Math Core

In the last lesson you found instantaneous rates of change one point at a time. That limit is so important that it gets its own name and notation: the derivative. This lesson gives the formal definition, shows how to compute a derivative as a whole new function, and connects it to the equation of a tangent line.

The derivative at a point

Definition

Derivative at a point

The derivative of $f$ at $x = a$ is

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

provided the limit exists. If it does, $f$ is differentiable at $a$ . The number $f'(a)$ is the instantaneous rate of change of $f$ at $a$ and the slope of the tangent line to the graph of $f$ at $\big(a, f(a)\big)$ .

There is a second, equivalent form. Instead of stepping a distance $h$ away from $a$ , let the second point be a general $x$ that approaches $a$ :

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

Both are slopes of secant lines with one endpoint fixed at $a$ . The AP exam uses both forms, so you should recognize each one.

The derivative as a function

If you compute the limit with a general $x$ in place of $a$ , the result is a formula that gives the slope at every point where the limit exists. That formula is the derivative function:

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

Several notations mean the same thing:

f'(x), \qquad y', \qquad \frac{dy}{dx}, \qquad \frac{d}{dx}\big[f(x)\big]

The Leibniz notation $\dfrac{dy}{dx}$ reminds you that a derivative is a limit of $\dfrac{\Delta y}{\Delta x}$ . To write the derivative at a specific point, use $f'(3)$ or $\left.\dfrac{dy}{dx}\right|_{x = 3}$ .

Worked example: A derivative from the definition

Use the definition to find $f'(x)$ for $f(x) = 3x^2 - x$ .

First build $f(x + h)$ by replacing every $x$ with $x + h$ :

f(x + h) = 3(x + h)^2 - (x + h) = 3x^2 + 6xh + 3h^2 - x - h

Subtract $f(x)$ ; the terms without $h$ cancel:

f(x + h) - f(x) = 6xh + 3h^2 - h

Divide by $h$ and take the limit:

f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2 - h}{h} = \lim_{h \to 0} (6x + 3h - 1) = 6x - 1

So the slope of $f$ at any $x$ is $6x - 1$ . For instance, $f'(2) = 11$ .

The pattern is always the same: expand, cancel, factor out $h$ , divide it away, and only then let $h \to 0$ . If the terms without $h$ don't all cancel, look for an algebra mistake.

Worked example: Using a conjugate

Find $f'(x)$ for $f(x) = \sqrt{x}$ , $x > 0$ .

\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} \\ &= \lim_{h \to 0} \frac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} \\ &= \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{2\sqrt{x}} \end{aligned}

Multiplying by the conjugate is the same trick you used on limits with radicals in Unit 1.

Tangent lines

Once you know $f(a)$ and $f'(a)$ , you have a point and a slope, which is everything you need for a line.

Equation of the tangent line

The tangent line to the graph of $f$ at $x = a$ is

y - f(a) = f'(a)(x - a)

Point-slope form is completely acceptable on the AP exam; you don't need to rearrange it.

Worked example: Writing a tangent line

Find the tangent line to $f(x) = x^2 + 1$ at $x = 2$ .

From the definition,

f'(2) = \lim_{h \to 0} \frac{(2 + h)^2 + 1 - 5}{h} = \lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0} (4 + h) = 4

The point is $\big(2, f(2)\big) = (2, 5)$ . The tangent line is $y - 5 = 4(x - 2)$ , or $y = 4x - 3$ .

The tangent line y = 4x − 3 touches y = x² + 1 at (2, 5).Open in grapher →

Recognizing a limit as a derivative

A favorite AP question hands you a limit and expects you to see a derivative in disguise. Match it to one of the two forms, identify $f$ and $a$ , and then find $f'(a)$ .

Worked example: A derivative in disguise

Evaluate $\displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ .

This is $\displaystyle \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ with $f(x) = x^3$ and $a = 2$ , since $f(2) = 8$ . So the limit equals $f'(2)$ , the slope of $y = x^3$ at $x = 2$ .

To evaluate it, factor the difference of cubes:

\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2} = \lim_{x \to 2} (x^2 + 2x + 4) = 12

In the next few lessons you'll learn rules that give $f'(2) = 12$ in one step.

Common mistake

In the definition, $h \to 0$ comes last. If you substitute $h = 0$ before canceling, you get $\dfrac{0}{0}$ , which tells you nothing. Also watch the parentheses in $f(x + h)$ : for $f(x) = 3x^2$ , $f(x + h) = 3(x + h)^2$ , not $3x^2 + h$ .

Practice

Practice 1

Use the definition of the derivative to find $f'(x)$ for $f(x) = 5x - 2$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Use the definition of the derivative to find $f'(x)$ for $f(x) = x^2 + 4x$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Let $f(x) = 2x^2 - 3x + 1$ . Use the definition to find $f'(x)$ , then evaluate $f'(-1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Use the definition of the derivative to find $f'(x)$ for $f(x) = \dfrac{1}{x}$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 5

Find an equation of the tangent line to $f(x) = x^2 - 3x$ at $x = 4$ . Enter it as $y = \ldots$ or in point-slope form.

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

$\displaystyle \lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h}$ is

Practice 7

Evaluate $\displaystyle \lim_{x \to 1} \frac{x^4 - 1}{x - 1}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A differentiable function $f$ has $f(2) = 5$ and $f'(2) = -3$ . Write the equation of the tangent line to $f$ at $x = 2$ , and use it to estimate $f(2.1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.