DP Maths AA HL - Calculus - Derivatives and Their Applications

DP Maths AA HL - Calculus - Derivatives and Their Applications - Full notes

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DP Maths AA HL | Derivatives and Their Applications

Derivatives and Their Applications

This topic contains 4 notes:

5.1 Introduction To Differentiation
5.6 Differentiation Rules
5.8 Applications of Differentiation
5.14 Implicit Differentiation

DP MATHEMATICS AA · TOPIC 5: CALCULUS

5.1

Introduction to Differentiation

Full notes for HL · Syllabus SL 5.1, 5.2, 5.3, 5.4

Contents

1	Rates of change and the idea of a limit
2	The derivative as a gradient function
3	Differentiation from first principles
4	The power rule and sums of terms
5	Interpreting derivatives: increasing and decreasing functions
6	Tangents and normals
7	Derivatives in context
8	Average rate of change
9	Estimating the gradient numerically
10	First principles for more functions
11	Power rule practice
12	Gradients at points and points with given gradients
13	Tangents and normals: further practice
14	Increasing and decreasing: sign of the derivative
15	Sketching f′ from f
16	Rates of change in context
17	Mixed review: worked problems
18	Additional exam-style practice
19	Technology for derivatives
20	Limits and continuity (informal)
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
understand the derivative as a limit of gradients of chords and as an instantaneous rate of change
use the notation f′(x), dy/dx and d/dx
derive simple derivatives from first principles
differentiate axn for rational n, and sums and multiples of such terms
identify where a function is increasing or decreasing from the sign of f′(x)
find the equations of tangents and normals
interpret derivatives as rates of change in context

This booklet covers SL 5.1, 5.2, 5.3, 5.4, which is common to SL and HL. HL students extend this topic in booklets 5.12, 5.14.

Guiding questions

How can we measure how fast something is changing at a single instant? What does the derivative tell us about the shape of a graph?

How this topic is examined

Introductory differentiation appears on both papers: differentiate polynomials and powers, find the gradient at a point, find the equation of a tangent or normal, determine intervals where a function is increasing, and interpret rates of change in context.

Rewrite roots and fractions as powers before differentiating, and use the derivative's sign, not just its value, to describe behaviour.

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 Rates of change and the idea of a limit

From average to instantaneous.

Think about it

A car travels 120 km in 2 hours, so its average speed is 60 km/h. But the speedometer shows a speed at each instant. How could we define speed at one instant using only distances and times?

The average rate of change of f between x = a and x = b is the gradient of the chord joining the two points: (f(b) − f(a))/(b − a). To find the rate of change at a single point, let the second point slide closer and closer to the first. The chord gradients approach a limit: the gradient of the tangent, the instantaneous rate of change.

h	Gradient of chord from x = 1 to x = 1 + h for y = x2
1	(4 − 1)/1 = 3
0.1	(1.21 − 1)/0.1 = 2.1
0.01	2.01
0.001	2.001
→ 0	→ 2

Figure 1. Chords from (1, 1) approach the tangent (dashed), whose gradient is 2.

2 The derivative as a gradient function

A new function from the old.

The derivative of f, written f′(x) or dy/dx, gives the gradient of the tangent at every point. It is itself a function. dy/dx is read 'the derivative of y with respect to x' and suggests 'small change in y over small change in x'.

Where the graph of f is increasing, f′(x) > 0; where it is decreasing, f′(x) < 0; at a turning point, f′(x) = 0. You can sketch f′ from the shape of f by reading off gradients.

Figure 2. f′ is zero at the turning points of f and negative where f decreases.

3 Differentiation from first principles

The definition of the derivative.

WORKED EXAMPLE 1 First principles (standard)

Differentiate f(x) = x2 + 3x from first principles.

Solution

f(x + h) − f(x) = x2 + 2xh + h2 + 3x + 3h − x2 − 3x = 2xh + h2 + 3hM1 A1

Divide by h: 2x + h + 3A1

Let h → 0: f′(x) = 2x + 3A1

Practice exercise 1

1. Differentiate f(x) = x3 from first principles.

2. Differentiate f(x) = 5x − 2 from first principles and interpret the answer.

4 The power rule and sums of terms

Bring the power down, reduce it by one.

The derivative of a sum is the sum of the derivatives, and constants multiply through. Before differentiating, rewrite every term as a power of x: √x = x1/2, 1/x3 = x−3, 4/√x = 4x−1/2. Expand products and split fractions with a single-term denominator.

WORKED EXAMPLE 2 Rewriting first (standard)

Differentiate y = 3√x − 2/x2 + 5.

Solution

y = 3x1/2 − 2x−2 + 5M1

dy/dx = (3/2)x−1/2 + 4x−3A1

= 3/(2√x) + 4/x3A1

WORKED EXAMPLE 3 Expanding and dividing (exam level)

Differentiate y = (x2 + 4)(x − 1)/x.

Solution

Expand: (x3 − x2 + 4x − 4)/x = x2 − x + 4 − 4x−1M1

dy/dx = 2x − 1 + 4/x2A1 A1

Common mistake

Differentiating a product or quotient term by term, as if d/dx[(x2 + 4)(x − 1)] = (2x)(1). Expand first, or use the product rule (5.6).

Practice exercise 2

1. Differentiate y = 4x5 − 3x2 + 7x.

2. Differentiate f(x) = 6/x − ∛x.

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

4. Find f′(4) for f(x) = x√x.

5. Differentiate y = (x3 − 5x)/√x.

5 Interpreting derivatives: increasing and decreasing functions

Use the sign of f′.

f is increasing on an interval where f′(x) > 0 and decreasing where f′(x) < 0. To find these intervals, solve f′(x) = 0 and test the sign of f′ between the solutions.

WORKED EXAMPLE 4 Intervals (standard)

Find the intervals on which f(x) = x3 − 6x2 + 9x + 1 is increasing.

Solution

f′(x) = 3x2 − 12x + 9 = 3(x − 1)(x − 3)M1 A1

f′ > 0 outside the rootsR1

x < 1 or x > 3A1

Practice exercise 3

1. Find where f(x) = x2 − 8x + 3 is decreasing.

2. Show that f(x) = x3 + 3x − 2 is increasing for all x.

6 Tangents and normals

Lines through a point on the curve.

At the point (a, f(a)), the tangent has gradient m = f′(a): y − f(a) = m(x − a). The normal is perpendicular to the tangent, with gradient −1/m.

WORKED EXAMPLE 5 Tangent and normal (standard)

Find the equations of the tangent and the normal to y = x3 − 2x at x = 2.

Solution

y(2) = 4; dy/dx = 3x2 − 2 = 10A1 A1

Tangent: y − 4 = 10(x − 2), y = 10x − 16A1

Normal: gradient −1/10: y = −x/10 + 4.2A1

WORKED EXAMPLE 6 Given gradient (exam level)

Find the points on y = x3 − 3x2 + 1 where the tangent is parallel to y = 9x + 2.

Solution

3x2 − 6x = 9M1

x2 − 2x − 3 = 0, x = 3 or −1A1

(3, 1) and (−1, −3)A1

Practice exercise 4

1. Find the tangent to y = √x at x = 4.

2. Find the normal to y = 2/x at (1, 2).

3. The tangent to y = x2 + k at x = 1 passes through the origin. Find k.

7 Derivatives in context

Rates of change with units.

If y depends on t, dy/dt is the rate of change of y with respect to t, with units 'units of y per unit of t'. For example, if V(t) litres is the volume in a tank after t minutes, V′(t) is in litres per minute; a negative value means the volume is decreasing.

WORKED EXAMPLE 7 Rate of change (standard)

The population of a town is P(t) = 2000 + 150t + 3t2, t in years. Find P′(10) and interpret it.

Solution

P′(t) = 150 + 6tA1

P′(10) = 210: after 10 years the population is increasing by about 210 people per yearA1 R1

Practice exercise 5

1. A balloon's radius is r = 3 + 0.5t cm after t seconds. Find dV/dt when t = 2, where V = (4/3)πr3. (Expand or use the chain rule after 5.6.)

2. C(x) = 500 + 20x − 0.01x2 is the cost of making x items. Find and interpret C′(200).

8 Average rate of change

Gradient of a chord.

The average rate of change of f between x = a and x = b is (f(b) − f(a))/(b − a): the gradient of the chord. In context it has units of 'output per input', such as metres per second or dollars per item.

WORKED EXAMPLE 8 Average speed (basic)

s(t) = t3 + 2t metres. Find the average velocity between t = 1 and t = 3.

Solution

(33 − 3)/2M1

= 15 m/sA1

Practice exercise 6

1. Find the average rate of change of f(x) = √x from x = 4 to x = 9.

2. A population grows from 12 000 to 15 600 in 6 years. Find the average rate of growth.

9 Estimating the gradient numerically

Shrinking the interval.

The gradient of the tangent at x = a is the limit of chord gradients as the second point approaches a. A table of chord gradients for h = 0.1, 0.01, 0.001 shows the limit clearly.

h	1	0.1	0.01	0.001
((2 + h)3 − 8)/h	19	12.61	12.06	12.006

WORKED EXAMPLE 9 Numerical gradient (standard)

Estimate the gradient of y = x3 at x = 2.

Solution

Chord gradients approach 12M1

Gradient = 12 (= 3(2)2)A1

Practice exercise 7

1. Estimate the gradient of y = 2x at x = 0 using h = 0.001.

10 First principles for more functions

Algebra of limits.

f′(x) = limh→0 (f(x + h) − f(x))/h. Expand, simplify so that h cancels, then let h → 0.

WORKED EXAMPLE 10 Quadratic (standard)

Differentiate f(x) = 3x2 − 5x from first principles.

Solution

f(x + h) − f(x) = 6xh + 3h2 − 5hM1 A1

Divide by h: 6x + 3h − 5 → 6x − 5A1

WORKED EXAMPLE 11 Reciprocal (exam level)

Differentiate f(x) = 1/x from first principles.

Solution

(1/(x + h) − 1/x)/h = −1/(x(x + h))M1 A1

→ −1/x2A1

Practice exercise 8

1. Differentiate f(x) = x3 + x from first principles.

2. Differentiate f(x) = 4 − x2 from first principles.

11 Power rule practice

Rewrite, then differentiate.

Rewrite every term as axn (roots as fractional powers, reciprocals as negative powers) before applying d/dx(axn) = anxn−1. Expand products and split fractions with single-term denominators.

WORKED EXAMPLE 12 Mixed powers (standard)

Differentiate y = 4x3 − 2/x2 + 6√x − 7.

Solution

y = 4x3 − 2x−2 + 6x1/2 − 7M1

dy/dx = 12x2 + 4x−3 + 3x−1/2A2

WORKED EXAMPLE 13 Quotient split (standard)

Differentiate y = (x2 − 3x + 1)/x.

Solution

y = x − 3 + x−1M1

dy/dx = 1 − 1/x2A1

Practice exercise 9

1. Differentiate y = (2x − 1)2.

2. Differentiate y = 5/√x.

3. Differentiate y = x2(x − 4).

12 Gradients at points and points with given gradients

Substitute or solve.

To find the gradient at x = a, substitute into f′(x). To find where the gradient equals m, solve f′(x) = m and find the corresponding y-values.

WORKED EXAMPLE 14 Given gradient (standard)

Find the points on y = x3 − 6x2 + 4 where the gradient is −9.

Solution

3x2 − 12x = −9: x2 − 4x + 3 = 0M1

x = 1, 3: (1, −1) and (3, −23)A1 A1

Practice exercise 10

1. Find the gradient of y = 8/x at x = 2.

2. Find where y = x2 + 2x has gradient 10.

13 Tangents and normals: further practice

Lines touching curves.

Tangent at (a, f(a)): y − f(a) = f′(a)(x − a). Normal: gradient −1/f′(a). Questions often ask where a tangent meets an axis or where a normal meets the curve again.

WORKED EXAMPLE 15 Tangent and axes (exam level)

Find the area of the triangle formed by the tangent to y = 4/x at x = 2 and the coordinate axes.

Solution

y(2) = 2; gradient −1A1

Tangent y = −x + 4: intercepts (4, 0), (0, 4)A1

Area = 8A1

Practice exercise 11

1. Show that the tangent to y = k/x at any point forms a triangle of area 2k with the axes.

2. Find where the normal to y = x2 at (1, 1) meets the curve again.

14 Increasing and decreasing: sign of the derivative

Intervals.

Solve f′(x) > 0 for increasing and f′(x) < 0 for decreasing intervals. A sign diagram of f′ is the clearest method.

WORKED EXAMPLE 16 Cubic (standard)

Find the intervals where f(x) = 2x3 − 3x2 − 12x increases.

Solution

f′ = 6(x − 2)(x + 1)M1 A1

x < −1 or x > 2A1

Practice exercise 12

1. Find where f(x) = x4 − 8x2 is decreasing.

2. Show that f(x) = x3 + 3x is always increasing.

15 Sketching f′ from f

Read gradients from the graph.

Where f has a turning point, f′ = 0 (f′ crosses the axis). Where f increases, f′ > 0. Where f is steepest, f′ has a maximum or minimum. A cubic f has a quadratic f′.

Figure 3. f and its derivative f′.

WORKED EXAMPLE 17 Reading (standard)

From the graph, state where f′(x) < 0 and where f has its steepest decrease.

Solution

0 < x < 2A1

At x = 1 (minimum of f′)A1

16 Rates of change in context

Interpret with units.

dV/dt is the rate of change of V with respect to t, with units of V per unit of t. A negative value means V is decreasing. Always interpret in a sentence.

WORKED EXAMPLE 18 Tank (standard)

V(t) = 500 − 20t + 0.2t2 litres for 0 ≤ t ≤ 50 minutes. Find V′(10) and interpret.

Solution

V′ = −20 + 0.4t: V′(10) = −16A1

At 10 minutes, the volume is decreasing at 16 litres per minuteR1

Practice exercise 13

1. C(x) = 2000 + 15x − 0.01x2. Find C′(300) and interpret.

17 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

The curve y = ax2 + bx has gradient 7 at (1, 4). Find a and b, and the equation of the normal at this point.

Solution

a + b = 4; 2a + b = 7M1

a = 3, b = 1A1

Normal: y − 4 = −(1/7)(x − 1)A1

Practice exercise 14

1. Find the point on y = x2 − 4x + 1 where the tangent is parallel to y = 2x − 3.

18 Additional exam-style practice

Paper 1.

WORKED EXAMPLE 20 Paper 1 question (exam level)

f(x) = x3 − 3x2 − 9x + 11. (a) Find f′(x). (b) Find where f is decreasing. (c) Find the equation of the tangent at x = 0.

Solution

(a) 3x2 − 6x − 9A1

(b) 3(x − 3)(x + 1) < 0: −1 < x < 3M1 A1

Practice exercise 15

1. Find the tangent to y = 2√x at x = 9.

19 Technology for derivatives

Numerical derivatives and graphs.

The GDC calculates f′(a) numerically (nDeriv, or dy/dx on a graph) and can graph y = f′(x). Use these to check answers and to find where f′ = 0 for complicated functions.

WORKED EXAMPLE 21 Check (standard)

Use technology to find the gradient of y = x2e−x at x = 1.

Solution

0.368 (= 1/e)A1

GDC tip

Graph y = nDeriv(f(x), x, x) to see the gradient function, then use the zero tool to locate stationary points.

20 Limits and continuity (informal)

The idea behind derivatives.

limx→a f(x) = L means f(x) gets arbitrarily close to L as x approaches a. Tables of values and graphs give estimates. A function is continuous at a if its graph has no break there. Derivatives exist only where the function is continuous and smooth.

WORKED EXAMPLE 22 Limit from a table (standard)

Estimate limx→0 (ex − 1)/x.

Solution

x = 0.1: 1.0517; 0.01: 1.0050; 0.001: 1.0005M1

Limit 1A1

Practice exercise 16

1. Estimate limx→3 (x2 − 9)/(x − 3).

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) = x3 − 3x2 − 9x + 5.

(a) Find f′(x). [2]

(b) Find the intervals on which f is decreasing. [3]

(d) Find the x-coordinate of the other point where the tangent in (c) meets the curve. [3]

Markscheme

Part	Answer	Marks
(a)	3x2 − 6x − 9	A2
(b)	3(x − 3)(x + 1) = 0	M1
	f′ < 0 between the roots	R1
	−1 < x < 3	A1
(c)	f(1) = −6; f′(1) = −12	A1
	y + 6 = −12(x − 1)	M1
	y = −12x + 6	A1
(d)	x3 − 3x2 − 9x + 5 = −12x + 6	M1
	x3 − 3x2 + 3x − 1 = (x − 1)3 = 0	A1
	no other point: x = 1 is a triple root (the tangent crosses at a point of inflexion)	R1

Examiner's comments. Part (d) rewards recognizing the perfect cube; the tangent meets the curve only at x = 1.

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

Use first principles to find the derivative of f(x) = 2x2 − x. [4]

Markscheme

Part	Answer	Marks
	f(x + h) = 2x2 + 4xh + 2h2 − x − h	A1
	(4xh + 2h2 − h)/h	M1
	= 4x + 2h − 1	A1
	→ 4x − 1 as h → 0	A1

Examiner's comments. The limit must be stated explicitly.

22 Common misconceptions and links

Students often think…	Actually…
d/dx(√x) = √x/2	rewrite as x1/2: (1/2)x−1/2
differentiate products term by term	expand or use the product rule
the normal gradient is −m	it is −1/m
f′(x) = 0 means a maximum	it may be a minimum or an inflexion

How 5.1 connects to the rest of the course

Topic	Connection
5.6 Differentiation rules	chain, product, quotient
5.8 Applications	optimization, kinematics
2.1 Lines	tangent equations
5.12 Limits (HL)	formal limits

23 Check your understanding

Section A: multiple choice

1. d/dx(x5) =

A. 5x4 B. x4 C. 5x5 D. 4x5

2. d/dx(1/x) =

A. 1/x2 B. ln x C. −1/x2 D. −1/x

3. If f′(x) < 0 on an interval, f is

A. increasing B. decreasing C. constant D. at a maximum

4. A normal is

A. parallel to the tangent B. horizontal C. the tangent D. perpendicular to the tangent

5. d/dx(7) =

A. 7 B. 0 C. 1 D. 7x

Section B: structured

6. Find the gradient of y = 4x3/2 at x = 9. [2]

7. Find the equation of the tangent to y = x2 − 4x + 1 parallel to the x-axis. [3]

Answers

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

6. 6x1/2 = 18.

7. x = 2, y = −3: y = −3.

24 Summary and topic map

Idea	Remember
Derivative	limit of chord gradients; instantaneous rate
First principles	lim (f(x + h) − f(x))/h
Power rule	d/dx(axn) = anxn−1
Increasing	f′(x) > 0; decreasing f′(x) < 0
Tangent	y − f(a) = f′(a)(x − a)
Normal	gradient −1/f′(a)
Context	units: y-units per x-unit

Before the exam, make sure you…

•rewrite roots and fractions as powers
•expand before differentiating
•use the sign of f′
•state units in context

Introduction to 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 Introduction to Differentiation.

25 Answers to practice exercises

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

Practice exercise 1

1. (3x2h + 3xh2 + h3)/h → 3x2

2. 5: a straight line has constant gradient

Practice exercise 2

1. 20x4 − 6x + 7

2. −6/x2 − (1/3)x−2/3

3. 8x + 4

4. (3/2)√4 = 3

5. (5/2)x3/2 − (5/2)x−1/2

Practice exercise 3

1. x < 4

2. f′(x) = 3x2 + 3 > 0 for all x

Practice exercise 4

1. y = x/4 + 1

2. tangent gradient −2, so normal y = x/2 + 3/2

3. tangent y = 2x + (k − 1): through the origin when k = 1

Practice exercise 5

1. r = 4: dV/dt = 4πr2(0.5) = 32π cm3/s

2. C′(200) = 16: the cost increases by about $16 per extra item

Practice exercise 6

1. 1/5

2. 600 per year

Practice exercise 7

1. (20.001 − 1)/0.001 = 0.693

Practice exercise 8

1. 3x2 + 1

2. −2x

Practice exercise 9

1. 8x − 4

2. −5/(2x3/2)

3. 3x2 − 8x

Practice exercise 10

1. −2

2. (4, 24)

Practice exercise 11

1. tangent at a: intercepts 2a and 2k/a: area 2k

2. (−3/2, 9/4)

Practice exercise 12

1. x < −2 or 0 < x < 2

2. f′ = 3x2 + 3 > 0

Practice exercise 13

1. 9: the 301st item costs about $9

Practice exercise 14

1. 2x − 4 = 2: (3, −2)

Practice exercise 15

1. y = x/3 + 3

Practice exercise 16

1. 6

DP MATHEMATICS AA · TOPIC 5: CALCULUS

5.6

Differentiation Rules and the Second Derivative

Full notes for HL · 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	Implicit hints and preview
21	Technology for checking derivatives
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	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. HL students extend this topic in booklet 5.14.

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 Implicit hints and preview

Beyond explicit functions (HL).

When y is not given explicitly, differentiate each term with respect to x, using the chain rule on terms in y (d/dx(y2) = 2y dy/dx). This is implicit differentiation, developed in booklet 5.14.

WORKED EXAMPLE 24 Circle (standard)

Find dy/dx for x2 + y2 = 25.

Solution

2x + 2y dy/dx = 0M1

dy/dx = −x/yA1

21 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 25 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.

22 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.

23 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

24 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).

25 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.

26 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 HL · 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. HL students extend this topic in booklet 5.14.

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

DP MATHEMATICS AA · TOPIC 5: CALCULUS

5.14

Implicit Differentiation and Related Rates

Full notes for HL · Syllabus AHL 5.14, 5.15

Contents

1	Derivatives of further functions
2	Derivatives of inverse trigonometric functions
3	Implicit differentiation
4	Tangents and normals to implicit curves
5	Second derivatives implicitly
6	Related rates of change
7	Optimization with further functions
8	Implicit differentiation: more practice
9	Tangents and normals to implicit curves
10	Second derivatives implicitly
11	Derivatives of inverse trigonometric functions: practice
12	Derivatives of exponentials and logarithms with any base
13	Related rates: geometric shapes
14	Related rates: angles
15	Optimization with implicit and inverse functions
16	Mixed review: worked problems
17	Additional exam-style practice
18	Strategy for related rates
19	Implicit curves with technology
20	Rates of change of volumes and areas
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 sec x, cosec x, cot x, ax and loga x
differentiate arcsin x, arccos x and arctan x and composites of them
differentiate implicitly, finding dy/dx for curves not written as y = f(x)
find tangents, normals and stationary points of implicit curves
solve related rates problems using the chain rule
solve optimization problems involving further functions

This booklet covers Additional Higher Level (AHL) content (AHL 5.14, 5.15). It is studied by HL students only.

Guiding questions

How can we differentiate a curve that is not written as y = f(x)? How are the rates of change of connected quantities linked?

How this topic is examined

Implicit differentiation and related rates are frequent HL questions on both papers. Related rates questions (expanding circles, filling tanks, sliding ladders) require setting up an equation linking the variables, differentiating with respect to time, and substituting the given values at the correct moment.

In related rates, differentiate first, then substitute the values at the given instant.

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 further functions

Extending the table.

f(x)	f′(x)
sec x	sec x tan x
cosec x	−cosec x cot x
cot x	−cosec2x
ax	ax ln a
loga x	1/(x ln a)

Why it works

sec x = (cos x)−1, so by the chain rule its derivative is −(cos x)−2(−sin x) = sin x/cos2x = sec x tan x. For ax, write ax = ex ln a, whose derivative is ln a × ex ln a = ax ln a. For loga x = ln x/ln a, the derivative is 1/(x ln a).

WORKED EXAMPLE 1 Further derivatives (standard)

Differentiate (a) y = 52x; (b) y = log2(x2 + 1); (c) y = sec2x.

Solution

(a) 2 ln 5 × 52xA1

(b) 2x/((x2 + 1) ln 2)A1

Practice exercise 1

1. Differentiate y = cot(3x).

2. Differentiate y = x × 3x.

3. Differentiate y = cosec x + log10 x.

2 Derivatives of inverse trigonometric functions

Via implicit differentiation.

Why it works

Let y = arcsin x, so sin y = x. Differentiate implicitly: cos y (dy/dx) = 1, so dy/dx = 1/cos y = 1/√(1 − sin2y) = 1/√(1 − x2) (cos y ≥ 0 since −π/2 ≤ y ≤ π/2). Similarly, tan y = x gives sec2y (dy/dx) = 1, so dy/dx = 1/(1 + tan2y) = 1/(1 + x2).

WORKED EXAMPLE 2 Inverse trig with the chain rule (standard)

Differentiate (a) y = arctan(2x); (b) y = arcsin(x/3).

Solution

(a) 2/(1 + 4x2)A1

(b) (1/3)/√(1 − x2/9) = 1/√(9 − x2)M1 A1

Practice exercise 2

1. Differentiate y = x arctan x.

2. Differentiate y = arccos(ex).

3 Implicit differentiation

Differentiate both sides with respect to x.

Think about it

The circle x2 + y2 = 25 is not the graph of a single function y = f(x). How could we still find the gradient at (3, 4)?

An equation such as x2 + y2 = 25 defines y implicitly. To find dy/dx, differentiate every term with respect to x, treating y as a function of x. By the chain rule, d/dx(yn) = nyn−1(dy/dx), and products such as xy need the product rule: d/dx(xy) = y + x(dy/dx). Then collect the dy/dx terms and solve.

WORKED EXAMPLE 3 A circle (basic)

Find dy/dx for x2 + y2 = 25 and the gradient at (3, 4).

Solution

2x + 2y(dy/dx) = 0M1

dy/dx = −x/yA1

At (3, 4): −3/4A1

WORKED EXAMPLE 4 With a product term (standard)

Find dy/dx for x3 + xy + y3 = 7.

Solution

3x2 + y + x(dy/dx) + 3y2(dy/dx) = 0M1 A1

dy/dx(x + 3y2) = −3x2 − yM1

dy/dx = −(3x2 + y)/(x + 3y2)A1

Common mistake

Differentiating y2 as 2y and forgetting the factor dy/dx, or differentiating xy as 1 × dy/dx.

Practice exercise 3

1. Find dy/dx for x2 − 3xy + y2 = 1.

2. Find dy/dx for ey + xy = e.

3. Find dy/dx for sin(x + y) = y.

4 Tangents and normals to implicit curves

Gradient from both coordinates.

WORKED EXAMPLE 5 Tangent to an implicit curve (exam level)

Find the equation of the tangent to x2y + y3 = 10 at (3, 1).

Solution

2xy + x2(dy/dx) + 3y2(dy/dx) = 0M1

At (3, 1): 6 + 9(dy/dx) + 3(dy/dx) = 0, dy/dx = −1/2A1

y − 1 = −½(x − 3), or x + 2y = 5A1

WORKED EXAMPLE 6 Horizontal tangents (exam level)

Find the points on x2 + xy + y2 = 12 where the tangent is horizontal.

Solution

dy/dx = −(2x + y)/(x + 2y); horizontal when y = −2xM1 A1

x2 − 2x2 + 4x2 = 12, x2 = 4M1

(2, −4) and (−2, 4)A1

Practice exercise 4

1. Find the normal to x2 + y2 = 13 at (2, 3).

2. Find the points on x2 + y2 − 4x = 0 where the tangent is vertical.

5 Second derivatives implicitly

Differentiate dy/dx again.

WORKED EXAMPLE 7 Second derivative (exam level)

For x2 + y2 = 25, show that d2y/dx2 = −25/y3.

Solution

dy/dx = −x/yA1

d2y/dx2 = −(y − x(dy/dx))/y2 = −(y + x2/y)/y2M1 A1

= −(y2 + x2)/y3 = −25/y3AG

6 Related rates of change

Chain rule with time.

When several quantities change with time, their rates are linked by the chain rule. Method: (1) draw a diagram and define variables; (2) write an equation linking the quantities; (3) differentiate both sides with respect to time t; (4) substitute the known values at the instant in question; (5) solve for the unknown rate, with units.

WORKED EXAMPLE 8 Expanding circle (standard)

A circular ripple's radius increases at 3 cm/s. How fast is the area increasing when the radius is 10 cm?

Solution

A = πr2, dA/dt = 2πr (dr/dt)M1

= 2π(10)(3)A1

= 60π ≈ 188 cm2/sA1

WORKED EXAMPLE 9 Sliding ladder (exam level)

A 5 m ladder leans against a wall. The foot slides away at 0.4 m/s. How fast is the top sliding down when the foot is 3 m from the wall?

Solution

x2 + y2 = 25, so 2x(dx/dt) + 2y(dy/dt) = 0M1

x = 3, y = 4: 3(0.4) + 4(dy/dt) = 0A1

dy/dt = −0.3 m/s: the top slides down at 0.3 m/sA1

WORKED EXAMPLE 10 Conical tank (exam level)

Water flows into an inverted cone (radius 3 m at the top, height 6 m) at 2 m3/min. How fast is the depth rising when the depth is 4 m?

Solution

By similar triangles r = h/2, so V = (1/3)π(h/2)2h = πh3/12M1 A1

dV/dt = (πh2/4)(dh/dt)M1

2 = 4π(dh/dt), dh/dt = 1/(2π) = 0.159 m/minA1

Common mistake

Substituting the numerical values before differentiating, which turns variables into constants with zero derivative.

Practice exercise 5

1. A spherical balloon is inflated at 100 cm3/s. Find dr/dt when r = 5 cm.

2. A cube's side increases at 0.2 cm/s. Find the rate of increase of volume when the side is 4 cm.

3. A plane at height 6 km flies horizontally at 600 km/h directly over a radar. How fast is its distance from the radar increasing when that distance is 10 km?

7 Optimization with further functions

Trigonometric and inverse trigonometric models.

WORKED EXAMPLE 11 Viewing angle (exam level)

A painting 2 m tall hangs with its bottom edge 1 m above eye level. How far from the wall should a viewer stand to maximize the viewing angle θ = arctan(3/x) − arctan(1/x)?

Solution

dθ/dx = −3/(x2 + 9) + 1/(x2 + 1)M1 A1

= 0 when x2 + 9 = 3x2 + 3, x2 = 3M1

x = √3 ≈ 1.73 mA1

8 Implicit differentiation: more practice

Every y term gets dy/dx.

Differentiate each term with respect to x: d/dx(yn) = nyn−1 dy/dx; products like xy need the product rule; composites like sin y or ey need the chain rule. Collect dy/dx terms and factorize.

WORKED EXAMPLE 12 Exponential (exam level)

Find dy/dx for exy = x + y.

Solution

exy(y + x dy/dx) = 1 + dy/dxM1 A1

dy/dx(xexy − 1) = 1 − yexyA1

dy/dx = (1 − yexy)/(xexy − 1)A1

WORKED EXAMPLE 13 Trigonometric (standard)

Find dy/dx for x sin y + y2 = 4.

Solution

sin y + x cos y dy/dx + 2y dy/dx = 0M1 A1

dy/dx = −sin y/(x cos y + 2y)A1

Practice exercise 6

1. Find dy/dx for x3 + y3 = 6xy.

2. Find dy/dx for ln(xy) = x + y.

9 Tangents and normals to implicit curves

Gradient at a point.

Substitute the coordinates into dy/dx to find the gradient, then write the tangent or normal. Horizontal tangents occur where the numerator of dy/dx is zero; vertical tangents where the denominator is zero.

WORKED EXAMPLE 14 Folium (exam level)

Find the tangent to x3 + y3 = 9xy at (2, 4).

Solution

dy/dx = (3y − x2)/(y2 − 3x) = (12 − 4)/(16 − 6) = 4/5M1 A1

y − 4 = (4/5)(x − 2)A1

Practice exercise 7

1. Find the points on x2 + xy + y2 = 3 with horizontal tangents.

2. Find the normal to x2 + y2 = 25 at (3, −4).

10 Second derivatives implicitly

Differentiate dy/dx again.

Differentiate dy/dx with respect to x (quotient rule), then substitute the expression for dy/dx. Simplify using the original equation where possible.

WORKED EXAMPLE 15 Second derivative (exam level)

For y2 = 4x, find d2y/dx2 in terms of y.

Solution

2y dy/dx = 4: dy/dx = 2/yA1

d2y/dx2 = −2y−2 dy/dx = −4/y3M1 A1

Practice exercise 8

1. For xy = 1, find d2y/dx2 by implicit differentiation.

11 Derivatives of inverse trigonometric functions: practice

Chain rule with arcsin, arccos, arctan.

d/dx arcsin(u) = u′/√(1 − u2); d/dx arccos(u) = −u′/√(1 − u2); d/dx arctan(u) = u′/(1 + u2).

WORKED EXAMPLE 16 Composite (standard)

Differentiate y = arctan(x2) and y = arcsin(√x).

Solution

2x/(1 + x4)A1

(1/(2√x))/√(1 − x) = 1/(2√(x(1 − x)))M1 A1

Practice exercise 9

1. Differentiate y = x arcsin x.

2. Differentiate y = arctan(1/x) and simplify.

12 Derivatives of exponentials and logarithms with any base

Changing base.

d/dx(ax) = ax ln a and d/dx(loga x) = 1/(x ln a). Logarithmic differentiation (take ln of both sides, then differentiate implicitly) handles functions like xx.

WORKED EXAMPLE 17 Logarithmic differentiation (exam level)

Differentiate y = xx.

Solution

ln y = x ln xM1

(1/y) dy/dx = ln x + 1A1

dy/dx = xx(ln x + 1)A1

Practice exercise 10

1. Differentiate y = 32x.

2. Differentiate y = xsin x.

13 Related rates: geometric shapes

Chain rule with time.

Write an equation connecting the variables, differentiate with respect to t, and substitute values at the instant required. Use similar triangles or geometry to eliminate extra variables first.

WORKED EXAMPLE 18 Shadow (exam level)

A 1.8 m person walks away from a 6 m lamp post at 1.5 m/s. How fast is the shadow lengthening?

Solution

Similar triangles: s/1.8 = (x + s)/6: s = 3x/7M1 A1

ds/dt = (3/7)(1.5) = 0.643 m/sA1

WORKED EXAMPLE 19 Square area (standard)

The side of a square increases at 0.3 cm/s. Find the rate of increase of the area when the side is 8 cm.

Solution

A = x2: dA/dt = 2x dx/dtM1

= 4.8 cm2/sA1

Practice exercise 11

1. A circle's area increases at 20 cm2/s. Find dr/dt when r = 5.

2. A cube's volume increases at 30 cm3/s. Find the rate of increase of the surface area when the side is 5 cm.

14 Related rates: angles

Trigonometric relationships.

When an angle changes, differentiate a trigonometric equation with respect to time, using the chain rule: d/dt(tan θ) = sec2θ dθ/dt.

WORKED EXAMPLE 20 Rocket (exam level)

A rocket rises vertically at 50 m/s, observed from 300 m away. Find the rate of change of the angle of elevation when the rocket is at 400 m.

Solution

tan θ = h/300: sec2θ dθ/dt = (1/300) dh/dtM1 A1

sec2θ = 1 + (4/3)2 = 25/9A1

dθ/dt = (50/300)(9/25) = 0.06 rad/sA1

Practice exercise 12

1. A 10 m ladder slides down a wall; the foot moves away at 1 m/s. Find the rate of change of the angle with the ground when the foot is 6 m from the wall.

15 Optimization with implicit and inverse functions

Further applications.

Optimization problems may involve inverse trigonometric functions (viewing angles) or implicit relationships. Differentiate, set equal to zero, and justify.

WORKED EXAMPLE 21 Viewing angle (exam level)

A painting 2 m tall hangs with its base 1 m above eye level. Find the distance x maximizing θ = arctan(3/x) − arctan(1/x).

Solution

dθ/dx = −3/(x2 + 9) + 1/(x2 + 1) = 0M1 A1

x2 + 9 = 3x2 + 3: x = √3 mA1

Practice exercise 13

1. Find the maximum value of θ in the example.

16 Mixed review: worked problems

Bringing the ideas together.