2.2 - Defining the Derivative of a Function and Using Derivative Notation

What is a Derivative?

A derivative is an expression that calculates the instantaneous rate of change of a function. In other words, it gives you the slope of the tangent line to a function at any given

x

-value.

Visual Understanding

Consider the function

f(x) = x^2

, whose derivative is

f'(x) = 2x

2.2 - Defining the Derivative of a Function and Using Derivative Notation

At $x = 1$ :

f'(1) = 2(1) = 2

, so the tangent line has slope 2.

At $x = 2$ :

f'(2) = 2(2) = 4

, so the tangent line has slope 4.

At $x = -2$ :

f'(-2) = 2(-2) = -4

, so the tangent line has slope -4.

Key insight: The derivative function tells you the slope of the original function at any point you choose.

Interpreting Derivatives

Example 1: Function Values vs. Derivative Values

Given only the derivative

g'(x) = \frac{3}{x - 1}

, find

g'(3)

g'(3) = \frac{3}{3 - 1} = \frac{3}{2}

Interpretation: The slope of the tangent line to

g(x)

x = 3

\frac{3}{2}

Notice we never needed to know what

g(x)

actually is. The derivative gives us the slope directly.

Example 2: Real-World Context

Let

f(x)

represent the distance in meters you have run, where

x

is time in minutes.

Interpret $f(8) = 1200$ :

After 8 minutes, I have run 1200 meters. This is a function value, so it tells us position at a specific time.

Interpret $f'(5) = 180$ :

At the 5-minute mark, I was running at a rate of 180 meters per minute. This is a derivative value, so it tells us the rate of change (speed) at a specific instant.

Critical distinction: -

f(x)

gives position -

f'(x)

gives the rate of change of position (velocity)

The derivative always represents a rate, so the units should include “per”: meters per minute, miles per hour, and so on.

Derivative Notation

There are several ways to write derivatives, and you should recognize all of them.

Notation	Name	Read as
$f'(x)$	Lagrange notation	“f prime of x”
$y'$	Prime notation	“y prime”
$\frac{dy}{dx}$	Leibniz notation	“dee y dee x” or “the derivative of y with respect to x”

When to Use Each

Lagrange notation

f'(x)

: when your function is written as

f(x) = \ldots

Prime notation

y'

: when your function is written as

y = \ldots

Leibniz notation

\frac{dy}{dx}

: emphasizes that you’re finding how

y

changes with respect to

x

. Later you’ll see variations like

\frac{dr}{dt}

, which is how the radius changes with respect to time.

All three mean the same thing: the derivative.

At a specific point: to write the derivative at one

x

-value, plug that value in, like

f'(3)

. In Leibniz notation you’ll see

\left.\frac{dy}{dx}\right|_{x = 3}

, read “dy/dx at

x = 3

.” Both mean the slope of the tangent line at

x = 3

The

\frac{d}{dx}

symbol:

\frac{d}{dx}

by itself means “take the derivative, with respect to

x

, of whatever comes next.” So

\frac{d}{dx}[x^2] = 2x

just says the derivative of

x^2

2x

. You’ll see this a lot starting with the power rule in Topic 2.5.

The Definition of the Derivative

The derivative of a function

f

at any point

x

is defined as:

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

What it’s saying: - Start at the point

(x, f(x))

- Move a tiny distance

h

to the right, to

(x + h, f(x + h))

- Calculate the slope between those two points - Let

h

approach zero, making the points infinitely close

This works provided the limit exists. That’s the difference from Topic 2.1: there,

a

was one fixed number, so the limit gave one slope,

f'(a)

. Here,

x

stays a variable, so the limit gives a whole new function,

f'

, that outputs the slope of

f

at every

x

where the limit exists. Those

x

-values make up the domain of

f'

, which can be smaller than the domain of

f

(more on that in Topic 2.4).

From Secant Line to Tangent Line

Secant line (red): connects two points, slope

= \frac{f(x+h) - f(x)}{h}

Tangent line (green): touches at one point, slope

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

h \to 0

, the secant line approaches the tangent line.

Finding Derivatives Using the Definition

Example 3: Polynomial Function

Find the derivative of

f(x) = 3x^2 - 5x + 2

using the definition.

Step 1: Write the definition

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

Step 2: Find

f(x + h)

f(x + h) = 3(x + h)^2 - 5(x + h) + 2

= 3(x^2 + 2xh + h^2) - 5x - 5h + 2

= 3x^2 + 6xh + 3h^2 - 5x - 5h + 2

Step 3: Set up the difference quotient

f'(x) = \lim_{h \to 0} \frac{[3x^2 + 6xh + 3h^2 - 5x - 5h + 2] - [3x^2 - 5x + 2]}{h}

Step 4: Simplify the numerator

= \lim_{h \to 0} \frac{6xh + 3h^2 - 5h}{h}

Step 5: Factor out

h

= \lim_{h \to 0} \frac{h(6x + 3h - 5)}{h}

Step 6: Cancel

h

= \lim_{h \to 0} (6x + 3h - 5)

Step 7: Evaluate the limit

= 6x + 0 - 5 = 6x - 5

Answer:

f'(x) = 6x - 5

This derivative gives the slope of

f(x)

at any

x

-value.

Example 4: Rational Function

Find the derivative of

y = \frac{1}{x^2}

using the definition.

Step 1: Set up the definition

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

Step 2: Substitute

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

Step 3: Find a common denominator in the numerator

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

Step 4: Expand

(x + h)^2

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

Step 5: Multiply by the reciprocal of

h

= \lim_{h \to 0} \frac{-2xh - h^2}{x^2(x + h)^2} \cdot \frac{1}{h} = \lim_{h \to 0} \frac{-2xh - h^2}{hx^2(x + h)^2}

Step 6: Factor

h

from the numerator

= \lim_{h \to 0} \frac{h(-2x - h)}{hx^2(x + h)^2}

Step 7: Cancel

h

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

Step 8: Evaluate the limit

= \frac{-2x - 0}{x^2(x + 0)^2} = \frac{-2x}{x^4} = \frac{-2}{x^3}

Answer:

\frac{dy}{dx} = -\frac{2}{x^3}

Good News: Shortcuts Are Coming

The process above is algebraically brutal. The good news is that you’ll learn shortcut rules soon that make finding derivatives way faster.

So why learn the long way first? - Understanding the limit definition helps you recognize derivative expressions on exams - It connects what you learned about limits to the concept of derivatives - Some AP exam questions ask you to identify the original function from a limit expression

For the rest of the year: you’ll use shortcut rules, not this lengthy limit process.

Equation of a Tangent Line

Finding the equation of a tangent line is one of the most common applications of derivatives.

Point-Slope Form

y - y_1 = m(x - x_1)

where

(x_1, y_1)

is a point on the line and

m

is the slope.

For Tangent Lines

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

where: -

a

is the

x

-value where you want the tangent line -

f(a)

is the

y

-coordinate of the point on the curve -

f'(a)

is the slope of the tangent line, from the derivative

Three things you need: 1. The

x

-value:

a

2. The

y

-value:

f(a)

3. The slope:

f'(a)

Example 5: Writing a Tangent Line Equation

The point

(4, -3)

lies on the curve

g(x)

, and

g'(4) = 5

. Write the equation of the tangent line at

x = 4

What we have: -

a = 4

, the

x

-value -

g(4) = -3

, the

y

-value -

g'(4) = 5

, the slope

Apply point-slope form:

y - (-3) = 5(x - 4) \quad \Rightarrow \quad y + 3 = 5(x - 4)

Answer:

y + 3 = 5(x - 4)

Note: Unless you’re specifically asked to solve for

y

(slope-intercept form), you can leave your answer in point-slope form.

Example 6: The Complete Process

Given

h(x) = \frac{x^3 - 1}{x}

and

h'(x) = \frac{2x^3 + 1}{x^2}

, find the equation of the tangent line at

x = 2

Step 1: Find the

y

-coordinate

h(2) = \frac{(2)^3 - 1}{2} = \frac{8 - 1}{2} = \frac{7}{2}

Step 2: Find the slope

h'(2) = \frac{2(2)^3 + 1}{(2)^2} = \frac{2(8) + 1}{4} = \frac{17}{4}

Step 3: Write the equation

y - \frac{7}{2} = \frac{17}{4}(x - 2)

Answer:

y - \frac{7}{2} = \frac{17}{4}(x - 2)

Reading Derivative Graphs

Sometimes you’re given the graph of the derivative

f'(x)

instead of the original function

f(x)

Key Understanding

If you’re looking at the graph of

f'(x)

: - The

y

-values on that graph tell you slopes of

f(x)

- To find

f'(3)

, look at the point where

x = 3

on the derivative graph

Example 7: Using a Derivative Graph

The graph above shows

f'(x)

, the derivative. Given that

f(3) = 8

, write the equation of the tangent line to

f(x)

x = 3

Step 1: We already know

f(3) = 8

, since it’s given.

Step 2: Find the slope from the derivative graph. Look at

x = 3

on the

f'(x)

graph. The

y

-value there is

-2

, so

f'(3) = -2

Step 3: Write the tangent line equation

y - 8 = -2(x - 3)

Answer:

y - 8 = -2(x - 3)

Important: When you’re reading from a derivative graph, remember you’re reading slope values, not function values.

Four Ways to Represent a Derivative

A derivative can show up in four forms, and AP questions switch between them all the time. They all describe the same thing: how fast the function is changing.

Representation	What it looks like	Example
Graphical	The slope of the tangent line on a graph of $f$ , or the $y$ -value on a graph of $f'$	Example 7: $f'(3) = -2$ , read from the graph of $f'$
Numerical	A table of values, where you estimate the slope from nearby points	Topic 2.3: $\frac{h(8) - h(2)}{8 - 2}$ estimates $h'(5)$
Analytical	A formula, from the definition or the derivative rules	Example 3: $f'(x) = 6x - 5$
Verbal	A sentence with units that describes a rate	Example 2: at the 5-minute mark, you’re running 180 meters per minute

Practice Problems

Problem 1

Find the derivative of

f(x) = 4x^2 - 3x

using the definition.

Answer:

f'(x) = \lim_{h \to 0} \frac{[4(x+h)^2 - 3(x+h)] - [4x^2 - 3x]}{h}

After simplification:

f'(x) = 8x - 3

Problem 2

g(5) = 12

and

g'(5) = -3

, write the equation of the tangent line at

x = 5

Answer:

y - 12 = -3(x - 5)

Problem 3

A car’s position in miles after

t

hours is given by

s(t)

. Interpret

s'(2) = 65

Answer: After 2 hours, the car is traveling at 65 miles per hour.

Key Takeaways

The derivative gives the slope of the tangent line at any point

Three notations, $f'(x)$ , $y'$ , and $\frac{dy}{dx}$ , all mean the same thing

Derivative = instantaneous rate of change

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

Derivatives represent rates, so include “per” in the units

Tangent line equation: $y - f(a) = f'(a)(x - a)$

You need three things for a tangent line: the $x$ -value, the $y$ -value, and the slope

When reading derivative graphs, $y$ -values represent slopes of the original function

Finding derivatives by definition is algebraically intensive, but shortcuts are coming

Understanding the definition helps you recognize derivative expressions on exams

Leave tangent line equations in point-slope form unless told otherwise

The derivative can be shown four ways: graphically, numerically, analytically, and verbally

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