DP Maths AA SL - Calculus - Derivatives and Their Applications

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DP Maths AA SL | Derivatives and Their Applications

Derivatives and Their Applications

This topic contains 2 notes:

5.6 Differentiation Rules
5.8 Applications of Differentiation

DP MATHEMATICS AA · TOPIC 5: CALCULUS

5.6

Differentiation Rules and the Second Derivative

Full notes for SL · Syllabus SL 5.6, 5.7

Contents

1	Derivatives of standard functions
2	The chain rule
3	The product rule
4	The quotient rule
5	Combining the rules
6	The second derivative
7	Concavity and points of inflexion
8	Chain rule: more practice
9	Product rule: more practice
10	Quotient rule: more practice
11	Derivatives of exponentials and logarithms
12	Derivatives of trigonometric functions
13	Choosing the rule
14	Second derivatives: more practice
15	Concavity and points of inflexion: more practice
16	Graphs of f, f′ and f″
17	Tangents and normals with harder functions
18	Mixed review: worked problems
19	Additional exam-style practice
20	Technology for checking derivatives
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
differentiate sin x, cos x, tan x, ex and ln x
use the chain rule for composite functions
use the product rule and the quotient rule
combine rules for complicated functions
find second derivatives and interpret them
use the second derivative to determine concavity and points of inflexion

This booklet covers SL 5.6, 5.7, which is common to SL and HL.

Guiding questions

How can complicated functions be differentiated by breaking them into simpler pieces? What does the second derivative reveal about the shape of a graph?

How this topic is examined

These rules are tested on every paper, usually inside longer questions: differentiate a product or composite, find a gradient, then use it for tangents, stationary points or kinematics. Second derivatives appear in classifying stationary points and finding inflexion points.

Identify the structure of the function (product, quotient, composite) before choosing a rule, and simplify the result where asked.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Derivatives of standard functions

Results to know.

Think about it

The graph of y = ex has a special property: at every point, the gradient equals the height. What does that say about its derivative?

f(x)	f′(x)
xn	nxn−1
sin x	cos x
cos x	−sin x
tan x	1/cos2x
ex	ex
ln x	1/x

These require x in radians. The derivative of ex is itself, which is why e is the natural base. The derivative of ln x, the inverse of ex, is 1/x (for x > 0).

Figure 1. The gradient of sin x at each point equals cos x.

Practice exercise 1

1. Differentiate y = 3 sin x − 4ex.

2. Differentiate y = 5 ln x + 2 cos x.

3. Find the gradient of y = tan x at x = π/4.

2 The chain rule

Differentiate the outside, multiply by the derivative of the inside.

Useful special cases: d/dx[(ax + b)n] = an(ax + b)n−1; d/dx[eg(x)] = g′(x)eg(x); d/dx[ln g(x)] = g′(x)/g(x); d/dx[sin(g(x))] = g′(x)cos(g(x)).

Why it works

A small change δx causes a change δu in u, which causes a change δy in y. Then δy/δx = (δy/δu)(δu/δx). Letting δx → 0 (so δu → 0) gives dy/dx = (dy/du)(du/dx).

WORKED EXAMPLE 1 Chain rule (basic)

Differentiate (a) y = (3x2 − 1)5; (b) y = e−2x; (c) y = ln(x2 + 4).

Solution

(a) 5(3x2 − 1)4 × 6x = 30x(3x2 − 1)4A1

(b) −2e−2xA1

WORKED EXAMPLE 2 Trig and roots (standard)

Differentiate (a) y = sin3x; (b) y = √(1 + cos 2x).

Solution

(a) 3sin2x cos xA1

(b) ½(1 + cos 2x)−1/2 × (−2 sin 2x)M1

= −sin 2x/√(1 + cos 2x)A1

Common mistake

Forgetting to multiply by the derivative of the inside: d/dx(sin 3x) = 3 cos 3x, not cos 3x.

Practice exercise 2

1. Differentiate y = (2 − 5x)4.

2. Differentiate y = ex².

3. Differentiate y = cos(4x + 1).

4. Differentiate y = ln(3x).

5. Differentiate y = 1/(2x + 3)2.

3 The product rule

For y = uv.

WORKED EXAMPLE 3 Product rule (standard)

Differentiate y = x2e3x.

Solution

u = x2, v = e3x; u′ = 2x, v′ = 3e3xM1

dy/dx = 3x2e3x + 2xe3xA1

= xe3x(3x + 2)A1

WORKED EXAMPLE 4 Product with trig (standard)

Differentiate y = x sin 2x and find the gradient at x = π/4.

Solution

dy/dx = 2x cos 2x + sin 2xM1 A1

At π/4: 2(π/4)(0) + 1 = 1A1

Practice exercise 3

1. Differentiate y = x3 ln x.

2. Differentiate y = ex cos x.

3. Differentiate y = (x + 1)√(x − 2).

4 The quotient rule

For y = u/v.

WORKED EXAMPLE 5 Quotient rule (standard)

Differentiate y = (2x + 1)/(x2 + 3).

Solution

u′ = 2, v′ = 2xA1

dy/dx = (2(x2 + 3) − (2x + 1)(2x))/(x2 + 3)2M1

= (6 − 2x − 2x2)/(x2 + 3)2A1

WORKED EXAMPLE 6 Deriving the derivative of tan x (exam level)

Show that d/dx(tan x) = 1/cos2x.

Solution

tan x = sin x/cos xM1

(cos x cos x − sin x(−sin x))/cos2xA1

= (cos2x + sin2x)/cos2x = 1/cos2xAG

Common mistake

Reversing the order in the numerator: it is v u′ − u v′, with v first.

Practice exercise 4

1. Differentiate y = ln x/x.

2. Differentiate y = e2x/(x + 1).

3. Differentiate y = x/(1 − x).

5 Combining the rules

Work from the outside in.

WORKED EXAMPLE 7 Product with a chain (exam level)

Differentiate y = x2(3x − 1)4 and factorize.

Solution

2x(3x − 1)4 + x2 × 4(3x − 1)3 × 3M1 A1

= 2x(3x − 1)3[(3x − 1) + 6x]M1

= 2x(3x − 1)3(9x − 1)A1

Practice exercise 5

1. Differentiate y = e−xsin(2x).

2. Differentiate y = ln(x2 + 1)/x.

6 The second derivative

The rate of change of the gradient.

Differentiating f′(x) gives the second derivative f″(x), or d2y/dx2. It measures how the gradient is changing. In kinematics, if s is displacement, v = ds/dt and a = dv/dt = d2s/dt2 (5.8).

WORKED EXAMPLE 8 Second derivative (standard)

Find f″(x) for f(x) = x4 − 2x3 + 5x and for g(x) = xex.

Solution

f′ = 4x3 − 6x2 + 5; f″ = 12x2 − 12xA1

g′ = ex(x + 1); g″ = ex(x + 2)A1 A1

Practice exercise 6

1. Find d2y/dx2 for y = sin 3x.

2. Show that y = e2x satisfies d2y/dx2 − 4y = 0.

7 Concavity and points of inflexion

Which way the curve bends.

A graph is concave up where f″(x) > 0 (it bends upward, like a cup; the gradient is increasing) and concave down where f″(x) < 0. A point of inflexion is where the concavity changes: f″(x) = 0 and f″ changes sign. If, in addition, f′ = 0 there, it is a stationary (horizontal) point of inflexion.

Figure 2. y = x3 − 3x2 + 2 changes from concave down to concave up at its point of inflexion.

WORKED EXAMPLE 9 Inflexion point (standard)

Find the point of inflexion of y = x3 − 3x2 + 2.

Solution

y″ = 6x − 6 = 0, x = 1M1 A1

y″ changes sign at x = 1 (negative to positive)R1

Point (1, 0)A1

Common mistake

Assuming f″(x) = 0 always gives an inflexion. For y = x4, f″(0) = 0 but there is no change of concavity (it is a minimum).

Practice exercise 7

1. Find the intervals where y = x4 − 6x2 is concave up.

2. Find the points of inflexion of y = xe−x.

8 Chain rule: more practice

Outside derivative times inside derivative.

For y = f(g(x)), dy/dx = f′(g(x)) × g′(x). Identify the inside function first. Nested composites need the rule more than once.

WORKED EXAMPLE 10 Nested (exam level)

Differentiate y = sin2(3x).

Solution

y = (sin 3x)2: 2 sin 3x × cos 3x × 3M1 A1

= 6 sin 3x cos 3x = 3 sin 6xA1

WORKED EXAMPLE 11 Exponential (standard)

Differentiate y = e√x.

Solution

e√x × 1/(2√x)M1

= e√x/(2√x)A1

Practice exercise 8

1. Differentiate y = (x2 + 1)−3.

2. Differentiate y = ln(cos x).

3. Differentiate y = √(1 − 4x).

9 Product rule: more practice

u′v + uv′.

For y = uv, dy/dx = u′v + uv′. Factorize the result where possible: common factors such as ex or (x + 1)n make later steps (stationary points) easier.

WORKED EXAMPLE 12 Factorizing (standard)

Differentiate y = x3e2x and factorize.

Solution

3x2e2x + 2x3e2xM1 A1

= x2e2x(3 + 2x)A1

WORKED EXAMPLE 13 With a chain (exam level)

Differentiate y = x√(2x + 1).

Solution

√(2x + 1) + x/√(2x + 1)M1 A1

= (3x + 1)/√(2x + 1)A1

Practice exercise 9

1. Differentiate y = x2 ln x.

2. Differentiate y = ex sin x.

3. Differentiate y = (x + 2)3(x − 1).

10 Quotient rule: more practice

(vu′ − uv′)/v2.

For y = u/v, dy/dx = (vu′ − uv′)/v2. Simplify the numerator. Sometimes rewriting u/v as uv−1 and using the product rule is easier.

WORKED EXAMPLE 14 Quotient (standard)

Differentiate y = (x2 + 1)/(x − 2).

Solution

((x − 2)(2x) − (x2 + 1))/(x − 2)2M1 A1

= (x2 − 4x − 1)/(x − 2)2A1

Practice exercise 10

1. Differentiate y = ex/x.

2. Differentiate y = sin x/(1 + cos x).

3. Differentiate y = ln x/x2.

11 Derivatives of exponentials and logarithms

Special cases.

d/dx(ekx) = kekx; d/dx(ln(kx)) = 1/x; d/dx(ln g(x)) = g′(x)/g(x). Use the laws of logarithms to simplify before differentiating: ln(x2√(x + 1)) = 2 ln x + ½ ln(x + 1).

WORKED EXAMPLE 15 Simplify first (exam level)

Differentiate y = ln((x2 + 1)/√x).

Solution

y = ln(x2 + 1) − ½ ln xM1

dy/dx = 2x/(x2 + 1) − 1/(2x)A1 A1

Practice exercise 11

1. Differentiate y = ln(3x4).

2. Differentiate y = 5e−0.2x + ln(2x + 1).

12 Derivatives of trigonometric functions

Radians required.

d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = 1/cos2x. With a linear inside function, multiply by its coefficient: d/dx(cos(4x + 1)) = −4 sin(4x + 1).

WORKED EXAMPLE 16 Trig derivative (standard)

Differentiate y = 3 tan(2x) − cos2x.

Solution

6/cos2(2x) + 2 cos x sin xM1 A1

= 6/cos2(2x) + sin 2xA1

Practice exercise 12

1. Differentiate y = sin3x.

2. Find the gradient of y = x cos x at x = π.

13 Choosing the rule

Structure first.

Function	Structure	Rule
(3x + 1)5	function of a function	chain
x2ex	product	product
ex/(x + 1)	quotient	quotient
sin(x2)	composite	chain
x ln(x2 + 1)	product with a composite	product + chain

WORKED EXAMPLE 17 Combined (exam level)

Differentiate y = x2 sin(3x).

Solution

Product: 2x sin 3x + x2(3 cos 3x)M1 A1

= x(2 sin 3x + 3x cos 3x)A1

14 Second derivatives: more practice

Differentiate twice.

f″(x) is the derivative of f′(x). Simplify f′ before differentiating again to reduce errors.

WORKED EXAMPLE 18 Second derivative (standard)

Find f″(x) for f(x) = e−x cos x.

Solution

f′ = −e−x(cos x + sin x)A1

f″ = e−x(cos x + sin x) − e−x(cos x − sin x) = 2e−x sin xM1 A1

Practice exercise 13

1. Find d2y/dx2 for y = ln(x2 + 1).

2. Show that y = e2xsin x satisfies y″ − 4y′ + 5y = 0.

15 Concavity and points of inflexion: more practice

Sign changes of f″.

f is concave up where f″ > 0 and concave down where f″ < 0. A point of inflexion requires f″ = 0 and a sign change. Check with a sign diagram.

Figure 3. Maximum and inflexion of y = xe−x.

WORKED EXAMPLE 19 Inflexion (exam level)

Find the point of inflexion of y = xe−x.

Solution

y′ = e−x(1 − x); y″ = e−x(x − 2)A1 A1

y″ changes sign at x = 2: (2, 2e−2)R1 A1

Practice exercise 14

1. Find the intervals of concavity of f(x) = x4 − 4x3.

2. Find the inflexion points of y = ln(x2 + 1).

16 Graphs of f, f′ and f″

Reading information across graphs.

Zeros of f′ give stationary points of f; zeros of f″ (with sign change) give inflexion points of f, which are turning points of f′. If f′ is increasing, f is concave up.

WORKED EXAMPLE 20 Interpreting (standard)

The graph of f′ has a maximum at x = 2 and crosses zero at x = −1 and x = 5 (from + to − at 5). What can you say about f?

Solution

f has a local maximum at x = 5 and a local minimum at x = −1 (− to +)A1

f has an inflexion at x = 2A1

17 Tangents and normals with harder functions

Combining rules.

Use the appropriate rule to find the gradient, then write the tangent or normal through the point. Exact answers are expected on Paper 1.

WORKED EXAMPLE 21 Tangent (exam level)

Find the tangent to y = x ln x at x = e.

Solution

y(e) = e; y′ = ln x + 1 = 2A1 A1

y = 2x − eA1

Practice exercise 15

1. Find the normal to y = e2x at x = 0.

2. Find the tangent to y = sin 2x at x = π/4.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 22 Mixed problem (exam level)

f(x) = (ln x)/x, x > 0. (a) Find f′(x). (b) Find the maximum point. (c) Find f″(x) and the point of inflexion.

Solution

(a) (1 − ln x)/x2A1

(b) ln x = 1: (e, 1/e)A1

Practice exercise 16

1. Differentiate y = ex².

2. Find the stationary point of y = x2e−x other than the origin.

19 Additional exam-style practice

Paper 1.

WORKED EXAMPLE 23 Paper 1 question (exam level)

y = (2x + 3)4(x − 1). (a) Show that dy/dx = (2x + 3)3(10x − 5). (b) Find the x-coordinates of the stationary points.

Solution

(a) 8(2x + 3)3(x − 1) + (2x + 3)4 = (2x + 3)3(8x − 8 + 2x + 3)M1 A1 AG

(b) x = −3/2, x = 1/2A1 A1

Practice exercise 17

1. Find the gradient of y = (x2 + 1)/ex at x = 0.

20 Technology for checking derivatives

Numerical comparison.

Compare your derivative with the GDC's numerical derivative at one or two points. If they agree, your algebra is probably correct.

WORKED EXAMPLE 24 Checking (standard)

Check that d/dx(x2ex) = xex(x + 2) at x = 1.

Solution

Formula: 3e = 8.155; GDC: 8.155 ✓A1

GDC tip

Evaluate your f′(a) and nDeriv(f(x), x, a) at a convenient value such as a = 1.3; they should agree to several decimal places.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

Let f(x) = x2e−x.

(a) Show that f′(x) = x(2 − x)e−x. [3]

(b) Find the x-coordinates of the stationary points. [2]

(d) Hence find the x-coordinates of the points of inflexion. [2]

Markscheme

Part	Answer	Marks
(a)	2xe−x − x2e−x	M1 A1
	= x(2 − x)e−x	AG A1
(b)	x = 0, x = 2	A2
(c)	(2 − 2x)e−x − (2x − x2)e−x	M1 A1
	= (x2 − 4x + 2)e−x	AG A1
(d)	x2 − 4x + 2 = 0	M1
	x = 2 ± √2	A1

Examiner's comments. In (d), e−x is never zero, so only the quadratic factor matters.

Question 2 (Paper 1 style) (about 8 minutes)

Find the gradient of y = ln(x2 + 1)/(x + 1) at x = 0. [4]

Markscheme

Part	Answer	Marks
	quotient rule	M1
	numerator (x + 1)(2x/(x2 + 1)) − ln(x2 + 1)	A1
	at x = 0: (1)(0) − 0	M1
	gradient 0	A1

Examiner's comments. Substituting early simplifies the arithmetic.

22 Common misconceptions and links

Students often think…	Actually…
d/dx(sin 3x) = cos 3x	3 cos 3x
d/dx(uv) = u′v′	uv′ + vu′
f″ = 0 means inflexion	f″ must change sign
d/dx(ln 5x) = 5/x	1/x

How 5.6 connects to the rest of the course

Topic	Connection
5.1 Differentiation	power rule
5.8 Applications	optimization and kinematics
2.5 Composite functions	chain rule structure
5.14 Further differentiation (HL)	implicit differentiation

23 Check your understanding

Section A: multiple choice

1. d/dx(e3x) =

A. 3e3x B. e3x C. 3xe3x D. e3

2. d/dx(cos x) =

A. sin x B. −cos x C. −sin x D. cos x

3. d/dx(ln(2x + 1)) =

A. 1/(2x + 1) B. 2/(2x + 1) C. 2x + 1 D. ln 2

4. Concave up means

A. f′(x) > 0 B. f(x) > 0 C. f″(x) < 0 D. f″(x) > 0

5. The product rule gives d/dx(x sin x) =

A. cos x B. sin x + x cos x C. x cos x D. sin x − x cos x

Section B: structured

6. Differentiate y = e2xsin x. [3]

7. Find the point of inflexion of y = x3 + 3x2 − 9x. [3]

Answers

1. A 2. C 3. B 4. D 5. B

6. e2x(2 sin x + cos x).

7. y″ = 6x + 6 = 0: (−1, 11).

24 Summary and topic map

Idea	Remember
Standard	sin → cos, cos → −sin, ex → ex, ln x → 1/x
Chain rule	f′(g(x))g′(x)
Product rule	uv′ + vu′
Quotient rule	(vu′ − uv′)/v2
Second derivative	rate of change of gradient
Concavity	f″ > 0 up; f″ < 0 down
Inflexion	f″ = 0 and changes sign

Before the exam, make sure you…

•identify the structure first
•multiply by the inner derivative
•keep v first in the quotient rule
•check sign change for inflexion

Differentiation Rules on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 4. Topic map for Differentiation Rules.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 3 cos x − 4ex

2. 5/x − 2 sin x

3. 2

Practice exercise 2

1. −20(2 − 5x)3

2. 2xex²

3. −4 sin(4x + 1)

4. 1/x

5. −4(2x + 3)−3

Practice exercise 3

1. 3x2 ln x + x2

2. ex(cos x − sin x)

3. √(x − 2) + (x + 1)/(2√(x − 2)) = (3x − 3)/(2√(x − 2))

Practice exercise 4

1. (1 − ln x)/x2

2. e2x(2x + 1)/(x + 1)2

3. 1/(1 − x)2

Practice exercise 5

1. e−x(2 cos 2x − sin 2x)

2. [2x2/(x2 + 1) − ln(x2 + 1)]/x2

Practice exercise 6

1. −9 sin 3x

2. 4e2x − 4e2x = 0

Practice exercise 7

1. y″ = 12x2 − 12 > 0: x < −1 or x > 1

2. y″ = e−x(x − 2) = 0: (2, 2e−2)

Practice exercise 8

1. −6x(x2 + 1)−4

2. −tan x

3. −2/√(1 − 4x)

Practice exercise 9

1. 2x ln x + x

2. ex(sin x + cos x)

3. (x + 2)2(4x − 1)

Practice exercise 10

1. ex(x − 1)/x2

2. 1/(1 + cos x)

3. (1 − 2 ln x)/x3

Practice exercise 11

1. 4/x

2. −e−0.2x + 2/(2x + 1)

Practice exercise 12

1. 3sin2x cos x

2. cos π − π sin π = −1

Practice exercise 13

1. 2(1 − x2)/(x2 + 1)2

2. substitute y′ = e2x(2 sin x + cos x), y″ = e2x(3 sin x + 4 cos x)

Practice exercise 14

1. f″ = 12x(x − 2): concave up for x < 0 or x > 2

2. x = ±1: (±1, ln 2)

Practice exercise 15

1. y = −x/2 + 1

2. y = 1 (horizontal)

Practice exercise 16

1. 2xex²

2. (2, 4e−2)

Practice exercise 17

1. −1

DP MATHEMATICS AA · TOPIC 5: CALCULUS

5.8

Applications of Differentiation

Full notes for SL · Syllabus SL 5.8, 5.9

Contents

1	Stationary points
2	Classifying stationary points
3	Global maxima and minima on an interval
4	Sketching curves using derivatives
5	Optimization problems
6	Kinematics: displacement, velocity and acceleration
7	Distance travelled and speed
8	Stationary points of harder functions
9	Second derivative test: more practice
10	Global maximum and minimum on an interval
11	Optimization: geometric problems
12	Optimization: cost, revenue and distance
13	Kinematics: velocity and acceleration
14	Kinematics with trigonometric and exponential models
15	Curve sketching with calculus
16	Mixed review: worked problems
17	Additional exam-style practice
18	Optimization checklist
19	Rates of change in context
20	Exam-style questions with markscheme
21	Common misconceptions and links
22	Check your understanding
23	Summary and topic map
24	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
find stationary points by solving f′(x) = 0
classify stationary points using the first or second derivative test
find global maxima and minima on a closed interval, including endpoints
sketch graphs using derivative information
set up and solve optimization problems in context
use v = ds/dt and a = dv/dt in kinematics
distinguish displacement from distance, and velocity from speed

This booklet covers SL 5.8, 5.9, which is common to SL and HL.

Guiding questions

How can calculus find the best possible outcome? How are position, velocity and acceleration connected?

How this topic is examined

Optimization and kinematics are among the most common long questions on both papers. Optimization: build a function of one variable from a context, differentiate, solve, justify the nature, and answer in context. Kinematics: find velocity and acceleration, when a particle is at rest or changes direction, and the total distance travelled.

Always justify that a stationary point is a maximum or minimum, and check the endpoints of any interval.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Stationary points

Where the gradient is zero.

Think about it

A ball is thrown upwards. At its highest point, what is its vertical velocity? What does that suggest about how to find the maximum of any function?

A stationary point is a point where f′(x) = 0: the tangent is horizontal. Stationary points can be a local maximum, a local minimum, or a stationary point of inflexion.

WORKED EXAMPLE 1 Finding stationary points (basic)

Find the stationary points of y = 2x3 − 9x2 + 12x − 4.

Solution

dy/dx = 6x2 − 18x + 12 = 6(x − 1)(x − 2) = 0M1 A1

x = 1: y = 1; x = 2: y = 0A1

(1, 1) and (2, 0)A1

2 Classifying stationary points

First and second derivative tests.

First derivative test: examine the sign of f′ just before and after the point. + to − gives a maximum; − to + a minimum; no change in sign a stationary inflexion.

Second derivative test: if f″(a) < 0, a local maximum; if f″(a) > 0, a local minimum; if f″(a) = 0, the test is inconclusive (use the first derivative test).

f′ before	f′ after	f″(a)	Nature
+	−	< 0	local maximum
−	+	> 0	local minimum
same sign	same sign	0 (usually)	stationary inflexion

WORKED EXAMPLE 2 Second derivative test (standard)

Classify the stationary points of y = 2x3 − 9x2 + 12x − 4.

Solution

y″ = 12x − 18A1

x = 1: y″ = −6 < 0, maximum (1, 1)R1

x = 2: y″ = 6 > 0, minimum (2, 0)R1

WORKED EXAMPLE 3 When the second derivative fails (exam level)

Classify the stationary point of y = x4 − 4x3.

Solution

y′ = 4x3 − 12x2 = 4x2(x − 3) = 0: x = 0, 3M1 A1

x = 3: y″ = 12(9) − 24(3) = 36 > 0, minimum (3, −27)R1

x = 0: y″ = 0; y′ < 0 on both sides, so stationary inflexion (0, 0)R1

Practice exercise 1

1. Find and classify the stationary points of y = x3 − 12x + 5.

2. Find and classify the stationary point of y = xe−x.

3. Find and classify the stationary points of y = x + 4/x.

3 Global maxima and minima on an interval

Check endpoints too.

On a closed interval [a, b], the global (absolute) maximum or minimum occurs either at a stationary point inside the interval or at an endpoint. Evaluate f at all stationary points in the interval and at both endpoints, and compare.

WORKED EXAMPLE 4 Global extremes (standard)

Find the maximum and minimum values of f(x) = x3 − 3x on −2 ≤ x ≤ 3.

Solution

f′(x) = 3x2 − 3 = 0: x = ±1M1

f(−2) = −2, f(−1) = 2, f(1) = −2, f(3) = 18A1

Maximum 18 (at x = 3); minimum −2 (at x = −2 and x = 1)A1

Practice exercise 2

1. Find the maximum of f(x) = 4x − x2 on [0, 5] and on [3, 5].

2. Find the minimum of f(x) = x − ln x on [0.5, 3].

4 Sketching curves using derivatives

Put all the information together.

To sketch y = f(x): find intercepts; find stationary points and classify them; find points of inflexion; consider end behaviour and asymptotes; then draw a smooth curve through the key points, labelling them.

Figure 1. A sketch with stationary points and the point of inflexion labelled.

5 Optimization problems

Model, differentiate, justify, interpret.

Method: (1) Define variables and draw a diagram. (2) Write the quantity to optimize as a function of the variables. (3) Use a constraint to write it as a function of one variable. (4) Differentiate and set the derivative to zero. (5) Justify the nature (second derivative or sign test). (6) Answer the question in context, with units, and check the domain.

WORKED EXAMPLE 5 Open box (standard)

Squares of side x cm are cut from the corners of a 24 cm × 24 cm sheet, which is folded into an open box. Find the maximum volume.

Solution

V = x(24 − 2x)2, 0 < x < 12M1 A1

V′ = (24 − 2x)2 − 4x(24 − 2x) = (24 − 2x)(24 − 6x) = 0: x = 4 (x = 12 excluded)M1 A1

V″(4) < 0 (maximum); V = 4(16)2 = 1024 cm3R1 A1

WORKED EXAMPLE 6 Minimum surface area (exam level)

A closed cylinder has volume 500 cm3. Find the radius that minimizes its surface area.

Solution

πr2h = 500, so h = 500/(πr2)M1

A = 2πr2 + 2πrh = 2πr2 + 1000/rA1

A′ = 4πr − 1000/r2 = 0: r3 = 250/π, r = 4.30 cmM1 A1

A″ = 4π + 2000/r3 > 0: minimumR1

WORKED EXAMPLE 7 Distance to a curve (exam level)

Find the point on y = √x closest to (4, 0).

Solution

D2 = (x − 4)2 + x (minimize D2)M1

d/dx: 2(x − 4) + 1 = 0, x = 3.5A1

Point (3.5, √3.5), distance √(0.25 + 3.5) = 1.94A1

Common mistake

Stopping at x = 4 without justifying it is a maximum, or forgetting to give the volume itself when asked for the maximum volume.

Practice exercise 3

1. A rectangle has perimeter 60 cm. Find the maximum area.

2. A farmer fences a rectangular field against a river with 400 m of fencing on three sides. Find the maximum area.

3. Find two positive numbers with product 64 whose sum is minimal.

4. Profit P(x) = −0.02x2 + 30x − 2000. Find the number of units for maximum profit.

6 Kinematics: displacement, velocity and acceleration

Derivatives with respect to time.

For a particle moving in a straight line, s(t) is its displacement from a fixed origin. Velocity v is the rate of change of displacement and acceleration a the rate of change of velocity. The particle is at rest when v = 0, and changes direction when v changes sign. When v and a have the same sign, the particle is speeding up; opposite signs, slowing down.

WORKED EXAMPLE 8 A particle (standard)

s(t) = t3 − 6t2 + 9t, t ≥ 0, in metres and seconds. Find (a) when the particle is at rest; (b) the acceleration at those times.

Solution

v = 3t2 − 12t + 9 = 3(t − 1)(t − 3) = 0: t = 1 s and t = 3 sM1 A1

a = 6t − 12: a(1) = −6 m/s2, a(3) = 6 m/s2A1 A1

Figure 2. Displacement and velocity; the particle is at rest where v = 0 (t = 1 and t = 3).

7 Distance travelled and speed

Speed is |v|; distance counts both directions.

Speed is the magnitude of velocity, |v|. Displacement is the change in position; distance travelled adds up the lengths of each stage of the journey, so when the particle changes direction you must split the journey at those times. (With integration, distance = ∫|v| dt; see 5.5.)

WORKED EXAMPLE 9 Distance travelled (exam level)

For s(t) = t3 − 6t2 + 9t, find the distance travelled in the first 4 seconds.

Solution

s(0) = 0, s(1) = 4, s(3) = 0, s(4) = 4A1

Stages: 0 → 4, 4 → 0, 0 → 4M1

Distance = 4 + 4 + 4 = 12 m; displacement only 4 mA1

Practice exercise 4

1. s(t) = 20t − 5t2. Find the maximum height and the time it occurs.

2. v(t) = t2 − 4t + 3. Find the times when the particle is speeding up for 0 ≤ t ≤ 4.

8 Stationary points of harder functions

Combining rules.

Differentiate with the appropriate rules, factorize f′(x), and solve f′(x) = 0. Exponential factors ekx are never zero, so only the other factors matter.

WORKED EXAMPLE 10 Exponential product (exam level)

Find and classify the stationary points of y = x2e−x.

Solution

y′ = xe−x(2 − x) = 0: x = 0, 2M1 A1

Sign of y′: −, +, −M1

Minimum (0, 0); maximum (2, 4e−2)A1

Practice exercise 5

1. Find the stationary point of y = x ln x and its nature.

2. Find the stationary points of y = (x2 − 3)ex.

9 Second derivative test: more practice

Sign of f″ at the stationary point.

At a stationary point a: f″(a) < 0 means a maximum, f″(a) > 0 a minimum. If f″(a) = 0, use the first derivative test.

WORKED EXAMPLE 11 Trigonometric (standard)

Classify the stationary points of f(x) = x + 2 cos x on [0, 2π].

Solution

f′ = 1 − 2 sin x = 0: x = π/6, 5π/6A1

f″ = −2 cos x: f″(π/6) < 0 max; f″(5π/6) > 0 minM1 A1

Practice exercise 6

1. Classify the stationary points of f(x) = x4 − 2x2.

2. Show that f(x) = x4 has a minimum at 0 although f″(0) = 0.

10 Global maximum and minimum on an interval

Endpoints matter.

Evaluate f at the stationary points inside the interval and at both endpoints; the largest value is the global maximum and the smallest the global minimum.

WORKED EXAMPLE 12 Global extremes (exam level)

Find the global maximum and minimum of f(x) = x3 − 6x2 + 9x on [0, 5].

Solution

Stationary x = 1 (f = 4), x = 3 (f = 0)A1

Endpoints f(0) = 0, f(5) = 20A1

Max 20 at x = 5; min 0 at x = 0 and x = 3A1

Practice exercise 7

1. Find the global maximum of f(x) = sin x + cos x on [0, π].

11 Optimization: geometric problems

Areas and volumes.

Set up the quantity to optimize in one variable using the constraint, state the domain, differentiate, solve, justify, and answer in context.

WORKED EXAMPLE 13 Window (exam level)

A window is a rectangle topped by a semicircle, with perimeter 6 m. Find the width that maximizes its area.

Solution

Width 2r, height h: 2r + 2h + πr = 6A1

A = 2rh + πr2/2 = 6r − 2r2 − πr2/2M1 A1

A′ = 6 − 4r − πr = 0: r = 6/(4 + π) = 0.840: width 1.68 mM1 A1

Practice exercise 8

1. Find the maximum area of a rectangle inscribed under y = 12 − x2 with its base on the x-axis.

2. A cylinder is inscribed in a sphere of radius 10. Find the maximum volume.

12 Optimization: cost, revenue and distance

Applied models.

Minimize cost, maximize profit, or minimize distance. For distance to a curve, minimize the square of the distance to avoid square roots.

WORKED EXAMPLE 14 Pipeline (exam level)

A pipe runs from A on one bank of a 300 m wide river to B, 1000 m downstream on the other bank. Laying pipe costs $50/m on land and $80/m under water. Find the minimum cost.

Solution

Cross to a point x m downstream: C = 80√(3002 + x2) + 50(1000 − x)M1 A1

C′ = 0: 80x/√(90 000 + x2) = 50: x = 240.2M1

C ≈ $68 735A1

Practice exercise 9

1. Find the point on y = x2 closest to (0, 3).

13 Kinematics: velocity and acceleration

Derivatives of displacement.

v = ds/dt and a = dv/dt. The particle is at rest when v = 0, speeding up when v and a have the same sign, and changes direction when v changes sign.

WORKED EXAMPLE 15 Particle (exam level)

s(t) = t3 − 9t2 + 24t, t ≥ 0. Find when the particle is at rest and when it is speeding up.

Solution

v = 3(t − 2)(t − 4): rest at t = 2, 4A1

a = 6t − 18: a > 0 for t > 3A1

Speeding up for 2 < t < 3 and t > 4R1 A1

Practice exercise 10

1. Find the distance travelled by this particle between t = 0 and t = 5.

14 Kinematics with trigonometric and exponential models

Non-polynomial motion.

Oscillating motion is often modelled by s = A sin(ωt) or damped motion by s = Ae−ktsin(ωt). Differentiate carefully with the chain and product rules.

WORKED EXAMPLE 16 Oscillation (standard)

s = 4 sin(2t). Find the maximum speed and when it first occurs.

Solution

v = 8 cos(2t): maximum speed 8A1

At t = 0 (and every π/2)A1

Practice exercise 11

1. s = 5e−0.1t. Find the velocity and acceleration at t = 0.

15 Curve sketching with calculus

Complete analysis.

Find: domain, intercepts, asymptotes, stationary points and their nature, points of inflexion, and end behaviour. Then sketch, labelling all key points.

Figure 3. A complete sketch using calculus.

WORKED EXAMPLE 17 Sketch (exam level)

Analyze y = x/(x2 + 1).

Solution

Odd; asymptote y = 0; through originA1

y′ = (1 − x2)/(x2 + 1)2: max (1, ½), min (−1, −½)A1 A1

Inflexions at x = 0, ±√3A1

16 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 18 Mixed problem (exam level)

A closed box has a square base of side x cm and volume 500 cm3. The base and top cost 3 cents/cm2 and the sides 2 cents/cm2. Find x for minimum cost.

Solution

h = 500/x2; C = 6x2 + 8xh = 6x2 + 4000/xM1 A1

C′ = 12x − 4000/x2 = 0: x3 = 333.3M1

x = 6.93 cm; C″ > 0: minimumA1 R1

Practice exercise 12

1. Find the minimum cost for this box.

17 Additional exam-style practice

Paper 2.

WORKED EXAMPLE 19 Paper 2 question (exam level)

The concentration of a drug is C(t) = 20t/(t2 + 4) mg/L. (a) Find C′(t). (b) Find the maximum concentration. (c) Find when the concentration is decreasing fastest.

Solution

(a) 20(4 − t2)/(t2 + 4)2A1 A1

(b) t = 2: C = 5 mg/LA1

Practice exercise 13

1. Find when the concentration is increasing fastest.

18 Optimization checklist

A reliable method.

1. Draw a diagram and define variables. 2. Write the quantity to optimize. 3. Use the constraint to reduce to one variable. 4. State the domain. 5. Differentiate and solve. 6. Justify (second derivative or sign table, and endpoints). 7. Answer the question asked, with units.

WORKED EXAMPLE 20 Justification (standard)

Explain why checking endpoints is necessary in optimization on a closed interval.

Solution

The extreme value may occur at an endpoint rather than a stationary pointR1

19 Rates of change in context

Interpreting derivatives.

The derivative gives the instantaneous rate of change: for example, the marginal cost C′(x) approximates the cost of one more item, and the rate of flow dV/dt gives how fast a tank fills. Interpret signs and units.

WORKED EXAMPLE 21 Marginal profit (standard)

P(x) = −0.02x2 + 30x − 1500. Find P′(500) and interpret.

Solution

P′ = −0.04x + 30: P′(500) = 10A1

Selling one more item at x = 500 adds about $10 profitR1

Practice exercise 14

1. Find the production level for maximum profit in the example.

20 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 2 style) (about 15 minutes)

A rectangular box with a square base and no lid has volume 32 000 cm3. The base has side x cm and the height is h cm.

(a) Show that the external surface area is A = x2 + 128 000/x. [3]

(b) Find the value of x that minimizes A. [3]

(d) Find the minimum surface area. [1]

Markscheme

Part	Answer	Marks
(a)	x2h = 32 000, so h = 32 000/x2	M1
	A = x2 + 4xh	A1
	= x2 + 128 000/x	AG A1
(b)	A′ = 2x − 128 000/x2 = 0	M1
	x3 = 64 000	A1
	x = 40	A1
(c)	A″ = 2 + 256 000/x3	M1
	> 0 at x = 40: minimum	R1
(d)	4800 cm2	A1

Examiner's comments. Answers must be in context with units.

Question 2 (Paper 1 style) (about 12 minutes)

A particle moves with displacement s(t) = 2t3 − 15t2 + 24t, t ≥ 0.

(a) Find the times at which the particle is at rest. [3]

(b) Find the acceleration when t = 2. [2]

Markscheme

Part	Answer	Marks
(a)	v = 6t2 − 30t + 24 = 6(t − 1)(t − 4)	M1 A1
	t = 1, t = 4	A1
(b)	a = 12t − 30	M1
	−6	A1
(c)	s(0) = 0, s(1) = 11, s(4) = −16, s(5) = −5	A1 A1
	11 + 27 + 11	M1
	49	A1

Examiner's comments. Splitting the journey at t = 1 and t = 4 is essential.

21 Common misconceptions and links

Students often think…	Actually…
f′ = 0 always gives a maximum or minimum	it may be an inflexion
ignore endpoints when finding global extremes	evaluate them
distance = final displacement	add each stage separately
speed can be negative	speed = \|v\|

How 5.8 connects to the rest of the course

Topic	Connection
5.6 Rules	second derivative
5.5 Integration	displacement from velocity
2.6 Quadratics	vertex as optimum
5.14 Related rates (HL)	rates of change in context

22 Check your understanding

Section A: multiple choice

1. At a stationary point

A. f′(x) = 0 B. f(x) = 0 C. f″(x) = 0 D. f′(x) = 1

2. f″(a) > 0 at a stationary point means

A. local maximum B. inflexion C. local minimum D. no conclusion

3. Velocity is

A. dv/dt B. ds/dt C. ∫s dt D. d2s/dt2

4. A particle changes direction when

A. a = 0 B. s = 0 C. v is maximal D. v changes sign

5. Global extremes on [a, b] can occur at

A. stationary points only B. stationary points or endpoints C. endpoints only D. inflexions only

Section B: structured

6. Find the maximum value of y = x2e−x for x > 0. [3]

7. A particle has v = 8 − 2t. Find when it is at rest and its acceleration. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 4e−2 at x = 2.

7. t = 4; a = −2.

23 Summary and topic map

Idea	Remember
Stationary points	f′(x) = 0
Classify	f″ < 0 max, > 0 min; else sign test
Global extremes	compare stationary points and endpoints
Optimization	one-variable function, differentiate, justify, interpret
Kinematics	v = ds/dt, a = dv/dt
Rest / direction	v = 0; v changes sign
Distance	split at direction changes

Before the exam, make sure you…

•justify the nature of stationary points
•check endpoints
•answer in context with units
•split journeys when v changes sign

Applications of Differentiation on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 4. Topic map for Applications of Differentiation.

24 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. max (−2, 21), min (2, −11)

2. max (1, 1/e)

3. max (−2, −4), min (2, 4)

Practice exercise 2

1. 4 at x = 2; 3 at x = 3

2. 1 at x = 1

Practice exercise 3

1. 225 cm2

2. 20 000 m2

3. 8 and 8

4. 750 units

Practice exercise 4

1. 20 m at t = 2

2. v and a = 2t − 4 same sign: 1 < t < 2 or t > 3

Practice exercise 5

1. (1/e, −1/e): minimum

2. x2 + 2x − 3 = 0: x = 1 (min, −2e) and x = −3 (max, 6e−3)

Practice exercise 6

1. max (0, 0); min (±1, −1)

2. f′ changes from − to +

Practice exercise 7

1. √2 at x = π/4

Practice exercise 8

1. x = 2: area 2x(12 − x2) = 32

2. h = 20/√3: V = 4000π/(3√3) = 2418

Practice exercise 9

1. D2 = x2 + (x2 − 3)2: x2 = 2.5: (±1.58, 2.5)

Practice exercise 10

1. s(0) = 0, s(2) = 20, s(4) = 16, s(5) = 20: 20 + 4 + 4 = 28

Practice exercise 11

1. −0.5; 0.05

Practice exercise 12

1. 6(48.07) + 4000/6.93 = 865 cents = $8.65

Practice exercise 13

1. t = 0 (C′ is largest at t = 0)

Practice exercise 14

1. x = 750

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