Introduction to differentiation AI HL

IB Mathematics: Applications and Interpretation HL · Unit 5: Calculus

Lesson 5.1 · Introduction to Differentiation

Limits, gradients of curves, instantaneous rates of change and their meaning in context.

Go and share the good news

Jesus told his disciples, “Go ye into all the world, and preach the gospel to every creature” (Mark 16:15, KJV). His command reaches beyond familiar places and people. The good news is that Jesus Christ died for our sins and rose again; through him, people are invited to repent, believe and receive life in his name.

In Acts 8:26–40, Philip obeyed God’s direction, met an Ethiopian traveller and explained the Scripture to him. Philip began where the man was and “preached unto him Jesus” (Acts 8:35, KJV). We, too, can listen carefully, explain the gospel clearly and share it with every person God places in our path. As we study how a quantity changes at a particular moment, consider how a timely conversation can open an opportunity to tell someone about Christ.

Learning objectives

By the end of this lesson, you should be able to:

estimate a limit from a table or graph and distinguish it from the value of the function at that point;
calculate and interpret an average rate of change from two points;
estimate the gradient of a tangent and interpret it as an instantaneous rate of change;
recognise $f'(x)$ as the gradient function and use appropriate derivative notation;
use a graphical display calculator (GDC) to estimate a derivative, stating units and meaning.

1. What is a limit?

A limit describes the value a function approaches as its input moves closer to a particular number. We inspect values from both sides of that number. For example, let $g(x)=(x^2-9)/(x-3)$ for $x\neq3$ .

$x$	$2.9$	$2.99$	$2.999$	$3.001$	$3.01$	$3.1$
$g(x)$	$5.9$	$5.99$	$5.999$	$6.001$	$6.01$	$6.1$

As $x$ approaches $3$ , the output approaches $6$ . We write:

$\lim_{x\to3}g(x)=6$

Here $g(3)$ is undefined because the denominator would be zero. The limit still exists: a limit concerns nearby inputs, not necessarily the function’s value at the point. For this lesson, estimate limits from tables or graphs; formal analytic methods for evaluating limits are not required.

2. Average rate of change · A secant line

For points $(a,f(a))$ and $(b,f(b))$ on a curve, the straight line through them is a secant. Its gradient is the average rate of change over the interval:

$\text{average rate of change}=\frac{f(b)-f(a)}{b-a}$

If $s(t)$ measures displacement in metres and $t$ measures time in seconds, the average rate has units of metres per second.

Worked example · A moving object.

The displacement of an object is $s(t)=t^2+2t$ metres. Between $t=2$ and $t=4$ seconds, $s(2)=8$ m and $s(4)=24$ m. Therefore:

$\frac{s(4)-s(2)}{4-2}=\frac{24-8}{2}=8\text{ m s}^{-1}$

The object’s average velocity over these two seconds is $8\text{ m s}^{-1}$ . This does not tell us its exact velocity at every moment in the interval.

3. Instantaneous rate of change · A tangent line

To estimate the rate at $t=3$ , compare the displacement at $3$ seconds with displacement at a nearby time $3+h$ . As $h$ gets closer to zero, the secant gradient approaches the gradient of the tangent at $t=3$ :

$\frac{s(3+h)-s(3)}{h}=\frac{(3+h)^2+2(3+h)-15}{h}=8+h\qquad(h\neq0)$

$h$ (s)	$-0.1$	$-0.01$	$0.01$	$0.1$
Secant gradient ( $\text{m s}^{-1}$ )	$7.9$	$7.99$	$8.01$	$8.1$

The gradients approach $8$ from both sides. The derivative of $s$ at $t=3$ is $8\text{ m s}^{-1}$ , the instantaneous velocity at that time. Informally:

$s'(3)=\left.\frac{\mathrm{d}s}{\mathrm{d}t}\right|_{t=3}=8\text{ m s}^{-1}$

We can write the general idea as $f'(a)=\lim_{h\to0}[f(a+h)-f(a)]/h$ when that limit exists. Here it expresses the meaning of a derivative; formal limit calculations are not the focus of this lesson.

Reading derivatives correctly

Notation	Meaning	Typical interpretation
$f'(x)$ or $\dfrac{\mathrm{d}y}{\mathrm{d}x}$	Derivative or gradient function	Gradient at each suitable value of $x$
$f'(a)$	Derivative at $x=a$	Gradient of one tangent; one instantaneous rate
$\dfrac{\mathrm{d}s}{\mathrm{d}t}$	Rate of change of displacement with time	Velocity, in distance per unit time
$\dfrac{\mathrm{d}V}{\mathrm{d}t}$	Rate of change of volume with time	Volume per unit time

Positive derivative: the function is increasing locally. Negative derivative: it is decreasing locally. Zero derivative: the tangent is horizontal; a turning point is possible, but not guaranteed.

4. The gradient function and the tangent

For $s(t)=t^2+2t$ , the same nearby-time calculation at a general time gives a tangent gradient of $2t+2$ . Thus:

$s'(t)=2t+2$

The derivative is itself a function: at $t=1$ the instantaneous velocity is $4\text{ m s}^{-1}$ ; at $t=3$ it is $8\text{ m s}^{-1}$ . A tangent through $(a,f(a))$ with gradient $m=f'(a)$ has equation:

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

Worked example · Gradient and tangent equation.

For $f(x)=x^2-4x+1$ , use a GDC or a nearby-point table to estimate $f'(3)=2$ . Since $f(3)=-2$ , the tangent at $x=3$ passes through $(3,-2)$ :

$y+2=2(x-3)\quad\Longrightarrow\quad y=2x-8$

Check: the tangent and curve have the same height and gradient at $x=3$ .

Using a TI-84 Plus CE

Enter the function in Y= and set a window around the input you are investigating.
Press MATH → 8:nDeriv( and enter the function, variable and input, for example nDeriv(Y1,X,3). You can insert Y1 from VARS → Y-VARS → Function.
Alternatively, on the graph use 2nd → TRACE (CALC) → 6:dy/dx, then enter the required $x$ -value.
Interpret the result with its units. If the question asks for a tangent equation, also calculate the corresponding function value.

A numerical derivative is an approximation. If a graph has a sharp corner or a jump, inspect it carefully: the derivative at that point may not exist.

IB-style Paper 1 · Skill practice

Show your method and give units where appropriate. Use a GDC unless a question asks you to use a table. Mark allocations are shown in square brackets.

Question 1 · Limit from a table [4 marks]

For $x\neq3$ , let $g(x)=(x^2-9)/(x-3)$ . Values of $g(x)$ at $x=2.99$ , $2.999$ , $3.001$ and $3.01$ are $5.99$ , $5.999$ , $6.001$ and $6.01$ , respectively.

Estimate $\lim_{x\to3}g(x)$ . [2]
State whether $g(3)$ is defined and explain why your answer to part (a) is still possible. [2]

Show solution

(a) Values from both sides approach $6$ , so the estimated limit is $6$ . (b) $g(3)$ is undefined because the denominator is $0$ at $x=3$ . The limit concerns values arbitrarily close to $3$ , not the value at $3$ itself.

Question 2 · From secants to a tangent [7 marks]

Let $f(x)=x^2-3x+2$ .

Find $f(2)$ and show that $\dfrac{f(2+h)-f(2)}{h}=1+h$ for $h\neq0$ . [3]
Evaluate this secant gradient for $h=-0.01$ and $h=0.01$ . [2]
Estimate the tangent gradient at $x=2$ and find the equation of the tangent. [2]

Show solution

(a) $f(2)=0$ and $f(2+h)=(2+h)^2-3(2+h)+2=h+h^2$ ; dividing by $h$ gives $1+h$ . (b) The gradients are $0.99$ and $1.01$ . (c) They approach $1$ . The tangent passes through $(2,0)$ , so $y=x-2$ .

Question 3 · Gradient of a cubic [5 marks]

Let $p(x)=x^3-3x+1$ .

Use your GDC to estimate $p'(1)$ and $p'(2)$ . [2]
Find $p(1)$ and $p(2)$ . [1]
Explain what $p'(1)$ says about the tangent and find the tangent equation at $x=2$ . [2]

Show solution

(a) $p'(1)=0$ and $p'(2)=9$ (numerically). (b) $p(1)=-1$ and $p(2)=3$ . (c) The tangent at $(1,-1)$ is horizontal. At $x=2$ the tangent has gradient $9$ and passes through $(2,3)$ , so $y-3=9(x-2)$ , or $y=9x-15$ .

Question 4 · Average and instantaneous velocity [5 marks]

The displacement of a particle after $t$ seconds is $s(t)=4t+0.5t^2$ metres, $t\geq0$ .

Find the average velocity between $t=2$ and $t=4$ . [2]
Estimate its instantaneous velocity at $t=3$ using your GDC. [2]
State the units of both answers. [1]

Show solution

$s(2)=10$ m and $s(4)=24$ m. (a) Average velocity $(24-10)/(4-2)=7\text{ m s}^{-1}$ . (b) The numerical derivative at $t=3$ is $s'(3)=7\text{ m s}^{-1}$ . (c) Both are measured in metres per second. They happen to agree here because the interval is centred at $t=3$ for this quadratic model.

IB-style Paper 2 · Applications

Question 5 · Cooling liquid [8 marks]

The temperature of a liquid is modelled by $T(t)=20+60(0.8)^t$ degrees Celsius, where $t$ is the time in minutes after cooling begins, $0\leq t\leq10$ .

Find the initial temperature and the temperature after $2$ minutes. [2]
Calculate the average rate of change of temperature over the first $2$ minutes. [2]
Use your GDC to estimate $T'(2)$ , giving your answer to three significant figures. [2]
Interpret the sign and units of $T'(2)$ in this context. [2]

Show solution

(a) $T(0)=80^\circ\mathrm C$ and $T(2)=20+60(0.8)^2=58.4^\circ\mathrm C$ . (b) $(58.4-80)/2=-10.8^\circ\mathrm C\text{ per minute}$ . (c) A numerical derivative gives $T'(2)\approx-8.57^\circ\mathrm C\text{ per minute}$ . (d) At exactly $2$ minutes, the modelled temperature is falling at about $8.57^\circ\mathrm C$ per minute. The negative sign indicates cooling.

Question 6 · Marginal manufacturing cost [8 marks]

The cost of producing $q$ units in one day is modelled by $C(q)=500+12q+0.04q^2$ euros, for $0\leq q\leq100$ .

Calculate the average change in cost per additional unit from $q=50$ to $q=60$ . [2]
Use your GDC to find $C'(50)$ and state its meaning and units. [3]
Use $C'(50)$ to estimate the additional cost of making ten units after the first $50$ . Compare your estimate with the actual modelled additional cost. [3]

Show solution

(a) $C(50)=€1200$ and $C(60)=€1364$ , so the average change is $(1364-1200)/10=€16.40$ per unit. (b) $C'(50)=16$ euros per unit; near a production level of $50$ , the cost increases by about €16 for one additional unit. (c) The tangent estimate for ten additional units is $10(16)=€160$ . The actual modelled change is €164. The €4 difference arises because the rate increases as $q$ increases.

Question 7 · Interpreting measurements [6 marks]

Water is pumped into a tank. The table shows the volume $V$ , in litres, after $t$ minutes.

$t$ (min)	$4.8$	$4.9$	$5.0$	$5.1$	$5.2$
$V$ (L)	$192.16$	$196.04$	$200.00$	$204.04$	$208.16$

Calculate the average rate of change of volume from $t=4.8$ to $t=5.0$ , and from $t=5.0$ to $t=5.2$ . [3]
Use your answers to estimate $V'(5)$ , with units. [2]
Explain what your estimate describes physically. [1]

Show solution

(a) Before $t=5$ : $(200-192.16)/0.2=39.2$ L/min. After $t=5$ : $(208.16-200)/0.2=40.8$ L/min. (b) These nearby gradients bracket an estimate of $V'(5)\approx40.0$ L/min; equivalently, the centred estimate is $(204.04-196.04)/0.2=40.0$ L/min. (c) At $t=5$ minutes, the amount of water in the tank is increasing at approximately $40$ litres per minute.

Exam checklist

For an average rate, use two function values and divide by the change in input. For an instantaneous rate, find the tangent gradient or a numerical derivative at one input. Examine both sides when estimating a limit. State what your answer means, give its units, and check that it lies within the model’s domain.

Lesson 5.1 · Introduction to Differentiation

Go and share the good news

Learning objectives

1. What is a limit?

2. Average rate of change · A secant line

3. Instantaneous rate of change · A tangent line

Reading derivatives correctly

4. The gradient function and the tangent

Using a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Limit from a table [4 marks]

Question 2 · From secants to a tangent [7 marks]

Question 3 · Gradient of a cubic [5 marks]

Question 4 · Average and instantaneous velocity [5 marks]

IB-style Paper 2 · Applications

Question 5 · Cooling liquid [8 marks]

Question 6 · Marginal manufacturing cost [8 marks]

Question 7 · Interpreting measurements [6 marks]

Exam checklist

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Introduction to differentiation AI HL

Go and share the good news

Learning objectives

1. What is a limit?

2. Average rate of change · A secant line

3. Instantaneous rate of change · A tangent line

Reading derivatives correctly

4. The gradient function and the tangent

Using a TI-84 Plus CE

IB-style Paper 1 · Skill practice

Question 1 · Limit from a table [4 marks]

Question 2 · From secants to a tangent [7 marks]

Question 3 · Gradient of a cubic [5 marks]

Question 4 · Average and instantaneous velocity [5 marks]

IB-style Paper 2 · Applications

Question 5 · Cooling liquid [8 marks]

Question 6 · Marginal manufacturing cost [8 marks]

Question 7 · Interpreting measurements [6 marks]

Exam checklist

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