DP Maths AI HL 5.3 Derivatives and Their Applications

DP Maths AI HL 5.3 Derivatives and Their Applications - Full notes

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DP Maths AI HL | 5.3 Derivatives and Their Applications

DP MATHEMATICS AI · TOPIC 5: CALCULUS

5.3

Derivatives and Their Applications

Full notes for Applications and Interpretation HL

Contents

1	The derivative as a gradient function
2	The power rule
3	Sums, differences and constant multiples
4	Rewriting before differentiating
5	Gradient at a point
6	Finding where the gradient has a given value
7	Equations of tangents
8	Equations of normals
9	Increasing and decreasing functions
10	Stationary points
11	Classifying stationary points
12	Maximum and minimum values on an interval
13	Optimization: setting up a model
14	Optimization: area and volume
15	Optimization: cost, revenue and profit
16	Kinematics: velocity from displacement
17	Using technology for derivatives
18	Mixed review: worked problems
19	Further derivatives and the chain, product and quotient rules (HL)
20	The second derivative, concavity and acceleration (HL)
21	Investigation: the most economical can
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 practice exercise.

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 axn and sums of such terms
find the gradient at a point and the points with a given gradient
find equations of tangents and normals
find intervals where a function increases or decreases
find and classify stationary points
find maximum and minimum values on an interval
solve optimization problems in context
use v = ds/dt in kinematics

HL booklet: this booklet contains all the SL content plus derivatives of further functions, the chain, product and quotient rules, and the second derivative.

Guiding questions

How can calculus find the best possible outcome? What does the derivative tell us about the shape of a graph?

How this topic is examined

Differentiation and its applications appear in every AI examination. Paper 1: differentiate, find gradients, tangents and normals. Paper 2: optimization in context (boxes, cans, fences, profit), often with a GDC to find maximum values, and interpretation of rates of change.

Show the derivative and the equation you solve (f′(x) = 0), justify the nature of a stationary point, and answer the question in context with units.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

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, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
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
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), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 The derivative as a gradient function

Gradient at every point.

Think about it

A curve's gradient changes from point to point. Could there be a formula that gives the gradient at any x?

The derivative f′(x) (or dy/dx) gives the gradient of the tangent to y = f(x) at each x: the instantaneous rate of change. For y = x2, the gradient at x is 2x: at x = 3 the gradient is 6, at x = −1 it is −2.

Figure 1. The gradient function of y = x2 is 2x.

2 The power rule

Multiply by the power, reduce the power by one.

WORKED EXAMPLE 1 Power rule (basic)

Differentiate (a) y = x5; (b) y = 4x3; (c) y = 7x; (d) y = 12.

Solution

(a) 5x4; (b) 12x2A1

Practice exercise 1

1. Differentiate y = 3x4.

2. Differentiate f(x) = −2x6.

3. Differentiate y = 0.5x2.

3 Sums, differences and constant multiples

Term by term.

WORKED EXAMPLE 2 Polynomial (standard)

Differentiate f(x) = 2x3 − 5x2 + 4x − 9.

Solution

f′(x) = 6x2 − 10x + 4A2

Practice exercise 2

1. Differentiate y = x4 − 3x3 + x − 1.

2. Differentiate C(n) = 500 + 12n − 0.03n2.

4 Rewriting before differentiating

Negative powers and expanding.

Write fractions with x in the denominator as negative powers: 3/x2 = 3x−2. Expand brackets and split fractions with a single term in the denominator before differentiating.

WORKED EXAMPLE 3 Negative powers (standard)

Differentiate y = 5/x + 2/x3.

Solution

y = 5x−1 + 2x−3M1

dy/dx = −5x−2 − 6x−4A1

WORKED EXAMPLE 4 Expanding (standard)

Differentiate y = (2x − 1)(x + 3).

Solution

y = 2x2 + 5x − 3M1

dy/dx = 4x + 5A1

Common mistake

Differentiating a product as the product of derivatives: d/dx[(2x − 1)(x + 3)] is not 2 × 1.

Practice exercise 3

1. Differentiate y = 4/x2.

2. Differentiate y = (x3 + 2x)/x.

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

5 Gradient at a point

Substitute into f′.

WORKED EXAMPLE 5 Gradient (standard)

Find the gradient of y = x3 − 4x at x = 2.

Solution

dy/dx = 3x2 − 4A1

At x = 2: 8A1

Practice exercise 4

1. Find the gradient of y = 10/x at x = 5.

2. Find the gradient of f(x) = 2x2 − x at (3, 15).

6 Finding where the gradient has a given value

Solve f′(x) = m.

WORKED EXAMPLE 6 Given gradient (standard)

Find the points on y = x3 − 6x + 1 where the gradient is 6.

Solution

3x2 − 6 = 6, x2 = 4M1

x = ±2: (2, −3) and (−2, 5)A1 A1

Practice exercise 5

1. Find where y = x2 − 8x has gradient 0.

2. Find where y = 2x3 has gradient 24.

7 Equations of tangents

A line touching the curve.

At x = a, the tangent has gradient m = f′(a) and passes through (a, f(a)): y − f(a) = m(x − a).

WORKED EXAMPLE 7 Tangent (standard)

Find the tangent to y = x2 − 3x + 4 at x = 3.

Solution

y(3) = 4; dy/dx = 2x − 3 = 3A1 A1

y − 4 = 3(x − 3): y = 3x − 5A1

Figure 2. The tangent at x = 3.

Practice exercise 6

1. Find the tangent to y = 2x3 at x = 1.

2. Find the tangent to y = 6/x at x = 2.

8 Equations of normals

Perpendicular to the tangent.

The normal at a point is perpendicular to the tangent, so its gradient is −1/f′(a).

WORKED EXAMPLE 8 Normal (standard)

Find the normal to y = x2 − 3x + 4 at x = 3.

Solution

Normal gradient −1/3A1

y − 4 = −(1/3)(x − 3): x + 3y − 15 = 0M1 A1

Practice exercise 7

1. Find the normal to y = x2 at (2, 4).

2. Where does the normal to y = x2 at (1, 1) meet the curve again?

9 Increasing and decreasing functions

Sign of f′.

WORKED EXAMPLE 9 Intervals (standard)

Find where f(x) = x3 − 3x2 − 9x + 2 is decreasing.

Solution

f′(x) = 3x2 − 6x − 9 = 3(x − 3)(x + 1)M1 A1

f′ < 0 for −1 < x < 3A1

Practice exercise 8

1. Find where y = x2 − 10x is increasing.

2. Show that y = x3 + x is increasing for all x.

10 Stationary points

Where the gradient is zero.

At a stationary point, f′(x) = 0. Solve this equation to find the x-coordinates, then substitute into f(x) for the y-coordinates.

WORKED EXAMPLE 10 Stationary points (standard)

Find the stationary points of f(x) = x3 − 3x2 − 9x + 2.

Solution

3(x − 3)(x + 1) = 0M1

(−1, 7) and (3, −25)A1 A1

11 Classifying stationary points

Look at the sign of f′ either side.

If f′ changes from positive to negative, the point is a local maximum; from negative to positive, a local minimum; if f′ does not change sign, a stationary point of inflexion. A sign table is a clear way to show this.

x	−2	−1	0	3	4
f′(x)	+15	0	−9	0	+15
	increasing	max	decreasing	min	increasing

Figure 3. Local maximum and minimum.

Practice exercise 9

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

2. Find and classify the stationary point of y = x3.

12 Maximum and minimum values on an interval

Check the endpoints.

WORKED EXAMPLE 11 On an interval (exam level)

Find the greatest and least values of f(x) = x3 − 3x2 − 9x + 2 for −2 ≤ x ≤ 5.

Solution

Stationary values f(−1) = 7, f(3) = −25A1

Endpoints f(−2) = 0, f(5) = 7A1

Greatest 7, least −25A1

Practice exercise 10

1. Find the maximum of y = 6x − x2 for 0 ≤ x ≤ 2.

13 Optimization: setting up a model

From words to a function.

1. Draw a diagram and define variables. 2. Write the quantity to optimize. 3. Use the constraint to express it in terms of one variable. 4. State the domain. 5. Differentiate and solve f′(x) = 0 (or use a GDC). 6. Justify that it is a maximum or minimum. 7. Answer in context.

14 Optimization: area and volume

Fences and boxes.

WORKED EXAMPLE 12 Fencing (standard)

A farmer has 120 m of fencing to enclose a rectangular pen against a wall (three sides fenced). Find the maximum area.

Solution

Sides x, x and 120 − 2x: A = x(120 − 2x)M1 A1

A′ = 120 − 4x = 0: x = 30M1

A = 1800 m2 (max, since A′ changes from + to −)A1 R1

WORKED EXAMPLE 13 Open box (exam level)

Squares of side x cm are cut from the corners of an 18 cm square sheet to make an open box. Find the maximum volume.

Solution

V = x(18 − 2x)2 = 4x3 − 72x2 + 324xA1

V′ = 12x2 − 144x + 324 = 12(x − 3)(x − 9) = 0M1 A1

x = 3 (x = 9 gives no box): V = 432 cm3R1 A1

Practice exercise 11

1. A rectangle has perimeter 40 cm. Find its maximum area.

2. A box with square base and no lid has volume 32 000 cm3. Find the dimensions that minimize the surface area A = x2 + 128 000/x.

15 Optimization: cost, revenue and profit

Business models.

WORKED EXAMPLE 14 Maximum profit (exam level)

Revenue R(x) = 80x − 0.2x2 and cost C(x) = 1500 + 20x. Find the output that maximizes profit and the maximum profit.

Solution

P(x) = 60x − 0.2x2 − 1500A1

P′(x) = 60 − 0.4x = 0: x = 150M1 A1

P(150) = $3000A1

Practice exercise 12

1. The cost per item of making x items is A(x) = 0.01x2 − 2x + 150. Find the minimum average cost.

2. Revenue is R(p) = p(400 − 8p). Find the price that maximizes revenue.

16 Kinematics: velocity from displacement

v = ds/dt.

If s(t) is the displacement of an object, its velocity is v = ds/dt. The object is momentarily at rest when v = 0, moving forwards when v > 0 and backwards when v < 0.

WORKED EXAMPLE 15 Particle (standard)

s(t) = t3 − 6t2 + 9t m. Find the velocity at t = 2 and when the particle is at rest.

Solution

v = 3t2 − 12t + 9: v(2) = −3 m/sA1 A1

3(t − 1)(t − 3) = 0: t = 1 s and t = 3 sA1

Practice exercise 13

1. A ball's height is h = 20t − 5t2. Find its velocity at t = 1 and its maximum height.

17 Using technology for derivatives

Graphs and numerical derivatives.

GDC tip

Use dy/dx on a graph or nDeriv for gradients at a point. Graph f′(x) to find where it is zero (stationary points) or positive (increasing). For optimization, the 'maximum' tool on the graph of the model gives the answer directly; still state the model and domain.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 16 Mixed problem (exam level)

The tangent to y = ax2 + bx at x = 1 is y = 5x − 2. Find a and b.

Solution

Point (1, 3): a + b = 3A1

Gradient: 2a + b = 5A1

a = 2, b = 1A1

Practice exercise 14

1. Find the coordinates of the point on y = x2 + 2 where the tangent passes through the origin (x > 0).

19 Further derivatives and the chain, product and quotient rules (HL)

Beyond powers.

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

WORKED EXAMPLE 17 Chain and product (exam level)

Differentiate (a) y = e3x; (b) y = (2x + 1)5; (c) y = x2sin x; (d) y = ln x/x.

Solution

(a) 3e3x; (b) 10(2x + 1)4A1 A1

(d) (1 − ln x)/x2A1

Practice exercise 15

1. Differentiate y = 50e−0.2t.

2. Differentiate y = sin(3x) cos x.

20 The second derivative, concavity and acceleration (HL)

The rate of change of the gradient.

The second derivative f″(x) measures how the gradient changes. f″ > 0: concave up (local minimum if f′ = 0); f″ < 0: concave down (local maximum). A point of inflexion is where concavity changes. In kinematics, a = dv/dt = d2s/dt2.

WORKED EXAMPLE 18 Second derivative test (standard)

Use f″ to classify the stationary points of f(x) = x3 − 3x2 − 9x + 2.

Solution

f″(x) = 6x − 6A1

f″(−1) = −12 < 0: maximum; f″(3) = 12 > 0: minimumR1 R1

Practice exercise 16

1. Find the point of inflexion of y = x3 − 6x2 + 5.

2. s = t3 − 6t2 + 9t. Find the acceleration at t = 3.

21 Investigation: the most economical can

A Paper 3 style exploration.

A can (closed cylinder) must hold 330 cm3. 1. Show that the surface area is A = 2πr2 + 660/r. 2. Find the radius that minimizes A and the corresponding height. 3. Show that for the optimal can, the height equals the diameter. 4. Real cans have r ≈ 3.3 cm and h ≈ 11.5 cm. Compare and suggest reasons for the difference. 5. If the top and bottom cost twice as much per cm2 as the side, find the new optimal radius.

Answers. 1. h = 330/(πr2). 2. r = (165/π)1/3 = 3.74 cm, h = 7.49 cm. 3. h = 2r at the optimum. 4. Real cans are taller: easier to hold, stack and label. 5. Cost ∝ 4πr2 + 660/r: r = (82.5/π)1/3 = 2.97 cm.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

f(x) = 2x3 − 9x2 + 12x.

(a) Find f′(x). [2]

(b) Find the coordinates of the stationary points. [3]

Markscheme

Part	Answer	Marks
(a)	6x2 − 18x + 12	A2
(b)	6(x − 1)(x − 2) = 0	M1
	(1, 5)	A1
	(2, 4)	A1
(c)	sign of f′ changes + to − at x = 1: max	R1
	− to + at x = 2: min	R1

Examiner's comments. A sign table is a clear justification.

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

A closed box has a square base of side x cm and height h cm. Its volume is 1000 cm3.

(a) Show that the surface area is A = 2x2 + 4000/x. [3]

(b) Find dA/dx. [2]

Markscheme

Part	Answer	Marks
(a)	x2h = 1000	M1
	A = 2x2 + 4xh	A1
	substitute h = 1000/x2	AG A1
(b)	4x − 4000x−2	A2
(c)	4x = 4000/x2: x3 = 1000	M1
	x = 10	A1
	A = 600 cm2	A1

Examiner's comments. The optimal box is a cube.

Question 3 (Paper 1 style) (about 8 minutes)

The curve y = x2 − 4x + 1.

(a) Find the gradient at x = 3. [2]

(b) Find the equation of the normal at x = 3. [3]

Markscheme

Part	Answer	Marks
(a)	2x − 4	M1
	2	A1
(b)	point (3, −2); gradient −1/2	A1
	y + 2 = −½(x − 3)	M1
	y = −x/2 − 1/2	A1

Examiner's comments. The normal gradient is the negative reciprocal.

23 Common misconceptions and links

Students often think…	Actually…
d/dx(5) = 5	0
normal gradient = −f′(a)	−1/f′(a)
f′ = 0 always gives a max	classify it
ignore endpoints	check them for global extremes

How 5.3 connects to the rest of the course

Topic	Connection
5.1 Limits	the derivative as a limit
5.5 Integration	reverse process
2.2 Functions	lines and gradients
5.14 Differential equations	rates of change

24 Check your understanding

Section A: multiple choice

1. d/dx(4x3) =

A. 12x2 B. 4x2 C. 12x3 D. x4

2. d/dx(3/x) =

A. 3/x2 B. 3 ln x C. −3/x2 D. −3/x

3. At a stationary point

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

4. Velocity is

A. dv/dt B. s/t C. ∫s dt D. ds/dt

5. A tangent at (2, 5) with gradient 3 is

A. y = 3x + 5 B. y = 3x − 1 C. y = 5x − 3 D. y = −x/3 + 17/3

Section B: structured

6. Find the stationary point of y = x2 − 6x + 11 and its nature. [3]

7. Find where y = x3 − 27x is increasing. [2]

Answers

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

6. (3, 2): minimum

7. x < −3 or x > 3

25 Summary and topic map

Idea	Remember
Power rule	anxn−1
Tangent	y − f(a) = f′(a)(x − a)
Normal	gradient −1/f′(a)
Increasing	f′ > 0; decreasing f′ < 0
Stationary points	f′ = 0; classify with sign or f″
Optimization	model, differentiate, justify, interpret
Kinematics	v = ds/dt

Before the exam, make sure you…

•rewrite as powers first
•justify maxima and minima
•check domain and endpoints
•answer in context with units

Derivatives and Their Applications 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 Derivatives and Their Applications.

26 Answers to practice exercises

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

Practice exercise 1

1. 12x3

2. −12x5

3. x

Practice exercise 2

1. 4x3 − 9x2 + 1

2. 12 − 0.06n

Practice exercise 3

1. −8x−3

2. 2x

3. 2x + 4

Practice exercise 4

1. −0.4

2. 11

Practice exercise 5

1. (4, −16)

2. x = ±2: (2, 16) and (−2, −16)

Practice exercise 6

1. y = 6x − 4

2. y = −1.5x + 6

Practice exercise 7

1. y = −x/4 + 4.5

2. y = −x/2 + 1.5; x2 + x/2 − 1.5 = 0: x = −1.5, point (−1.5, 2.25)

Practice exercise 8

1. x > 5

2. dy/dx = 3x2 + 1 > 0

Practice exercise 9

1. max (−2, 16), min (2, −16)

2. (0, 0): stationary inflexion

Practice exercise 10

1. 8 at x = 2 (endpoint)

Practice exercise 11

1. 100 cm2

2. x = 40 cm, height 20 cm

Practice exercise 12

1. $50 at x = 100

2. $25

Practice exercise 13

1. 10 m/s; 20 m at t = 2

Practice exercise 14

1. tangent at a: y = 2ax − a2 + 2 through (0, 0): a = √2, point (√2, 4)

Practice exercise 15

1. −10e−0.2t

2. 3cos 3x cos x − sin 3x sin x

Practice exercise 16

1. (2, −11)

2. a = 6t − 12 = 6 m/s2

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