(x).$$

Lagrange's prime notation $f'(x)$ is compact. Leibniz's $\dfrac{dy}{dx}$ recalls where the derivative comes from, $\dfrac{\Delta y}{\Delta x}$, and makes the units visible. The derivative at a point can be written $f'(a)$ or $\dfrac{dy}{dx}\Big|_{x = a}$. The symbol $\dfrac{d}{dx}$ is an operator: an instruction to take the derivative of whatever follows, as in $\dfrac{d}{dx}\left(x^2\right) = 2x$.

Recognizing a derivative in disguise

Every limit of the form $\displaystyle\lim_{h \to 0}\frac{f(a + h) - f(a)}{h}$ is a derivative, even when it isn't labeled as one. To evaluate

$$\lim_{h \to 0}\frac{(2 + h)^4 - 16}{h},$$

notice that $16 = 2^4$. This is the difference quotient for $f(x) = x^4$ at $a = 2$, so the limit is $f'(2)$. Once you know derivative rules (section 05.03 shows $\frac{d}{dx}x^4 = 4x^3$), it is simply $4 \cdot 2^3 = 32$, with no expansion of $(2 + h)^4$ needed.

To identify $f$ and $a$: the part that depends on $h$ is $f(a + h)$, and the constant subtracted is $f(a)$. Check that they match. In $\lim_{h \to 0}\frac{\sqrt{9 + h} - 3}{h}$, we have $f(x) = \sqrt{x}$ and $a = 9$, since $\sqrt{9} = 3$.