The Derivative and Its Notation — AP® Calculus AB

Part 1 · Hook

Why this matters

A speedometer shows one number, how fast you are going right now, even though "right now" lasts no time at all. Behind that number is the limit from the last topic. Give it a name, the derivative, and a picture, the tangent line, and you have the central idea of the whole course.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The instantaneous rate of change of $f$ at $x = a$ is

$\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$
$\frac{f(a + h) - f(a)}{h}$ with $h = 0$
$\lim_{h \to 0} \left( f(a + h) - f(a) \right)$
$\frac{f(a)}{a}$

Show the answer

It is the limit of average rates on shorter and shorter intervals.

Correct: $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ :
$\frac{f(a + h) - f(a)}{h}$ with $h = 0$ :
$\lim_{h \to 0} \left( f(a + h) - f(a) \right)$ :
$\frac{f(a)}{a}$ :

2. A line passes through $(2, 5)$ with slope $-3$ . Its equation is

$y - 5 = -3(x - 2)$
$y - 2 = -3(x - 5)$
$y + 5 = -3(x + 2)$
$y = -3x + 5$

Show the answer

Point-slope form: $y - y_1 = m(x - x_1)$ .

Correct: $y - 5 = -3(x - 2)$ :
$y - 2 = -3(x - 5)$ :
$y + 5 = -3(x + 2)$ :
$y = -3x + 5$ :

3. For $x \ne 4$ , $\frac{\sqrt{x} - 2}{x - 4}$ equals

$\frac{1}{\sqrt{x} + 2}$
$\sqrt{x} + 2$
$\frac{1}{\sqrt{x} - 2}$
$\frac{1}{x + 4}$

Show the answer

Factor $x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$ and cancel.

Correct: $\frac{1}{\sqrt{x} + 2}$ :
$\sqrt{x} + 2$ :
$\frac{1}{\sqrt{x} - 2}$ :
$\frac{1}{x + 4}$ :

Part 4 · See it

See it first

The curve y = f(x) with a point at x = a and a second point at x = a + h. A dashed secant line joins them; the run h and the rise f(a + h) minus f(a) are marked. A solid tangent line touches the curve at x = a. The secant slope is the average rate on [a, a + h]; as h shrinks to 0 it approaches the tangent slope f prime of a. — The secant line through $(a, f(a))$ and $(a + h, f(a + h))$ has slope $\frac{f(a + h) - f(a)}{h}$ . As $h \to 0$ it turns into the tangent line at $x = a$ , whose slope is $f'(a)$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The limit $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ existsits value is called the derivative of $f$ at $a$ , written $f'(a)$
Secant lines through $(a, f(a))$ pivot as $h \to 0$ they approach the tangent line at $x = a$ , whose slope is $f'(a)$
The derivative is found at every input where the limit existsthe result is a new function, $f'(x)$ , also written $y'$ or $\frac{dy}{dx}$
You know $f(a)$ and $f'(a)$ the tangent line is $y = f(a) + f'(a)(x - a)$

Part 6 · Key ideas

Key ideas

The derivative of $f$ at $a$ is $f'(a)$ $= \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ , or equivalently $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ . Both are the same limit of secant slopes.
The tangent line at $x = a$ passes through $(a, f(a))$ with slope $f'(a)$ : $y = f(a) + f'(a)(x - a)$ .
Done at every $x$ , the limit gives the derivative function $f'(x)$ . In Leibniz notation, $y = f(x)$ has derivative $\frac{dy}{dx}$ , and $\frac{d}{dx}$ means "the derivative of".
A derivative has the units of a rate: output units per input unit.

Part 7 · Misconception

A common mistake

The wrong idea: $\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}$ is a strange limit to evaluate by plugging in, so it is $\frac{0}{0}$ and undefined.

What actually happens: Read its shape. It is $\lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}$ with $f(x) = x^2$ , so it is $f'(3)$ , the slope of $y = x^2$ at $x = 3$ . Simplify and it equals 6. A limit of a difference quotient is a derivative at a point.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A curve and one of its tangent lines

The graph shows

y = f(x)

, where

f(x) = 4 - \frac{(x - 1)^2}{2}

, and the line tangent to it at

x = 3

.

1. Use the tangent line in the graph to find $f'(3)$ .

$-2$
$2$
$-\frac{1}{2}$
$8$

Show the answer

$f'(3)$ is the slope of the tangent line at $x = 3$ . It passes through $(2, 4)$ and $(4, 0)$ : slope $\frac{0 - 4}{4 - 2} = -2$ .

Correct: $-2$ : Rise $-4$ over run 2 on the tangent line.
$2$ : That is $f(3)$ , the height of the point. The derivative is a slope, and this line falls.
$-\frac{1}{2}$ : That is run over rise, upside down.
$8$ : That is where the tangent line crosses the $y$ -axis, not its slope.

2. Which is an equation of the line tangent to the graph of $f$ at $x = 3$ ?

$y = 2 - 2(x - 3)$
$y = -2 + 2(x - 3)$
$y = 2 - 2(x + 3)$
$y = 2 - 2x$

Show the answer

Point $(3, f(3)) = (3, 2)$ and slope $f'(3) = -2$ : $y = f(3) + f'(3)(x - 3)$ .

Correct: $y = 2 - 2(x - 3)$ : It uses the point $(3, 2)$ and the slope $-2$ .
$y = -2 + 2(x - 3)$ : The value and the slope are swapped: $f(3) = 2$ is the height and $f'(3) = -2$ the slope.
$y = 2 - 2(x + 3)$ : Point-slope form subtracts the $x$ -coordinate: $x - 3$ , not $x + 3$ .
$y = 2 - 2x$ : At $x = 3$ this line gives $-4$ , not 2: it misses the point of tangency.

3. Which limit equals the slope of the tangent line shown?

$\lim_{x \to 3} \frac{f(x) - 2}{x - 3}$
$\lim_{x \to 3} \left( f(x) - 2 \right)$
$\lim_{x \to 3} \frac{f(x)}{x}$
$\lim_{h \to 0} \frac{f(3 + h) - f(h)}{h}$

Show the answer

Since $f(3) = 2$ , this is $\lim_{x \to 3} \frac{f(x) - f(3)}{x - 3}$ $= f'(3)$ , the second form of the definition.

Correct: $\lim_{x \to 3} \frac{f(x) - 2}{x - 3}$ : Secant slopes from $(3, 2)$ to $(x, f(x))$ , with $x \to 3$ .
$\lim_{x \to 3} \left( f(x) - 2 \right)$ : Without dividing by $x - 3$ this is a change in height; its limit is 0.
$\lim_{x \to 3} \frac{f(x)}{x}$ : A height over an input is the slope of a line through the origin, not a tangent slope.
$\lim_{h \to 0} \frac{f(3 + h) - f(h)}{h}$ : Both points must be near $x = 3$ : the numerator needs $f(3)$ , not $f(h)$ .

4. Which correctly orders $f'(-1)$ , $f'(1)$ and $f'(3)$ ?

$f'(3) < f'(1) < f'(-1)$
$f'(-1) < f'(1) < f'(3)$
$f'(-1) = f'(3) < f'(1)$
$f'(1) < f'(3) < f'(-1)$

Show the answer

At $x = -1$ the graph rises (slope 2), at $x = 1$ its tangent line is level (slope 0), and at $x = 3$ it falls (slope $-2$ ).

Correct: $f'(3) < f'(1) < f'(-1)$ : Falling, level, rising: $-2 < 0 < 2$ .
$f'(-1) < f'(1) < f'(3)$ : This orders them backward; the graph rises on the left and falls on the right.
$f'(-1) = f'(3) < f'(1)$ : This compares heights: $f(-1)$ $= f(3)$ $= 2 < f(1)$ $= 4$ . Derivatives compare slopes.
$f'(1) < f'(3) < f'(-1)$ : At the top of the graph the slope is 0, which is greater than the negative slope at $x = 3$ .

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

Type an expression for $f'(x) =$ .

Show the answer

$\frac{f(x + h) - f(x)}{h}$ $= \frac{3x^2 + 6xh + 3h^2 - x - h - 3x^2 + x}{h}$ $= 6x$ $+ 3h - 1 \to 6x - 1$ .

Answer: $f'(x) =$ $6x - 1$

6. What does $\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}$ represent?

$f'(3)$ for $f(x) = x^2$ , which equals 6
$f(3)$ for $f(x) = x^2$ , which equals 9
$f'(9)$ for $f(x) = x^2$ , which equals 18
The average rate of change of $x^2$ on $[3, 3 + h]$ , which equals $6 + h$

Show the answer

Since $9 = 3^2$ , the limit is $\lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}$ with $f(x) = x^2$ : the derivative at 3. It equals $\lim_{h \to 0} (6 + h)$ $= 6$ .

Correct: $f'(3)$ for $f(x) = x^2$ , which equals 6: The limit of the difference quotient of $x^2$ at $a = 3$ .
$f(3)$ for $f(x) = x^2$ , which equals 9: The 9 is $f(3)$ , subtracted inside the quotient; the whole limit is a slope.
$f'(9)$ for $f(x) = x^2$ , which equals 18: The point is $a = 3$ , from $3 + h$ ; the 9 is the output $f(3)$ .
The average rate of change of $x^2$ on $[3, 3 + h]$ , which equals $6 + h$ : That is the quotient before the limit; the limit sends $h$ to 0.

7. The line $y = 4x - 7$ is tangent to the graph of $g$ at $x = 3$ . What are $g(3)$ and $g'(3)$ ?

$g(3) = 5$ and $g'(3) = 4$
$g(3) = 4$ and $g'(3) = 5$
$g(3) = -7$ and $g'(3) = 4$
$g(3) = 5$ and $g'(3) = 3$

Show the answer

The tangent line touches the graph at $x = 3$ , so $g(3) = 4(3) - 7 = 5$ , and its slope is $g'(3) = 4$ .

Correct: $g(3) = 5$ and $g'(3) = 4$ : The point of tangency and the slope of the line.
$g(3) = 4$ and $g'(3) = 5$ : The value and the slope are swapped.
$g(3) = -7$ and $g'(3) = 4$ : That is the line's $y$ -intercept; the graph meets the line at $x = 3$ , not $x = 0$ .
$g(3) = 5$ and $g'(3) = 3$ : The 3 is the input; the slope is the coefficient of $x$ .

Part 9 · Summary

Summary

The derivative

f'(a)

is the limit of the difference quotient as

h \to 0

: the instantaneous rate of change of

f

at

a

and the slope of the tangent line there. Found at every

x

, it is the function

f'(x) = \frac{dy}{dx}

. The tangent line at

a

is

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

.

Part 10 · Up next

What comes next

Part 11 · Connections