Module 36 of 46Lesson 1 of 12 · AA 5.1 — Limits and the derivative as a gradient and rate of change

Course map · Topic 5: Calculus

5.1 Limits and the derivative as a gradient and rate of change

SL & HL11 worked questions

The derivative tells you how steep a curve is at one point, or how fast something is changing at one instant. It comes from a limit: the gradient of a chord as the second point slides closer and closer to the first.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[3 marks]

Find

\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}

Show the full step-by-step solution

Step 1🧠 ThinkWhy we cannot just put x = 3

If we put

x=3

we get

\dfrac{0}{0}

, which tells us nothing. A limit asks what the value gets close to, so we tidy the fraction up first.

Step 2✍️ By handFactorise the top (difference of two squares) and cancel

\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}

=x+3\quad (x\ne3)

\displaystyle\lim_{x\to3}(x+3)=6

✅ Check it At home you can check this on the GDC:

\dfrac{3.01^2-9}{3.01-3}=6.01

, very close to 6.

Final answer

6

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

The points

P(3,9)

and

Q(3+h,(3+h)^2)

lie on the curve

y=x^2

, where

h\ne0

(a) Show that the gradient of the chord $PQ$ is $6+h$ .[3]
(b) Hence write down the gradient of the curve at $P$ .[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkGradient = rise ÷ run

The chord is a straight line joining two points of the curve. Its gradient is the change in

y

divided by the change in

x

Step 2✍️ By handSubstitute the two points

m_{PQ}=\dfrac{(3+h)^2-9}{(3+h)-3}

=\dfrac{9+6h+h^2-9}{h}=\dfrac{6h+h^2}{h}

=\dfrac{h(6+h)}{h}=6+h

Part (b)

Step 1🧠 ThinkSlide Q towards P

Q

moves towards

P

h\to0

, and the chord becomes the tangent at

P

Step 2✍️ By hand

\displaystyle\lim_{h\to0}(6+h)=6

Final answer (a) shown (b) gradient at

P

6

Question 3●○○ Warm-up✍️ By hand · Paper 1 · no calculator[3 marks]

Water is drained from a tank in Francistown. The volume of water in the tank is

V

litres,

t

minutes after the tap is opened. When

t=5

\dfrac{dV}{dt}=-12

(a) Interpret $\dfrac{dV}{dt}=-12$ in context, including units.[2]
(b) State whether the volume is increasing or decreasing at $t=5$ . Give a reason.[1]

Show the full step-by-step solution

Step 1🧠 ThinkRead the notation aloud

\dfrac{dV}{dt}

is “the rate of change of

V

with respect to

t

”: how many litres the volume changes by per minute, at that exact moment.

Part (a)

Step 1✍️ By hand

t=5

minutes, the volume of water is changing at

-12

litres per minute,A1

i.e. the tank is losing 12 litres per minute at that instant.A1

Part (b)

Step 1✍️ By hand

Decreasing, because

\dfrac{dV}{dt}\lt 0

.R1

Final answer (a) At

t=5

the volume is falling at 12 litres per minute. (b) Decreasing, since

\dfrac{dV}{dt}\lt 0

⚠️ Common mistake Saying “12 litres are left”. The derivative is a rate, not an amount.

Question 4●●○ Standard🖩 GDC · Paper 2[4 marks]

Let

f(x)=x^3

. The gradient of the chord from

x=2

x=2+h

\dfrac{f(2+h)-f(2)}{h}

(a) Complete a table of chord gradients for $h=0.1,\ 0.01,\ 0.001$ .[3]
(b) Hence estimate $f'(2)$ .[1]

Show the full step-by-step solution

Step 1🧠 ThinkThe idea

Each value of

h

gives a chord that is a bit closer to the tangent at

x=2

. If the gradients settle down to one number, that number is the derivative.

Part (a)

Step 1🖩 On the GDCLet the spreadsheet build the table

ctrl+doc💻 Ctrl+I›=((2+h)^3-8)/h›enter💻 Enter

Add a Lists & Spreadsheet page. Name column A h and type 0.1, 0.01, 0.001. Name column B grad and in its grey formula cell type the chord formula (

f(2)=8

chord gradients 12.61, 12.0601, 12.006

Part (b)

Step 1✍️ By handSee where the numbers are heading

12.61,\ 12.0601,\ 12.006,\ \ldots\to 12

f'(2)\approx12

✅ Check it Later (5.3) you will learn

f'(x)=3x^2

, and

3\times2^2=12

. ✔

Final answer (a)

12.61,\ 12.0601,\ 12.006

(b)

f'(2)\approx 12

Question 5●●○ Standard🖩 GDC · Paper 2[3 marks]

Use your GDC to estimate

\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}

, where

x

is in radians.

Show the full step-by-step solution

Step 1🧠 ThinkWhy the GDC?

x=0

the fraction is

\dfrac{0}{0}

and it does not factorise. So we look at values of

x

close to 0 on both sides and see what the fraction approaches. Set the angle mode to Radian first (doc → Settings → Document Settings → Angle).

Step 2🖩 On the GDCDefine g and evaluate near 0

doc💻 File menu›Define g(x)=sin(x)/x›enter💻 Enter›g(0.1)›enter💻 Enter›g(0.01)›enter💻 Enter›g(-0.01)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ sin, cos, tan: click trig on the keypad, or type sin(

values on both sides approach 1

Step 3✍️ By handConclude

From both sides the values approach 1, soM1

\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1

Final answer

1

⚠️ Common mistake Working in degrees. In degree mode the values approach

\dfrac{\pi}{180}\approx0.01745

, which is not what IB means.

Question 6●●○ Standard✍️ By hand · Paper 1 · no calculator[4 marks]

The diagram shows the graph of

y=f(x)

and the tangent to the curve at

P(2,2)

. The tangent passes through

(1,0)

and

(3,4)

. The curve has its lowest point at

x=0

(a) Find $f'(2)$ .[2]
(b) Write down the value of $f'(0)$ .[1]
(c) State whether $f'(-1)$ is positive, negative or zero.[1]

Show the full step-by-step solution

Step 1🧠 Thinkf′(2) is the gradient of the tangent at x = 2

We cannot measure the curve's steepness directly, but the tangent is a straight line, so we use two points on it.

Part (a)

Step 1✍️ By hand

f'(2)=\dfrac{4-0}{3-1}

=2

Part (b)

Step 1✍️ By hand

At the lowest point the tangent is horizontal.

f'(0)=0

Part (c)

Step 1✍️ By hand

x=-1

the curve is going downhill (from left to right).

f'(-1)

is negative.A1

Final answer (a)

2

(b)

0

Question 7●●○ Standard🖩 GDC · Paper 2[3 marks]

Let

f(x)=x^3-4x

. Use your GDC to find the gradient of the curve at

x=1.5

, and state what this tells you about the curve there.

Show the full step-by-step solution

Step 1🧠 ThinkGradient of the curve = derivative at that point

The GDC can find the gradient at a point in two ways: on a Graphs page (dy/dx) or on a Calculator page (d/dx template).

Step 2🖩 On the GDCRoute 1: dy/dx on the graph

menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx›1.5›enter💻 Enter

Graph f1(x) = x³ − 4x. Choose dy/dx, click the curve, type 1.5 and press enter.

gradient 2.75 at x = 1.5

Step 3🖩 On the GDCRoute 2: the d/dx template on a Calculator page

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…›x›tab💻 Tab›1.5›enter💻 Enter›x^3-4x›enter💻 Enter

f′(1.5) = 2.75

Step 4✍️ By handInterpret

f'(1.5)=2.75

Positive, so the curve is increasing (going uphill) at

x=1.5

.R1

Final answer

f'(1.5)=2.75

; the curve is increasing there.

Question 8●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

Let

f(x)=x^3

. The points

A(1,1)

and

B(1+h, f(1+h))

lie on the curve,

h\ne0

(a) Show that the gradient of $AB$ is $3+3h+h^2$ .[4]
(b) Hence find $f'(1)$ .[1]
(c) Write down the gradient of the chord $AB$ when $h=0.1$ .[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkExpand (1 + h)³ carefully

Use the binomial expansion (Pascal's row 1, 3, 3, 1):

(1+h)^3=1+3h+3h^2+h^3

Step 2✍️ By hand

m_{AB}=\dfrac{(1+h)^3-1}{h}

=\dfrac{1+3h+3h^2+h^3-1}{h}

=\dfrac{3h+3h^2+h^3}{h}

=3+3h+h^2

Part (b)

Step 1✍️ By hand

f'(1)=\displaystyle\lim_{h\to0}(3+3h+h^2)=3

Part (c)

Step 1✍️ By hand

3+0.3+0.01=3.31

Final answer (a) shown (b)

3

(c)

3.31

Question 9●●● Exam-level🖩 GDC · Paper 2[7 marks]

Neo makes a cup of tea. Its temperature,

T

°C,

t

minutes later is

T(t)=20+60e^{-0.1t}

t\ge0

(a) Write down the initial temperature.[1]
(b) Find the average rate of change of $T$ for $0\le t\le10$ .[2]
(c) Find the rate of change of $T$ when $t=5$ .[2]
(d) Interpret the sign of your answer to (c).[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By hand

T(0)=20+60e^{0}=80\,^\circ\text{C}

Part (b)

Step 1🧠 ThinkAverage rate = chord gradient

Change in temperature ÷ change in time between

t=0

and

t=10

Step 2🖩 On the GDCDefine T and calculate

Define t1(x)=20+60e^(-0.1x)›enter💻 Enter›(t1(10)-t1(0))/10›enter💻 Enter

average rate −3.79272

Step 3✍️ By hand

\dfrac{T(10)-T(0)}{10-0}

=-3.79

°C per minuteA1

Part (c)

Step 1🧠 ThinkInstantaneous rate = derivative at t = 5

Use the d/dx template at x = 5 (the calculator uses x for the variable).

Step 2🖩 On the GDCd/dx template at x = 5

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…›x›tab💻 Tab›5›enter💻 Enter›t1(x)›enter💻 Enter

T′(5) = −3.63918

Step 3✍️ By hand

T'(5)=-3.64

°C per minuteA2

Part (d)

Step 1✍️ By hand

Negative, so the tea is cooling (its temperature is decreasing)R1

by about 3.64 °C per minute at

t=5

.A1

Final answer (a)

80

°C (b)

-3.79

°C/min (c)

-3.64

°C/min (d) the tea is cooling at about 3.64 °C per minute.

⚠️ Common mistake Giving the average rate in (c). “At

t=5

” means one instant, so it is the derivative.

Question 10●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

Let

f(x)=\dfrac{1}{x}

x\ne0

(a) Show that $\dfrac{f(2+h)-f(2)}{h}=-\dfrac{1}{2(2+h)}$ , for $h\ne0$ .[4]
(b) Hence find the gradient of the curve $y=f(x)$ at $x=2$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkFractions inside a fraction

First subtract

\dfrac{1}{2+h}-\dfrac{1}{2}

using a common denominator

2(2+h)

. Then divide by

h

Step 2✍️ By hand

f(2+h)-f(2)=\dfrac{1}{2+h}-\dfrac{1}{2}

=\dfrac{2-(2+h)}{2(2+h)}=\dfrac{-h}{2(2+h)}

Dividing by

h

\dfrac{-h}{2(2+h)}\times\dfrac{1}{h}

=-\dfrac{1}{2(2+h)}

Part (b)

Step 1✍️ By hand

\displaystyle\lim_{h\to0}\left(-\dfrac{1}{2(2+h)}\right)=-\dfrac{1}{2\times2}

=-\dfrac{1}{4}

✅ Check it At home you can check this on the GDC: d/dx(

\dfrac{1}{x}

) at x = 2 gives −0.25.

Final answer (a) shown (b)

-\dfrac{1}{4}

Question 11●●● Exam-level🖩 GDC · Paper 2[7 marks]

A car leaves a junction in Gaborone. Its distance from the junction after

t

seconds is

s(t)=0.6t^2+2t

metres,

0\le t\le10

(a) Find the average speed of the car between $t=2$ and $t=5$ .[2]
(b) Find the gradient of the chord from $t=4$ to $t=4+h$ for $h=0.1,\ 0.01,\ 0.001$ .[3]
(c) Hence estimate the speed of the car at $t=4$ and verify your answer with the GDC.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkAverage speed = distance travelled ÷ time taken

This is the gradient of the chord on the distance–time graph.

Step 2✍️ By hand

s(5)=15+10=25,\quad s(2)=2.4+4=6.4

\dfrac{25-6.4}{5-2}

=6.2\text{ m s}^{-1}

Part (b)

Step 1🖩 On the GDCChord gradients in a spreadsheet

ctrl+doc💻 Ctrl+I›=(0.6(4+h)^2+2(4+h)-17.6)/h›enter💻 Enter

s(4)=0.6\times16+8=17.6

. Column A h: 0.1, 0.01, 0.001. In the grey cell of column B type the formula with 17.6 in place of

s(4)

6.86, 6.806, 6.8006

Part (c)

Step 1✍️ By hand

The gradients approach 6.8, so the speed at

t=4

is about

6.8\text{ m s}^{-1}

Step 2🖩 On the GDCVerify with dy/dx on the graph

menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx›4›enter💻 Enter

gradient 6.8 at t = 4

Final answer (a)

6.2\text{ m s}^{-1}

(b)

6.86,\ 6.806,\ 6.8006

(c)

6.8\text{ m s}^{-1}

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

A limit $\displaystyle\lim_{x\to a} f(x)$ is the value $f(x)$ gets closer and closer to as $x$ gets closer to $a$ (from both sides). The function does not need to be defined at $a$ .
Gradient of a chord from $x=a$ to $x=a+h$ : $\dfrac{f(a+h)-f(a)}{h}$ . As $h\to 0$ , the chord turns into the tangent.
The derivative $f'(a)$ = gradient of the tangent at $x=a$ = $\displaystyle\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$ .
Notation: $\dfrac{dy}{dx}$ and $f'(x)$ mean the same thing — the rate of change of $y$ with respect to $x$ .
Rate of change: $\dfrac{dV}{dt}=-12$ means $V$ is decreasing by 12 units of $V$ per unit of $t$ at that instant. Units = (units of $V$ ) per (unit of $t$ ).
Average rate of change over an interval = chord gradient $\dfrac{f(b)-f(a)}{b-a}$ . Instantaneous rate = derivative.
A limit of the form $\dfrac{0}{0}$ is not “undefined”: simplify first (factorise and cancel), then substitute.

Derivative from first principles

f'(x)=\displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}

📘 In the formula booklet

Gradient of a chord

m=\dfrac{f(b)-f(a)}{b-a}

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Gradient at a point on a Graphs page: choose dy/dx, click the curve, type the x-value

menu💻 click menu on keypad›6: Analyze Graph›5: dy/dx›1.5›enter💻 Enter

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 5: dy/dx.
Press click on the curve, type the x-value and press enter. The gradient at that point is shown.
Type 1.5.
Press enter.

The menu path

gradient 2.75 at x = 1.5

Worked example: Question 7 — Route 1: dy/dx on the graph

Graph f1(x) = x³ − 4x. Choose dy/dx, click the curve, type 1.5 and press enter.

Derivative at a point on a Calculator page (pastes the d/dx template)

menu💻 click menu on keypad›4: Calculus›1: Numerical Derivative at a Point…

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 4: Calculus.
Choose 1: Numerical Derivative at a Point….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.

The menu path

f′(1.5) = 2.75

Worked example: Question 7 — Route 2: the d/dx template on a Calculator page

Table of chord gradients: Lists & Spreadsheet, formula in the grey cell

ctrl+doc💻 Ctrl+I›=((2+h)^3-8)/h›enter💻 Enter

Press ctrl then doc to add a new page.
Choose =((2+h)^3-8)/h.
Press enter.

chord gradients 12.61, 12.0601, 12.006

Worked example: Question 4 — Let the spreadsheet build the table

Add a Lists & Spreadsheet page. Name column A h and type 0.1, 0.01, 0.001. Name column B grad and in its grey formula cell type the chord formula (

f(2)=8

Get a decimal instead of an exact value

ctrl+enter💻 Ctrl+Enter

Press ctrl then enter to get the answer as a decimal.

values on both sides approach 1

Worked example: Question 5 — Define g and evaluate near 0

⚠️ Where students lose marks

Substituting $x=a$ straight away into a $\dfrac{0}{0}$ limit and writing “undefined”. Factorise and cancel first.
Mixing up the average rate (chord, over an interval) with the instantaneous rate (tangent, at one point).
Forgetting the sign when interpreting: a negative derivative means the quantity is decreasing.
Leaving out units in a rate of change, e.g. writing “−12” instead of “−12 litres per minute”.
Using degree mode for trig limits — calculus with trig always uses radians.

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL AA 5.1 — Limits and the derivative as a gradient and rate of change

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

5.1 Limits and the derivative as a gradient and rate of change

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks