The Definition of the Derivative

The derivative measures the instantaneous rate of change of a function, which is also the slope of its tangent line. It’s built from the limit of secant slopes you saw in average and instantaneous rates. This page gives the official definition, the notation, and how to estimate a derivative when all you have is a table or a graph.

Key ideas

The 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 this limit exists. An equivalent form uses a second point $x$ sliding toward $a$:

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

Both are limits of secant slopes. Use whichever makes the algebra easier.

The derivative as a function

If you keep $x$ as a variable instead of a number, you get a new function, the derivative function:

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

Plug in any $x$ to get the slope of $f$ at that point.

Notation

If $y = f(x)$, all of these mean the derivative:

Notation	Read as
$f'(x)$	“f prime of x”
$y'$	“y prime”
$\dfrac{dy}{dx}$	“dee y dee x”: the rate of change of $y$ with respect to $x$
$\dfrac{d}{dx}\big[f(x)\big]$	“the derivative of $f(x)$ with respect to $x$”
$f'(3)$ or $\left.\dfrac{dy}{dx}\right\rvert_{x = 3}$	the derivative evaluated at $x = 3$

The tangent line

The tangent line at $x = a$ passes through $(a, f(a))$ and has slope $f'(a)$. In point-slope form:

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

On the AP exam, it’s fine to leave a tangent line in point-slope form.

Spotting a derivative in disguise

AP questions often hide a derivative inside a limit. For example,

$$\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}$$

matches $\displaystyle\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ with $f(x) = x^2$ and $a = 3$, so it equals $f'(3)$. Match the pattern: what is being plugged in, and what number is $a$?

Estimating a derivative from a table or graph

With only a table of values, you can’t take a limit. Instead, estimate $f'(a)$ with the slope of a secant line through nearby points. The best choice is usually the closest points on both sides of $a$:

$$f'(a) \approx \frac{f(a + h) - f(a - h)}{2h}$$

This is called a symmetric difference quotient. If $a$ isn’t in the middle of two table values, just use the two closest values that surround it.

From a graph, estimate the derivative by sketching the tangent line at the point and finding its slope (rise over run) using two points on that line.

Worked examples

Example 1: A derivative at a point

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

Solution. With the $h$ form:

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

Check with the other form, factoring the difference of squares:

$$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 6$$

Example 2: The derivative function

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

Solution. First expand $f(x + h)$:

$$f(x + h) = 2(x + h)^2 - 5(x + h) = 2x^2 + 4xh + 2h^2 - 5x - 5h$$

Subtract $f(x) = 2x^2 - 5x$. Everything without an $h$ cancels:

$$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{4xh + 2h^2 - 5h}{h} \\ &= \lim_{h \to 0} (4x + 2h - 5) \\ &= 4x - 5 \end{aligned}$$

For example, the slope of the graph at $x = 2$ is $f'(2) = 3$.

Example 3: A tangent line

The derivative of $f(x) = x^2 - 4x + 5$ is $f'(x) = 2x - 4$ (you can check this with the definition, just like Example 2). Find the equation of the tangent line at $x = 3$.

Solution. You need a point and a slope.

• Point: $f(3) = 9 - 12 + 5 = 2$, so the point is $(3, 2)$.
• Slope: $f'(3) = 2(3) - 4 = 2$.

$$y - 2 = 2(x - 3) \quad\text{or}\quad y = 2x - 4$$

The tangent line $y = 2x - 4$ touches $f(x) = x^2 - 4x + 5$ at $(3, 2)$.

Example 4: Estimating from a table

A cup of coffee cools. Its temperature $H(t)$, in °C, after $t$ minutes is shown below.

$t$ (min)	$0$	$4$	$6$	$10$	$12$
$H(t)$ (°C)	$90$	$78$	$73$	$65$	$62$

Estimate $H'(5)$ and $H'(8)$, and explain what $H'(5)$ means.

Solution. The closest times around $t = 5$ are $4$ and $6$:

$$H'(5) \approx \frac{H(6) - H(4)}{6 - 4} = \frac{73 - 78}{2} = -2.5 \text{ °C/min}$$

Around $t = 8$, use $6$ and $10$:

$$H'(8) \approx \frac{H(10) - H(6)}{10 - 6} = \frac{65 - 73}{4} = -2 \text{ °C/min}$$

At $t = 5$ minutes, the coffee’s temperature is decreasing at about $2.5$ °C per minute.

Common mistakes

Forgetting that $f(a + h)$ means “replace every $x$ with $a + h$”. For $f(x) = 2x^2 - 5x$, $f(x + h)$ is $2(x + h)^2 - 5(x + h)$, not $2x^2 - 5x + h$.

Dropping the limit symbol too early. Keep writing $\displaystyle\lim_{h \to 0}$ on every line until you actually let $h \to 0$. AP graders look for correct notation.

Using $f'(a)$ where $f(a)$ belongs in the tangent line. The point on the line is $(a, f(a))$; the slope is $f'(a)$. Mixing them up is one of the most common tangent-line errors.

Picking far-away table values to estimate a derivative. Use the two closest values that surround the point. For $H'(5)$ above, $\tfrac{H(12) - H(0)}{12}$ gives a much worse estimate.

Saying "$f'(5) = -2.5$" for an estimate. A value from a table is an approximation. Write $\approx$, and include units in context.

Misreading a disguised derivative. In $\displaystyle\lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h}$, the function is $\sqrt{x}$ and $a = 9$ (not $a = 3$, which is $\sqrt{9}$).

Practice

1. (Warm-up) The limit $\displaystyle\lim_{h \to 0} \frac{5(1 + h)^2 - 5}{h}$ is $f'(a)$ for some function $f$ and number $a$. Name $f$ and $a$, then evaluate the limit.

Solution

$f(x) = 5x^2$ and $a = 1$ (note $f(1) = 5$).

(Another valid answer is $f(x) = 5(1 + x)^2$ with $a = 0$. It gives the same limit.)

$$\lim_{h \to 0} \frac{5(1 + 2h + h^2) - 5}{h} = \lim_{h \to 0} \frac{10h + 5h^2}{h} = \lim_{h \to 0} (10 + 5h) = 10$$

2. (Warm-up) Suppose $f(4) = 7$ and $f'(4) = -2$. Write the equation of the tangent line to $f$ at $x = 4$.

Solution

$$y - 7 = -2(x - 4) \quad\text{or}\quad y = -2x + 15$$

3. (Warm-up) A runner’s position is $s(t)$ metres from the start after $t$ seconds. What does $s'(4) = -3$ mean?

Solution

At $t = 4$ seconds, the runner’s position is decreasing at $3$ metres per second. In other words, the runner is moving back toward the start at $3$ m/s at that instant.

4. (Core) Use the definition to find $f'(x)$ for $f(x) = x^2 + 3x - 1$.

Solution

$$\begin{aligned} f(x + h) - f(x) &= (x + h)^2 + 3(x + h) - 1 - (x^2 + 3x - 1) \\ &= 2xh + h^2 + 3h \end{aligned}$$$$f'(x) = \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h} = \lim_{h \to 0} (2x + h + 3) = 2x + 3$$

5. (Core) Use the form $\displaystyle\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ to find $f'(2)$ for $f(x) = x^3$. (Hint: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$.)

Solution

$$f'(2) = \lim_{x \to 2} \frac{x^3 - 8}{x - 2} = \lim_{x \to 2} (x^2 + 2x + 4) = 4 + 4 + 4 = 12$$

6. (Core) Let $f(x) = \dfrac{1}{x + 1}$.

• (a) Use the definition to find $f'(x)$.
• (b) Find the equation of the tangent line at $x = 1$.

Solution

(a) Combine the fractions:

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

(b) $f(1) = \dfrac{1}{2}$ and $f'(1) = -\dfrac{1}{4}$:

$$y - \frac{1}{2} = -\frac{1}{4}(x - 1) \quad\text{or}\quad y = -\frac{1}{4}x + \frac{3}{4}$$

7. (Core) Use the table to estimate $g'(2)$ and $g'(2.75)$.

$x$	$1$	$1.5$	$2$	$2.5$	$3$
$g(x)$	$4.0$	$4.9$	$6.1$	$7.6$	$9.4$

Solution

Around $x = 2$, use $x = 1.5$ and $x = 2.5$ (a symmetric difference quotient):

$$g'(2) \approx \frac{7.6 - 4.9}{2.5 - 1.5} = 2.7$$

Around $x = 2.75$, use $x = 2.5$ and $x = 3$:

$$g'(2.75) \approx \frac{9.4 - 7.6}{3 - 2.5} = \frac{1.8}{0.5} = 3.6$$

8. (Challenge) For $f(x) = x^2$, the derivative is $f'(x) = 2x$. Find the equations of both tangent lines to $y = x^2$ that pass through the point $(2, 3)$, which is not on the curve.

Solution

The tangent line at $x = a$ goes through $(a, a^2)$ with slope $2a$:

$$y - a^2 = 2a(x - a) \quad\Rightarrow\quad y = 2ax - a^2$$

It must pass through $(2, 3)$:

$$3 = 4a - a^2 \quad\Rightarrow\quad a^2 - 4a + 3 = 0 \quad\Rightarrow\quad (a - 1)(a - 3) = 0$$

For $a = 1$: $y = 2x - 1$. For $a = 3$: $y = 6x - 9$.

Check: $2(2) - 1 = 3$ and $6(2) - 9 = 3$.

9. (Challenge) The function $f$ is continuous at $x = 2$, and $\displaystyle\lim_{x \to 2} \frac{f(x) - 5}{x - 2} = 3$. Find $f(2)$ and $f'(2)$, and write the tangent line at $x = 2$.

Solution

The denominator approaches $0$, so for the limit to be a finite number the numerator must also approach $0$: $\displaystyle\lim_{x \to 2} f(x) = 5$. Since $f$ is continuous at $2$, $f(2) = 5$.

Now the limit has the form $\displaystyle\lim_{x \to 2} \frac{f(x) - f(2)}{x - 2}$, which is $f'(2)$. So $f'(2) = 3$.

$$y - 5 = 3(x - 2) \quad\text{or}\quad y = 3x - 1$$