SL 5.1 Interpret limits and derivatives as gradients and rates of change | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

IB Mathematics: Analysis and Approaches SL — Study topic SL 5.1

A graph can show how one quantity changes as another changes. Between two points, its gradient describes an average change. At a single point, the tangent gradient describes the local change there. A limit explains how the average gradient approaches this tangent gradient as the two points move together. This lesson develops that idea using familiar algebra, tables, graphs, and a motion context. The focus is interpretation: what a gradient or rate means, how it is found, and how its units help explain an answer.

What you will learn

Distinguish the gradient of a secant from the gradient of a tangent.
Interpret a derivative as the limit of average rates of change.
Use a derivative to describe an instantaneous rate of change, including its units and meaning.
Connect algebraic calculations with numerical, graphical, and contextual interpretations.

1. Prior knowledge: gradients and average change

For two points on a graph, gradient is the change in the vertical coordinate divided by the change in the horizontal coordinate. If the points are

(x_1,y_1)

and

(x_2,y_2)

, their gradient is

(y_2-y_1)/(x_2-x_1)

, provided

x_2\ne x_1

. This is the gradient of the straight line through the points, called a secant.

If

y=f(x)

, changing the input from

a

to

a+h

changes the output from

f(a)

to

f(a+h)

. The average rate of change over that input interval is the output change divided by the input change. The number

h

is the size of the input change; it can be positive or negative, but it cannot be zero in this quotient.

Units follow the same division. If

f(x)

is measured in metres and

x

in seconds, a gradient is measured in metres per second. Before interpreting any rate, identify what the horizontal and vertical quantities represent.

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

A secant gradient uses two distinct points.
Average rate of change is change in output divided by change in input.
The units of a gradient are output units per input unit.

2. From secant gradients to a derivative

To describe change at exactly one input, consider a point

a

and a nearby point

a+h

. Their secant gradient is the average rate over that small interval. As

h

approaches zero, the nearby point approaches

a

. If the secant gradients approach a single finite value, that value is the gradient of the tangent at

x=a

.

The derivative of

f

at

a

, written

f'(a)

, is this limiting gradient. The limit notation means that we examine values of the quotient for nonzero

h

getting closer and closer to zero. It does not mean that we substitute

h=0

into the quotient, which would make its denominator zero.

On a graph, the tangent is the straight line that matches the curve’s direction at the point locally. Its gradient describes the derivative. A positive derivative means the graph is increasing locally; a negative derivative means it is decreasing locally; a zero derivative means the tangent is horizontal. These statements describe local behaviour and do not by themselves describe the entire graph.

A graphing calculator can support the interpretation. Plot the curve, zoom near the point, and use a tangent or numerical derivative feature if available. The displayed value is an estimate affected by calculator settings and rounding. Check that it agrees with secant gradients for nearby points, and keep the limit definition as the mathematical reason for the result.

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

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

A derivative is the limit of nearby secant gradients, when that limit exists.
The derivative is the tangent gradient and the instantaneous rate of change.
A calculator helps visualize and estimate; it does not replace the limit reasoning.

3. Reading rates in context

When a function models a situation, its derivative gives an instantaneous rate. For example, if position

s

is measured in metres and time

t

in seconds, the derivative at a time has units metres per second and represents instantaneous velocity. It is different from the average velocity across a time interval, which uses the total position change divided by the total time change.

The same interpretation applies to other quantities: the derivative’s units are always the units of the output divided by the units of the input. Include the point or input value and state whether the rate is positive, negative, or zero. A negative rate means the output is decreasing as the input increases; it does not mean that the output itself is negative.

A numerical table can reveal the approach to a derivative. Calculate secant gradients from points progressively nearer to the chosen input, using points on both sides when available. If these values settle near the same number, that supports the graphical and algebraic interpretation of the tangent gradient. Rounding and the choice of nearby points can affect an estimate, so retain enough precision during calculations.

\text{rate units}=\frac{\text{output units}}{\text{input units}}

Instantaneous rate and average rate refer to different intervals.
State the rate’s units and what is changing with respect to what.
Nearby numerical secant gradients can support an estimate of a derivative.

Nearby secant gradients for $f(x)=x^2$ at $x=1$

Nearby input $x$	Secant interval	Secant gradient
$0.9$	$0.9$ to $1$	$1.9$
$1.1$	$1$ to $1.1$	$2.1$
$0.99$	$0.99$ to $1$	$1.99$
$1.01$	$1$ to $1.01$	$2.01$

Worked example

Finding a tangent gradient from the limit

For

f(x)=x^2

, find the derivative at

x=3

using the limit definition, and interpret the result as a gradient.

Form the secant gradient
Use the point $x=3$ and a nearby point $x=3+h$ . Their output values are $9$ and $(3+h)^2$ . The quotient describes the average change between these two points.
$\frac{(3+h)^2-9}{h}$
Simplify for nonzero h
Expand the square and collect terms. Factoring out $h$ allows cancellation because the quotient is considered for $h\ne 0$ .
$\frac{9+6h+h^2-9}{h}=6+h$
Take the limit
As $h$ approaches zero, $6+h$ approaches $6$ . Therefore the tangent to the curve at $x=3$ has gradient $6$ .
$f'(3)=\lim_{h\to0}(6+h)=6$

Answer: The derivative at

x=3

is

6

, so the tangent gradient there is

6

.

Check: For small positive

h

, the secant gradient

6+h

is slightly greater than

6

; for small negative

h

, it is slightly less. Both approach

6

.

Worked example

Interpreting instantaneous velocity

A moving object has position

s(t)=t^2+2t

metres, where

t

is measured in seconds. Find its instantaneous velocity at

t=2

using a limit, and compare it with the average velocity from

t=2

to

t=2.5

.

Set up the nearby average velocity
The position at $t=2$ is $8$ metres. For a nearby time $2+h$ , the position is $(2+h)^2+2(2+h)$ . Divide the position change by the time change $h$ .
$\frac{(2+h)^2+2(2+h)-8}{h}$
Find the limiting rate
Expanding and simplifying gives $6+h$ . As the time interval shrinks toward zero, this approaches $6$ . Since position is in metres and time is in seconds, the rate is in metres per second.
$\lim_{h\to0}(6+h)=6\text{ m/s}$
Calculate the separate interval average
At $t=2.5$ , the position is $11.25$ metres. Over the $0.5$ -second interval, position increases by $3.25$ metres, giving an average velocity of $6.5$ metres per second.
$\frac{11.25-8}{2.5-2}=6.5\text{ m/s}$

Answer: The instantaneous velocity at

t=2

is

6

m/s. The average velocity from

t=2

to

t=2.5

is

6.5

m/s.

Check: The average is greater than the instantaneous value because the nearby secant gradient for this model is

6+h

, and here

h=0.5

.

Worked example

Estimating a derivative from nearby values

For

f(x)=x^2+1

, estimate the derivative at

x=1

using secant gradients from the left and right, then connect the estimate to a graphing-calculator check.

Choose nearby inputs
Use $x=0.9$ and $x=1.1$ , close to $1$ . The function values are $1.81$ , $2$ , and $2.21$ at these inputs respectively.
$f(0.9)=1.81,\quad f(1)=2,\quad f(1.1)=2.21$
Find left and right secant gradients
The left secant uses $0.9$ and $1$ ; the right secant uses $1$ and $1.1$ . Their gradients are close, suggesting that the tangent gradient is near $2$ .
$\frac{2-1.81}{1-0.9}=1.9,\quad \frac{2.21-2}{1.1-1}=2.1$
Check and interpret
Plot the curve and inspect it near $x=1$ ; a tangent or numerical derivative tool should give a value close to $2$ . The paired secant estimates approach this value as the chosen inputs move closer to $1$ . The exact value can also be confirmed by applying the limit definition.
$f'(1)=\lim_{h\to0}\frac{(1+h)^2+1-2}{h}=2$

Answer: The nearby secant gradients estimate the derivative at

x=1

as about

2

; the exact value is

2

.

Check: The left and right estimates,

1.9

and

2.1

, lie on opposite sides of

2

, consistent with the exact limit.

Common mistakes and how to avoid them

Substituting

h=0

into the difference quotient before simplifying.

Correction: The quotient is defined for nonzero

h

. Simplify it first, then consider what value it approaches as

h

tends to zero.

Calling a secant gradient an instantaneous rate.

Correction: A secant uses two distinct inputs and gives an average rate. The derivative is the limiting gradient as the inputs come together.

Giving a rate without units or without saying what it describes.

Correction: State the output units per input unit and interpret the sign at the specified input.

Assuming a calculator’s tangent display is the explanation.

Correction: Use the display to check a graph or numerical estimate, and explain the result through secant gradients and their limit.

Lesson summary

The gradient between two points is change in output divided by change in input.
A derivative at a point is the limit of nearby secant gradients and gives the tangent gradient.
In context, a derivative is an instantaneous rate of change with units.
Tables and graphing technology can support an estimate; algebraic limit reasoning explains why the estimate is the derivative.

Check your understanding

Question 1

For

f(x)=x^2

, the secant gradient from

x=2

to

x=2+h

is

4+h

. What is the limiting gradient at

x=2

?

2
4
$4+h$
0

Show answer and explanation

4

As

h

approaches zero,

4+h

approaches

4

, so the tangent gradient is

4

.

Question 2

A graph of a quantity against time has a tangent gradient of

-3

at a particular time. Which interpretation is correct?

The quantity is negative at that time.
The quantity is decreasing at 3 units per time unit at that instant.
The quantity has a total value of 3 units.
The average rate over every time interval is negative 3.

Show answer and explanation

The quantity is decreasing at 3 units per time unit at that instant.

The derivative describes instantaneous change. Its negative sign indicates a local decrease in the graph’s vertical quantity as time increases.

Question 3

What does the derivative represent on the graph of a function at an input where it exists?

The vertical coordinate of the point
The gradient of the tangent at that point
The gradient of every secant through that point
The horizontal coordinate of the point

Show answer and explanation

The gradient of the tangent at that point

The derivative at an input is the limiting secant gradient, interpreted geometrically as the tangent gradient.

Key terms

Secant: A straight line through two points on a curve.
Tangent gradient: The local gradient of a curve at a point, represented by the tangent line there.
Limit: The value that an expression approaches as its input approaches a specified value.
Derivative: The limiting gradient of nearby secants at a point; it represents the instantaneous rate of change when interpreted in context.

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SL 5.1 · Interpret limits and derivatives as gradients and rates of change

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — Study topic SL 5.1

What you will learn

1. Prior knowledge: gradients and average change

2. From secant gradients to a derivative

3. Reading rates in context

Nearby secant gradients for $f(x)=x^2$ at $x=1$

Finding a tangent gradient from the limit

Interpreting instantaneous velocity

Estimating a derivative from nearby values

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review