DP Maths AA SL - Calculus - Limits and Continuity

DP Maths AA SL - Calculus - Limits and Continuity - Full notes

RevisionPrep

DP Maths AA SL | Limits and Continuity

Limits and Continuity

This topic contains 1 note:

5.1 Introduction To Differentiation

DP MATHEMATICS AA · TOPIC 5: CALCULUS

5.1

Introduction to Differentiation

Full notes for SL · 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.

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

revisionprep.comIB-style notes · question bank · mock papers