MEI Core 3 - Woodhouse Collegevle.woodhouse.ac.uk/topicdocs/maths/ppANDms/C3NandE2011.pdf · MEI...

MEI Core 3

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Proof

Section 1: Methods of proof

Notes and Examples These notes have subsections on:

What is proof?

An interesting example: Fermat’s last theorem

Direct proof

Proof by contradiction

Proof by exhaustion

Disproof using a counter-example

What is proof? The word “proof” is used in many different contexts in everyday life. We often

hear that scientists have “proved” a previously unknown fact about the world we live in, or we hear that forensic evidence in a court case has “proved” the guilt or innocence of a defendant.

In reality these situations do not involve proof in the mathematical sense. Scientists and lawyers use evidence that they collect to make deductions. In a court of law the evidence needs only to show “beyond reasonable doubt” that

a defendant is guilty, but it can never be absolutely without doubt. DNA testing can offer overwhelmingly strong evidence, but even in these cases

there is a very small probability that two people could have very similar DNA, or that the sample was contaminated.

Similarly, evidence collected by a scientist from observations may be overwhelmingly in favour of a particular theory, but it is always possible that further observations might contradict the theory. To take a very simple

example, suppose you want to test whether or not a coin is biased. Suppose you tossed the coin 1000 times and got heads 1000 times. This seems to be very strong evidence that the coin is biased towards heads. However, it is

always possible (though extremely unlikely) that if you tossed the coin another 1000 times you would get tails 1000 times! (If you study Statistics, you will learn, or may have already learned, how hypothesis testing helps us

to make a decision about whether evidence is sufficient to support a theory. However, this does not prove that a theory is true!)

In real life, strong evidence for a particular claim is normally sufficient. In fact it has to be, since in the situations described above it is impossible to offer a completely watertight proof. However, a mathematician is never satisfied with

overwhelming evidence. A mathematician wants to prove that a particular result is always true, in all possible cases.

MEI Core 3 Proof Section 1 Notes and Examples

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To disprove a false conjecture, all you need to do is to find one counter-example for which the conjecture is not true. However, this may involve a lot of looking!

An interesting example: Fermat’s Last theorem Pierre de Fermat, a lawyer and amateur mathematician, was born in 1601.

After his death a note was found in the margin of one of his books stating a conjecture that is known as “Fermat’s last theorem”, and the words “I have discovered a truly marvellous proof of this, but this margin is too narrow to

contain it.” The conjecture itself is fairly straightforward to understand. You probably

know that the equation 2 2 2a b c

has solutions for which a, b and c are all integers, such as 3, 4, 5 (3² + 4² = 5²)

and 5, 12 , 13 (5² + 12² = 13²). These are known as Pythagorean triples as they

correspond to the length of the sides of a right-angled triangle, for which

Pythagoras’ theorem applies. Fermat’s conjecture was that there are no integer solutions to the equation

n n na b c for any values of n greater than 2.

Various mathematicians have proved that this conjecture is true for particular values of n, such as n = 3 and n = 4. Computers allow a large number of cases

to be checked in a short time, and by 1982 it had been shown that if a counter-example existed (i.e. a set of integers a, b, c and n for which

n n na b c ) then it involved values of a, b, c and n greater than 4,000,000.

Since the value of na for such numbers is greater than the total number of the

elementary particles in the known universe, it seemed unlikely that a counter-

example would be found! However, even this overwhelming evidence is not enough for a mathematician!

Fermat’s Last Theorem was finally proved in 1994 by Andrew Wiles of Cambridge, who had been obsessed since boyhood by the problem. Since his proof involved around 150 pages and used mathematical techniques

which would not have been available to Fermat in the 17th century, it seems unlikely that Wiles’ proof is the same as Fermat’s “truly marvellous proof”. It is now generally believed that Fermat had found a flaw in his own proof, as no

further references to it have been found in his work. However, we cannot be sure of this, so although the theorem has been proved, a mystery remains: what was Fermat’s “truly marvellous proof”, and was it valid?

In this section you will look at three main methods of proof: direct proof; proof by exhaustion, and proof by contradiction. You will also look at disproof by

use of a counter example.


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You will find examples of all these types of proof in the A2 Pure Mathematics textbook. Some additional notes and further examples of each type are given below.

Direct proof The main problem with direct proof is often deciding where to start. In many cases it will be useful to express the problem algebraically, and sometimes

simple algebraic manipulation is all that is needed. In geometrical proofs a diagram is likely to be an essential part of the proof. Often you will need to use results that you know, such as Pythagoras’ theorem or the quadratic

formula. Don’t be afraid to try something which you aren’t sure whether it will work: even if a particular approach doesn’t work, it may give you another idea.

It is important to be sure that each step that you make in your proof is valid. The web link below shows some incorrect proofs, for which you are invited to

find the invalid step. (One or two involve mathematics which you may not yet have covered, but there are several which you should find useful).

http://www.math.toronto.edu/mathnet/falseProofs/fallacies.html

Example 1 shows an example of a direct proof where a simple piece of algebraic manipulation is all that is needed.

Example 1

Prove that every odd integer is the difference of two perfect squares.

Solution

Any odd integer can be expressed in the form 2a + 1.

2a + 1 = a² + 2a + 1 – a²

= (a + 1)2 – a

2.

The note after Example 1.3 in the A2 Pure Mathematics textbook mentions two well-known results on divisibility. These can be proved using direct proof. This is shown in Example 2.

Example 2

Prove that:

(i) a number is divisible by 3 if and only if the sum of its digits is divisible by 3

(ii) a number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Solution

This conjecture involves numbers with any number of digits.

A number with n digits can be written as

1 2 3.... na a a a , where 1 2 3, , ,..., na a a a are all single digits.

Thinking about what you are aiming for helps to give the

idea for this step

http://www.math.toronto.edu/mathnet/falseProofs/fallacies.html


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This number can be expressed as 1 2

1 2 3 110 10 10 .... 10n n n

n na a a a a

1

1 1 2 2 1 1

1

1 2 1 1 2

(10 1) (10 1) .... 9

(10 1) (10 1) ... 9 ( ... )

n n

n n n

n n

n n

a a a a a a a

a a a a a a

All the numbers of the form 10 1n are numbers in which every digit is 9, so all these

numbers are divisible by both 3 and 9.

Therefore:

(i) the number 1 2 3.... na a a a is divisible by 3 if and only if

1 2 3 ... na a a a is

divisible by 3

(ii) the number 1 2 3.... na a a a is divisible by 9 if and only if

1 2 3 ... na a a a is

divisible by 9

Proof by contradiction In this method, you assume that the converse (i.e. the opposite) of the

conjecture is true, and then show that this leads to a contradiction. Examples 1.4 and 1.5 in the A2 Pure Mathematics textbook show this method. Here is a further example.

Example 3

Prove that there are infinitely many prime numbers.

Solution

Assume that there are only finitely many prime numbers. This means that all of them

can be listed as follows: p1, p2 ..., pn, where n is a finite number.

Now think about the number defined by q = p1p2... pn + 1. If you divide q by any of

the prime numbers p1, p2 ..., pn into q, there is a remainder of 1. So none of the prime

numbers are factors of q. Therefore q must be a prime number, not among the primes

listed above, contradicting the assumption that all primes are in the list p1, p2 ..., pn.

Therefore there are infinitely many prime numbers.

Proof by exhaustion

It is important to realise that proof by exhaustion may not necessarily involve testing every single case individually. In some cases it is possible to test a number of possible subsets of cases, as shown in the example below.

Example 4

Prove that no square number ends in an 8.


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Solution

Obviously it is not possible to test every square number! However, the final digit of

any square number is determined only by the final number of its square root, so you

can reduce the problem to considering 1-digit numbers.

0² = 0 so the square of any number ending in 0 ends in 0.










Therefore no square number ends in an 8.

Disproof using a counter-example This is often very simple. You may be able to find a counter-example very

quickly, and this is enough to disprove a conjecture. However, if you are unable to spot a counter-example, this does not mean that one does not exist. Using computers to check Fermat’s Last Theorem up to very large numbers,

as mentioned earlier, was an attempt to disprove the theorem rather that to prove it. However, there have been examples of conjectures for which counter-examples have been found after many years. In the 18th century Euler proposed a conjecture along similar lines to Fermat’s Last Theorem:

that the equation

1 2 3 1...n n n n n

n na a a a a

has no integer solutions for any n > 2. In this conjecture, notice that the

number of unknowns is equal to the degree of the equation. So this

conjecture states that there are no integer solutions to the equations 3 3 3a b c , 4 4 4 4a b c d , 5 5 5 5 5a b c d e etc. But in 1966, about 200

years after the conjecture was proposed, a counter example was found for

the case n = 5: 5 5 5 5 527 84 110 133 144 .

Example 5

It is suggested that for every prime number p, 2p + 1 is also prime. Prove that this is

false.

Solution

For p = 2, 2p + 1 = 5 which is prime.



For p = 7, 2p + 1 = 15 which is not prime.

The statement is false.

MEI Core 3

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Natural logarithms and exponentials

Section 1: Natural logarithms and exponentials

Notes and Examples These notes contain subsections on:

Natural logarithms

Solving equations using natural logarithms and exponentials

Natural logarithms

You have already met logarithms in Core 2 (chapter 11 of the AS Pure Mathematics textbook). You learnt that logarithms are related to indices:

logb

aa c c b

and that logarithms satisfy the following rules:

log log log

log log log

log log

c c c

c c c

n

c c

a b ab

aa b

b

a n a

In this chapter you are looking at natural logarithms, which are logarithms to base e, where e is a particular irrational number (e = 2.71828…. to 5 decimal

places). Working with natural logarithms is quite straightforward: they obey exactly the same rules as all other logarithms, so you already know quite a lot about them. Your calculator will be able to work out natural logarithms – you should have a button marked “ln” (note that this is ln not In; some people get

confused about this!)

You may be wondering (quite reasonably) what is so special about this number e, and why logarithms to this particular base are so useful. In fact, this

number has very many interesting properties, some of which you will learn about in this module and in others:

In the textbook you are shown how natural logarithms are related to

the area under the curve 1

yx

. You will find out more about this in

chapter 5. You can use the Geogebra resource Area under the

graph of y = 1/x to explore these ideas.

You will also learn more about the exponential function ex in chapters

4 and 5: one of its interesting characteristics is that the gradient of the

graph of exy is equal to the value of ex at all points.

MEI C3 Exp. & logs Section 1 Notes and Examples

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In Core 4 you will learn to form and solve simple differential equations (chapter 12 of the A2 Pure Mathematics textbook) and you will see that many “real-life” situations (for example, population growth; the

temperature of a hot object in a cool room) can be modelled using exponential functions.

If you go on to study Further Maths „A‟ level, you will find that the

number ex can be written as the infinite series 2 3 4

1 ...2! 3! 4!

x x xx

(try working out the sum of the first 10 or so terms of this series for x =

1 on your calculator, and see how close you get to the value of e).

But for now, all that is necessary is to accept that e, like , is a special and

useful number, and that this means that logarithms to base e are also

particularly useful.

Solving equations using natural logarithms and exponentials Most of the work in this chapter involves the same techniques as you used when studying logarithms in Core 2. You need to use the same rules of

logarithms, applied to natural logarithms:

ln ln ln

ln ln ln

ln lnn

a b ab

aa b

b

a n a

Also remember that since the log 1a a for any value of a, then

ln e 1

This means that you can use natural logarithms to solve equations involving exponentials, as in Example 1.

Example 1

Solve the equation 1 3e 5x .

Solution

Take natural logarithms of both sides: 1 3ln e ln5x

Using the laws of logarithms: (1 3 )lne ln5x

Since ln e = 1: 1 3 ln 5

1 ln 50.203

3

x

x

Similarly, the relationship between exponentials and logarithms

ln aa b e b

allows you to solve equations involving natural logarithms.

MEI C3 Exp. & logs Section 1 Notes and Examples

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Example 2

Solve the equation ln(2 1) 3x .

Solution

ln(2 1) 3x 3

3

2 1 e

19.54

2

x

ex

In Examples 1 and 2 above, you are using the fact that the exponential function and the natural logarithm function are inverses of one another, in the

same way that squaring and the square root function are inverses of one another. (You will learn more about functions and their inverses in chapter 3). In Example 1 you are “undoing” an exponential function by using natural

logarithms, and in Example 2 you are “undoing” a natural logarithm by using an exponential.

The inverse nature of these two functions can be summed up as follows:

ln

ln e

e

x

x

x

x

Examples 2.1 and 2.2 in the textbook show some “real-life” examples of the exponential function and the use of natural logarithms to solve problems involving exponential functions. Read these examples carefully. Make sure

that you know how to work out exponentials and logarithms on your calculator correctly – try using your calculator to check the examples in the book for yourself.

You can try some similar problems using the interactive questions Modelling population growth.

Example 2.3 in the textbook demonstrates rearranging a formula involving a

natural logarithm: the method used is similar to that used in Example 2 above.

In the first line of working both sides are written as powers of e, then lne x x

is used to simplify. You can miss out this line of working if you prefer, as in

Example 2 above.

MEI Core 3

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Functions

Section 1: Functions, graphs and transformations

Notes and Examples These notes contain subsections on:

The language of functions

Using transformations to sketch the graphs of functions

Summary of transformations of the graph of y = f(x)

The language of functions This first part of the chapter introduces the different types of mappings, using “real-life” contexts. Make sure that you understand the terminology introduced in this section before moving on to mathematical mappings. Here are some further practical examples of the different types of mappings: 1. Mapping from the set of children who live in a particular street to the set of

types of pet. This mapping is many-to-many since a particular child may have more than one pet, and also a particular type of pet can be owned by more than one child. The co-domain is the set of all possible pets, but as only four different types of pet are owned by this particular set of children, then the range of the mapping is the set {dog, cat, rabbit, fish}.

Domain: Children

Andy

Beth

Chris

Debbie

Eleanor

Co-domain: Types of pet

Dog

Cat

Rabbit

Fish

MEI C3 Functions Section 1 Notes and Examples

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2. Mapping from the set of mothers at a toddler group to the set of children

at the group.

This is a one-to-many mapping: a particular mother may have more than one child, but a particular child has only one mother. In this case the range is the same as the co-domain since the co-domain is defined as the children present at the group, rather than the set of all children.

3. Mapping from the set of children at a toddler group to their ages.

This is a many-to-one mapping: a particular child has only one age, but there may be more than one child of a particular age. In this case the co-domain is the set of possible ages for children attending a toddler group {0, 1, 2, 3, 4}, but as there are no 2-year-olds in this particular group, the range is the set {0, 1, 3, 4}

Domain: mothers

Pat

Susan

Becky

Michelle

Co-domain: children

Alex

Chloe

Molly

William

Thomas

Katie

Domain: children

Alex

Chloe

Molly

William

Thomas

Katie

Co-domain: ages

0

1

2

3

4


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4. Mapping from the set of women at a ballroom dancing class to the set of

their partners.

This is a one-to-one mapping: each woman has exactly one partner, and each man is the partner of exactly one woman.

In this case the range is equal to the co-domain. Notice that there are sometimes several different possible definitions for the co-domain: in example 3 the co-domain could have been defined as all possible ages (i.e. all whole numbers up to a reasonable maximum) rather than just the ages {0,1,2,3,4}. You may wonder why it is necessary to define a co-domain which is different from the range – why not just make the co-domain the same set as the range? The answer is that you can, but in some cases the range can only be defined by listing its values, and it is easier to give a co-domain which can be defined simply (see example 3 of the mathematical mappings below). The examples above can be helpful to familiarise yourself with the terminology associated with mappings, but in situations like these it is sometimes difficult to define the sets involved clearly and unambiguously (for example, what would you include in types of pet?). However, in mathematical mappings, sets can be defined precisely. Examples of possible domains and co-domains include the set of integers, the set of real numbers, or a restricted range of values such as the set of real numbers x for which 0 < x < 1. You may already be familiar with the mathematical “shorthand” that is often used to define particular sets of numbers. Here is a reminder of some of the notation which you may find useful:

means “is an element of” : means “such that”

denotes the set of integers denotes the set of rational numbers

denotes the set of real numbers denotes the set of positive real numbers (similarly for rational

numbers, integers etc.)

Domain: women

Liz

Jo

Melanie

Anna

Wendy

Co-domain: men

Mark

Paul

Rob

Phil

Steve


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So, for example,

x means that x is a real number : 0 10y y means that y is an integer such that y lies between

0 and 10. Here are some examples of mathematical mappings: 1. 1 3 ,x x x

This is a one-to-one mapping. For every value of x, there is one value of 1 – 3x, and no two objects map to the same image.

The co-domain of this mapping is also , and the range is equal to the

co-domain, since all rational numbers are the image of another rational number under this mapping. This mapping is also a function as there is only one possible image for each object.

2. 2 3,x x x

This is a many-to-one mapping, since each point has only one possible image, but, for example, -1 and 1 both have image 4. The co-domain of this mapping is also , but the image points are all greater or equal to 3, so the range can be written as : 3y y . You

could define the co-domain as this set if you wish. This mapping is also a function as there is only one possible image for each object.

3. ,x x x

This is a one-to-many mapping, since each object has two possible images, but no two objects map to the same image. The co-domain of this mapping is . The range is not the whole of , since there are plenty of real numbers which are not the square root of an

integer. The range can be written as 2:y y .

This mapping is not a function as there is more one possible image for each object.

4. 2 1,x x x

This is a many-to-many mapping, since each object has two possible

images (e.g. 1 maps to 2 and 2 ) and two objects map to each

image (e.g. 1 and -1 both map to 2 ). The co-domain and range of this mapping are both . This mapping is not a function as there is more one possible image for each object.

The Mathcentre video Introduction to functions covers definitions and terminology of functions.


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Using transformations to sketch the graphs of functions In C1 (chapter 3 of the AS Pure Mathematics textbook) you looked at the effect of a translation on the equation of a graph. You found that:

y = f(x – a) is the equation of the graph obtained when the graph of y = f(x) is moved a units in the positive x direction

y = f(x) + b is the equation of the graph obtained when the graph of y = f(x) is moved b units in the positive y direction

combining the above, y = f(x – a) + b is the equation of the graph

obtained when the graph of y = f(x) is translated through a

b

In C2 (chapter 10 of the AS Pure Mathematics textbook) you looked at the effect of a one-way stretch on the equation of a graph. You found that

y = af(x) is the equation of the graph obtained when the graph of y = f(x) is stretched with scale factor a parallel to the y axis

y = f(ax) is the equation of the graph obtained when the graph of y = f(x)

is stretched with scale factor 1

a parallel to the x axis

In this section you will look at more complicated examples, in which you combine translations and stretches. The main challenge is to make sure that you get the transformations in the correct order. Activity 3.1 in the textbook shows that results are sometimes different when you apply transformations in a different order. It is useful to look at the cases in this Activity, applied to a general graph y = f(x).

(i) (a) Translating the graph y = f(x) through 3

0

gives the graph y = f(x – 3).

Stretching the new graph y = f(x – 3) with a scale factor of 2 parallel to

the x axis gives the graph f 32

xy

(b) Stretching the graph y = f(x) with a scale factor of 2 parallel to the x

axis gives the graph f2

xy

. Translating the new graph f

2

xy

through 3

0

gives the graph 3

f2

xy

.

(ii) (a) Translating the graph y = f(x) through 3

0

gives the graph y = f(x – 3).

Stretching the new graph y = f(x – 3) with a scale factor of 2 parallel to the y axis gives the graph y = 2f(x – 3).


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(b) Stretching the graph y = f(x) with a scale factor of 2 parallel to the y axis gives the graph y = 2f(x). Translating the new graph y = 2f(x)

through 3

0

gives the graph y = 2f(x – 3).

(iii) (a) Translating the graph y = f(x) through 0

3

gives the graph y = f(x) + 3.

Stretching the new graph y = f(x) + 3 with a scale factor of 2 parallel to

the x axis gives the graph f 32

xy

.

(b) Stretching the graph y = f(x) with a scale factor of 2 parallel to the x


xy

. Translating the new graph f

2

xy

through 0

3

gives the graph f 32

xy

.

(iv) (a) Translating the graph y = f(x) through 0

3

gives the graph y = f(x) + 3.

Stretching the new graph y = f(x) + 3 with a scale factor of 2 parallel to the y axis gives the graph y = 2(f(x) + 3) = 2f(x) + 6

(b) Stretching the graph y = f(x) with a scale factor of 2 parallel to the y axis gives the graph y = 2f(x). Translating the new graph y = 2f(x)

through 0

3

gives the graph y = 2f(x) + 3.

(v) (a) Stretching the graph y = f(x) with a scale factor of 2 parallel to the x


xy

. Stretching the new graph f

2

xy

with a scale factor of 3 parallel to the y axis gives the graph

3f2

xy

.

(b) Stretching the graph y = f(x) with a scale factor of 3 parallel to the y axis gives the graph y = 3f(x). Stretching the new graph y = 3f(x) with a

scale factor of 2 parallel to the x axis gives the graph 3f2

xy

.

You can see that (i) and (iv) give different results when the transformations are applied in a different order, whereas in (ii), (iii) and (v) it makes no difference. Notice that the issue of order occurs when there is both a stretch and a translation involving the same direction: in such cases the order of transformation matters. A transformation involving, say, a stretch in the y direction and a translation in the x direction can be done in either order.


© MEI, 31/05/10 7/8

You can investigate the effect of two transformations using the Geogebra resource Combinations of transformations. Example 1

Explain how the graph y = 3sin(2x – 1) + 2 can be obtained from the graph y = sin x

using successive transformations.

Solution

The order of transformations is as follows:

1. Transform y = sin x to y = 3sin x Stretch scale factor 3

parallel to the y axis

2. Transform y = 3sin x to y = 3sin(x – 1) + 2 Translation of 1

2

3. Transform y = 3sin(x – 1) + 2 to y = 3sin(2x – 1) + 2 Stretch scale factor ½

parallel to the x axis

Example 2

The graph y = x² undergoes the following transformations:

One way stretch, scale factor 2, parallel to the x axis

Translation through 2

1

One way stretch, scale factor 3, parallel to the y axis Find the equation of the new graph.

Solution

First apply a one way stretch, scale factor 2, parallel to the x axis, to the graph y = x².

This results in the graph

2

2

xy

.

Next, apply a translation through 2

1

to the graph

2

2

xy

. This results in the

graph

22

12

xy

.

Finally, apply a one way stretch, scale factor 3, parallel to the y axis. This results in

the graph

22

3 12

xy

.

Since 2x – 1 means “multiply by 2 then subtract 1, it is very tempting to think that the stretch parallel to the x axis, which results in the 2x, must come before the translation in the x direction, which results in the - 1. But this order would result in

first y = sin(2x), then y = sin(2(x – 1)).


© MEI, 31/05/10 8/8

This can be rewritten as:

2

2

234

234

23 3

2

4 43 3

4

3 3 3

3

xy

x x

x x

x x

For additional practice, try the interactive resources Transformations of graphs.

In the final part of this section, you learn two further transformations of graphs:

y = f(x) is the equation of the graph obtained when the graph of y = f(x) is reflected in the x axis

y = f( x) is the equation of the graph obtained when the graph of y = f(x) is reflected in the y axis

In fact these are just special cases of the one-way stretches you already

know about. The equation y = f(x) represents a stretch with scale factor 1 parallel to the y axis, which is the same as reflection in the x axis. The

equation y = f( x) represents a stretch with scale factor 1 parallel to the x axis, which is the same as reflection in the y axis.

Summary of transformations of the graph of y = f(x)

Function Transformation f(x – a) + b

Translation through a

b

af(x) Stretch parallel to y axis, scale factor a f(ax)

Stretch parallel to x axis, scale factor 1

a

-f(x) Reflection in the x axis

f(-x) Reflection in the y axis

MEI Core 3

© MEI, 31/05/10 1/5

Functions

Section 2: Composite and inverse functions

Notes and Examples These notes contain subsections on

Composite functions

Inverse functions

The inverse trigonometric functions

Composite functions The important thing to remember when finding a composite function is the order in which the functions are written: fg(x) means first apply the function g to x, then apply the function f to the result. Example 1

The functions f, g and h are defined by:

2

f ( ) 1

g( )

h( ) 3

x x

x x

x x

Find the following composite functions:

(i) fg(x) (ii) gh(x) (iii) hgf(x) (iv) f²(x)

Solution

(i)

2

2

fg( ) f[g( )]

f ( )

1

x x

x

x

(ii)

2

2

gh( ) g[h( )]

g(3 )

(3 )

9

x x

x

x

x

(iii)

2

2

hgf ( ) hg[f ( )]

h[g( 1)]

h[( 1) ]

3( 1)

x x

x

x

x

Apply g followed by f; i.e. square, then add 1.

Apply h followed by g;

i.e. multiply by 3, then square.

Apply f followed by g followed by h;

i.e. add 1, then square, then multiply by 3.


© MEI, 31/05/10 2/5

(iv) 2f ( ) f[f ( )]

f ( 1)

( 1) 1

2

x x

x

x

x

Example 2

The functions f and g are defined as:

1f ( )

g( ) 2

xx

x x

Write the functions:

(i) 1

2x (ii)

2

x (iii) 4x

in terms of the functions f and g.

Solution

(i) This function is obtained by first multiplying by 2, then taking the reciprocal; i.e.

applying g followed by f. So this function is fg.

(ii) This function is obtained by first taking the reciprocal, then multiplying by 2; i.e.

applying f followed by g. So this function is gf.

(iii) This function is obtained by multiplying by 2 twice; i.e. applying g twice. So this

function is g².

You may find the Mathcentre video Composition of functions useful.

Inverse functions Example 3

The function f is defined by f(x) = 2x³ + 1. Find the inverse function f-1

.

Solution

The function can be written as: y = 2x³ + 1

Interchanging x and y: x = 2y³ + 1

Rearranging: x – 1 = 2y³

3

3

1

2

1

2

xy

xy

The inverse function is 1 31

f ( )2

xx .

Apply f twice;

i.e. add 1, then add 1.


© MEI, 31/05/10 3/5

Remember that to for a function to have an inverse function, it must be one-to-one (you can find the inverse of a many-to-one mapping, but this would be a one-to-many mapping, which is not a function). Example 4

The functions f and g are defined as:

2f ( ) 4 0

g( ) 3 3

x x x

x x x

(i) What is the range of each function?

(ii) Find the inverse function f-1

, stating its domain.

(iii) Find the inverse function g-1

, stating its domain.

(iv) Write down the range of f-1

and the range of g-1

.

Solution

(i) The range of f is f(x) -4.

The range of g is g(x) 0.

(ii) The function can be written as 2 4y x

Interchanging x and y: 2 4x y

Rearranging: 24

4

x y

y x

The domain of f-1

is the same as the range of f.

1f ( ) 4 4x x x

(iii) The function can be written as 3y x

Interchanging x and y: 3x y

Rearranging:

2

2

3

3

x y

y x

The domain of g-1

is the same as the range of g.

1 2g ( ) +3 0x x x

(iv) The range of f-1

is f-1

(x) 0 (the same as the domain of f)

The range of g-1

is g-1

(x) 3 (the same as the domain of g)

You can investigate inverse functions and their graphs using the Geogebra resource Inverse functions. You may also find the Mathcentre video Inverse functions useful.


© MEI, 31/05/10 4/5

The inverse trigonometric functions The trigonometric functions sin x, cos x and tan x are many-to-one functions, i.e. for any particular output of these functions, there is more than one input (in fact, for these functions there are an infinitely large number of inputs). For example, there are an infinite number of values of x for which sin x = 0.5. It is therefore not possible to find inverse functions for the trigonometric functions. However, it is possible to find inverse functions if the domains are restricted so that the functions are one-to-one. These inverse functions are arcsin x, arccos x and arctan x. In some textbooks these functions are called sin

-1x, cos

-1x and tan

-1x.

You can look at these inverse functions using the Geogebra resource Inverse trigonometric functions.

In the case of sin x, the restricted domain used is 1 12 2

x .

–¤/2 ¤/2

–2

–1

1

2

x

y

The domain of the function y = arcsin x is 1 1x , and its range is

1 12 2

y

In the case of cos x, a different restricted domain is needed, since the domain

1 12 2

x does not cover the whole range of cos x. The restricted domain

used is 0 x .

–¤ –¤/2 ¤/2 ¤

–3

–2

–1

1

2

3

x

y

The domain of the function y = arcos x is 1 1x , and its range is 0 y .

In the case of tan x, the restricted domain used is again 1 12 2

x .

y = sin x

y = arcsin x

y = cos x

y = arcos x


© MEI, 31/05/10 5/5

–2¤ –¤ ¤ 2¤

–4

–3

–2

–1

1

2

3

4

x

y

The domain of the function y = arctan x is x , and its range is 1 1

2 2y .

y = tan x

y = arctan x

MEI Core 3

© MEI, 25/03/09 1/7

Functions

Section 3: Types of function


Even, odd and periodic functions

The modulus function

Inequalities

Even, odd and periodic functions

A function is even if its graph has the y axis as a line of symmetry.

For an even function f(-x) = f(x).

A function is odd if its graph has rotational symmetry of order 2 about

the origin. For an odd function f(-x) = -f(x).

A function is periodic if its graph has a repeating pattern, i.e. if there is a value of k for which f(x + k) = f(x) for all x.

The smallest possible value of k is the period of the function.

You can investigate the graphs of odd and even functions using the Geogebra resources Odd functions and Even functions.

Example 1

(a) Classify each of the following functions as even, odd, or neither odd nor even.

(i) 2f ( ) 2x x (ii) f ( ) tan3x x

(iii) 3f ( ) 2 1x x x (iv) 2 1

f ( )x

xx

(v) 12

f ( ) cosx x (vi) 3

1f ( )

1

xx

x

(b) Which of the above functions are periodic? Give the period in each case.

Solution

(a) (i) 2 2f ( ) ( ) 2 2 f ( )x x x x

This is an even function.

(ii) f ( ) tan( 3 ) tan3 f ( )x x x x

This is an odd function.

(iii) 3 3f ( ) 2( ) ( ) 1 2 1x x x x x

This function is neither odd nor even.


© MEI, 25/03/09 2/7

(iv) 2 2( ) 1 1

f ( ) f ( )x x

x xx x

This is an odd function.

(v) 3 3

1 ( ) 1f ( )

1 ( ) 1

x xx

x x

This function is neither odd nor even.

(vi) 1 12 2

f ( ) cos( ) cos f ( )x x x x

This is an even function.

(b) (ii) is a periodic function. The period of tan x is , so the period of tan 3x is 3

.

(v) is a periodic function. The period of cos x is 2, so the period of 12

cos x is

1

2

24

Example 2

A periodic function with period 2 is defined as:

0 1

f ( )2 1 2

x xx

x x

Sketch the graph of y = f(x).

Solution

When x is between 0 and 1, the graph is a straight line through the origin with

gradient 1.

When x is between 1 and 2, the graph is a straight line with gradient 1 passing

through (1, 1) and (2, 0).

Since the function is periodic with period 2, the rest of the graph can be sketched

easily.

1 2 x

y

x

y

2 4 6 -2 -4


© MEI, 25/03/09 3/7

The modulus function

Example 3

Sketch the graphs of the functions:

(i) 2y x

(ii) 2 1 1y x

(iii) 22y x

Solution

(i)

(ii)

First sketch the graph of y = x – 2.

Reflect the negative parts of the

graph in the x-axis.

y = x – 2 y = |x – 2|

First sketch the graph of y = 2x + 1.

Reflect the negative parts of

the graph in the x-axis.

Finally translate the graph vertically

downwards by 1 unit.

y = 2x + 1 y = |2x + 1|

y = |2x + 1| – 1


© MEI, 25/03/09 4/7

(iii)

You can investigate more graphs involving a modulus function using the Geogebra resource Linear modulus functions.

Graphs can sometimes be useful if you need to solve an equation which involves a modulus function.

Example 4

(i) Sketch the graph of y = |2x + 1|.

(ii) Hence, or otherwise, solve the equation |2x + 1| = x + 2.

(iii) Solve the equation |2x + 1| = 2 – 3x.

Solution

(i)

First sketch the graph of y = 2 – x².

Reflect the negative parts of

the graph in the x-axis.

y = 2 – x²

y = |2 – x²|

y = |2x + 1|


© MEI, 25/03/09 5/7

2 1 2

1

x x

x

(2 1) 2

2 1 2

3 3

1

x x

x x

x

x

The solutions are x = 1 and x = 1.

15

2 1 2 3

5 1

x x

x

x

The solution is 15

x .

Part (iii) of the example above shows how mistakes can be made in solving this type of equation, particularly if a graph is not been used. A second solution could have been found by considering the case – (2x + 1) = 2 – 3x.

This would have generated the solution x = 1. By looking at the graph you can

see this is not a correct solution, but is the point where the left-hand side of the graph of y = |2x + 1| would meet the line y = 2 – 3x if it were extended.

Using a graph helps to avoid this kind of mistake. You can also check

solutions by substituting them back into the original equation.

(ii) By sketching the graphs of y = |2x + 1|

and y = x + 2 on the same axes, you can

see that there are two solutions.

One of the solutions involves the part of

y = |2x + 1| where 2x + 1 is positive,

i.e. where the graph is y = 2x + 1

The other solution involves the part of

y = |2x + 1| where 2x + 1 is negative,

i.e. where the graph is y = -(2x + 1)

(iii) By sketching the graphs of y = |2x + 1| and

y = 2 – 3x on the same axes, you can see

that there is only one solution, which

involves the part of y = |2x + 1| where

2x + 1 is positive. The line y = 2 – 3x will

not cross the part of y = |2x + 1| where

2x + 1 is negative, since the line y = 2 – 3x

has the steeper gradient.


© MEI, 25/03/09 6/7

Inequalities

To solve inequalities which involve a modulus sign, you need to consider whether the solution set involves two separate intervals or just one. An inequality of the form |f(x)| < a can be written as –a < f(x) < a. An inequality of the form |f(x)| > a can be written as f(x) < -a or f(x) > a.

The inequality can then be solved using the usual rules for inequalities.

Example 5

Solve the following inequalities.

(i) 2 1 7x

(ii) 2 2 1 3x

(iii) 1 3 4x

Solution

(i) 2 1 7

7 2 1 7

8 2 6

4 3

x

x

x

x

(ii) 2 2 1 3

2 2 2

2 1

2 1 or 2 1

1 or 3

x

x

x

x x

x x

(iii)

53

1 3 4

3 1 4

4 3 1 4

3 3 5

1

x

x

x

x

x

Any inequality of the form p < x < q can be written in the form |x – a| < b. This is

basically reversing the process of solving an inequality.

Example 6

Write the inequality 6 x 14 in the form |x – a| b.

Solution

|1 – 3x| is equal to |3x – 1|. It is easier to avoid having a

negative term in x.


© MEI, 25/03/09 7/7

x a b b x a b

b a x b a

Adding:

6

14

2 20

10

4

b a

b a

a

a

b

The inequality can be written as |x – 10| 4.

MEI Core 3

© MEI, 25/03/09 1/4

Techniques for differentiation

Section 1: The chain rule

Notes and Examples This chapter extends methods for differentiating functions.

These notes contain subsections on

The chain rule

Rates of change

The chain rule

Suppose you want to differentiate the function 3(2 5)y x . You could expand

the brackets: 3

3 2 2 3

3 2

(2 5)

(2 ) 3(2 ) .5 3(2 ).5 5

8 60 150 125

y x

x x x

x x x

Then you could differentiate term by term:

2

2

d(8 3 ) (60 2 ) 150

d

24 120 150

yx x

x

x x

This result can be factorised:

2

2

2

d24 120 150

d

6(4 20 25)

6(2 5)

yx x

x

x x

x

This suggests that there is perhaps a more efficient method. The expression for the derivative can be written as

3(2 5)y x

d

d

y

x = 2 × 3(2x + 5)

2

The derivative of (2x + 5) is 2

The derivative of (…)

3 is 3(…)

2

MEI C3 Differentiation Section 1 Notes and Examples

© MEI, 25/03/09 2/4

To make the method is clearer, let u = 2x + 5. Then y = u3

Now 2d

3d

y

uu

and d

2d

u

x

so 2 2d d d

2 3 2 5 3 2d d d

y y u

x ux u x

.

This is called the CHAIN RULE: d d d

d d d

y y u

x u x.

This result is proved formally on page 64 of the textbook.

Example 1

Differentiate y = (3x2 4)

7

Solution

Let 2 d3 4 6

d

uu x x

x

7 6d7

d

yy u u

u

Using the chain rule:

6

6

62

d d d

d d d

7 6

42

42 3 4

y y u

x u x

u x

xu

x x

After some practice, you may find that you can differentiate composite functions like the one in Example 1 without introducing the variable u.

7 6

2 2d3 4 7 3 4 6

d

yy x x x

x

Differentiate (…)7

to get 7(…)6…

…then multiply by the

derivative of (3x2 4)


© MEI, 25/03/09 3/4

Example 2

Differentiate y = 21 x

Solution

Let 2 d1 2

d

uu x x

x

1 12 21

2

d

d

yy u u u

u


121

2

2

d d d

d d d

2

1

y y u

x u x

u x

x

u

x

x

Alternatively, differentiate directly:

1221 y x

12

12

212

2

d1 2

d

1

yx x

x

x x

To practice using the chain rule, try the interactive questions The chain rule. You can also look at the chain rule video.

Rates of change The chain rule is also used to solve problems involving related rates of

change.

Example 3

A bank of a reservoir is modelled by the curve 2

250

xy for 10 x 20, where x is

measured in metres (see diagram).

Differentiate (…)1/2

to get ½ (…)1/2

…

…then multiply by the

derivative of (1 + x2)

10 15 x

20

y


© MEI, 25/03/09 4/4

The width of the reservoir is 15 m and is increasing at a rate of 0.2 m per day.

Calculate the rate of change of the depth of the reservoir.

Solution 2 d

250 d 25

x y x

yx

.

When x = 15, d 15 3

d 25 5

y

x

The rate of increase in width is given by d

d

x

t, so

d0.2

d

x

t.

The rate of change of depth is given by d

d

y

t.

By the chain rule:

d d d

d d d

30.2

5

0.12

y y x

t x t

So the rate of change of depth is 0.12 m per day.

MEI Core 3

© MEI, 25/03/09 1/5


Section 2: The product and quotient rules Notes and Examples

These notes contain subsections on

The product rule

The quotient rule

Inverse functions

The Product Rule

Suppose you want to differentiate 2 1 y x x . In this case, the function y is

the product of two simpler functions, 2u x and 1 v x . You can

differentiate u and v:

121

2

d d2 , (1 )

d d

u vx x

x x

You might think that the derivative of y is simply the product of the derivatives

of u and v. In fact, the formula is a bit more complicated:

If y = u × v, then: d d d

d d d

y v uu v

x x x.

This is called the product rule, and it is proved from first principles on page

69 of the textbook. Applying this formula to the example above:

12

2

2

12

1

d2

d

d1 (1 )

d

y x x

uu x x

x

vv x x

x

1 12 22 1

2

d d d

d d d

(1 ) (1 ) 2

y v uu v

x x x

x x x x

The tricky part of questions like this is to simplify your answer. In this case,

you can take out a factor of 121

2(1 )

x x :


© MEI, 25/03/09 2/5

1 12 2

12

12

212

12

12

d(1 ) 2 (1 )

d

(1 ) 4(1 )

(1 ) (5 4)

yx x x x

x

x x x x

x x x

Example 1

Given that 133(1 3 ) y x x , show that

232d

(1 3 ) (3 10 )d

yx x x

x. Hence find the

turning points of this curve.

Solution

y = uv where 133, (1 3 ) u x v x

3 2d3

d

uu x x

x

1 2 23 3 31

3

d1 3 (1 3 ) 3 (1 3 )

d

vv x x x

x

Using the product rule:

2 23 3

23

23

3 2

2

2

d d d

d d d

(1 3 ) (1 3 ) 3

(1 3 ) ( 3 9 )

(1 3 ) (3 10 ) as required.

y u vv u

x x x

x x x x

x x x x

x x x

The turning points occur when d

d

y

x = 0

232(1 3 ) (3 10 ) 0

x x x

x = 0 or x = 310

.

When x = 0, y = 0

When x = 0.3, y = 0.0125

So the turning points are (0, 0) and (0.3, 0.0125).

For extra practice in using the product rule, try the interactive questions The product rule.

You can also look at the product rule video.

using the chain rule

Take out 232 1 3

x x

as a factor


© MEI, 25/03/09 3/5

The Quotient Rule You have now learnt a technique for differentiating composite functions (the chain rule) and a technique for differentiating products. The third

differentiation technique deals with quotients. For a function y which can be expressed as a quotient of two simpler functions

u and v, there is a formula for the derivative of y:

If u

yv

then 2

d d

d d d

d

u vv u

y x x

x v

This is called the quotient rule and it is proved from first principles on page

71 of the textbook. You can also prove the quotient rule by thinking of the

quotient u

v as the product 1u v and applying the product rule.

Example 2

Differentiate 2

31 2

xy

x, simplifying your answer. Hence find the turning points of

this curve and determine their nature.

Solution

u

yv

where u = x2 and v = 1 + 2x3

2 d2

d

uu x x

x

3 2d1 2 6

d

vv x x

x

Using the quotient rule:

2

3 2 2

3 2

4 4

3 2

4

3 2

3

3 2

d d

d d d

d

(1 2 ) 2 6

(1 2 )

2 4 6

(1 2 )

2 2

(1 2 )

2 (1 )

(1 2 )

u vv u

y x x

x v

x x x x

x

x x x

x

x x

x

x x

x

It is best to factorise answers

as far as possible.


© MEI, 25/03/09 4/5

The turning points occur when d

d

y

x = 0

3

3 2

2 (1 )0 0 or 1

(1 2 )

x xx x

x

When x = 0, y = 0

When x = 1, y = 13

.

So (0, 0) is a minimum and (1, 13

) is a maximum.

For extra practice in using the quotient rule, try the interactive questions The quotient rule.

You can also look at the quotient rule video.

Inverse functions Try Activity 4.2 on page 77.

The first four parts should give you the following:

For the function 3y x , 2d

3d

y

xx

The function can be written in the form 13x y , which gives

23

23

13

2

d 1

d 3

1

3

xy

y y

x

In part 5, you should get similar results for the other functions suggested.

This suggests the relationship d 1

dd

d

x

yy

x

This result is used in the next example.

x x < 0 x = 0 0 < x < 1 x = 1 x > 1

d

d

y

x

–

\

0

+

/

0

–

\

To investigate the nature of the turning points,

you could find

2

2

d

d

y

x, but this would be

complicated. It is easier to investigate the sign of

d

d

y

x for values of x before and after the

stationary values. This can be done in a table like this:


© MEI, 25/03/09 5/5

Example 4

A circular ink blot is expanding. The area of the blot is increasing at a rate of 1 cm2

per second. Find the rate of change of the radius when the blot is 1 cm in diameter.

Solution

2 d2

d

d 1

d 2

AA r r

r

r

A r

Area is increasing at a rate of 1 cm² per second d

1d

A

t

By the chain rule:

d d d

d d d

11

2

1

2

r r A

t A t

r

r

When r = 1, d 1

0.318d 2

r

t

The rate of change of the radius is 0.318 cm s-1

.

MEI Core 3

© MEI, 25/03/09 1/5


Section 3: Differentiating logarithms and exponentials


The derivative of ax

The derivative of ln x

In the textbook (page 82) you are reminded that in chapter 2 you learned that

the integral of 1

x is ln x. This means that the derivative of ln x is

1

x, and you

can then use inverse functions to deduce the derivative of ex .

In these notes the opposite approach is used: first you will look at finding the derivative of exponential functions, and then use inverse functions to deduce the derivative of ln x.

You do not need to fully understand how these derivatives are found from first

principles, but it is useful to have some idea!

The derivative of ax

You can learn something about the derivative of exponential functions like

xy a by trying to differentiate using first principles:

Suppose we increase x by a small change x, resulting in a small change in y

y: x

x x

y a

y y a

Subtracting:

( 1)

x x x

x x x

x x

y a a

a a a

a a

Dividing by x: ( 1)x xy a a

x x

d

d

xyka

x where k is the limit of the expression

1xa

x

as x 0.

Putting x = 0 into the expression for y

x

tells you that k is actually the gradient

of the graph xy a at the origin.


© MEI, 25/03/09 2/5

It follows that if you can find a value for a so that this gradient is 1 then, for

this value of a, d

d

xya

x= . This value is just the constant e = 2.718…, which is

the base of natural logarithms. You can verify this result numerically in your

calculator, by finding the value of 0.0001e 1

0.0001

.

It follows that:

exy = d

ed

xy

x

You can explore the gradient of exponential functions using the Flash

resource The gradient graph of y = ax or the Geogebra resource Differentiating y = ax. Change the value of a to make the gradient as close

as possible to the value of y.

This result can be extended to differentiate ekxy , where k is a constant,

using the chain rule:

e

dLet

d

de e

d

kx

u u

y

uu kx k

x

yy

u

Using the chain rule: d d d

d d d

e

e

u

kx

y y u

x u x

k

k

d

e ed

kx kxyy k

x

Example 1

Differentiate:

(i) 2e xy x (ii)

2

31 2e x

xy

(iii)

2

exy

Solution

(i) Using the product rule with u = x and v = e2x

2 2

d1

d

de 2e

d

x x

uu x

x

vv

x


© MEI, 25/03/09 3/5

2 2

2

d d d

d d d

2e 1 e

(2 1)e

x x

x

y u vv u

x x x

x

x

.

(ii) Using the quotient rule with u = x2 and v = 1 + 2e

3x,

2

3 3

d2

d

d1 2e 6e

d

x x

uu x x

x

vv

x

2

3 2 3

3 2

3 2 3

3 2

d d

d d d

d

(1 2e )2 6e

(1 2e )

2 4 e 6 e

(1 2e )

x x

x

x x

x

u vv u

y x x

x v

x x

x x x

(iii) Using the chain rule with 2u x

2 d2

d

de e

d

u u

uu x x

x

yy

u

2

d d d

d d d

e 2

2 e

u

x

y y u

x u x

x

x

The derivative of ln x

You can now differentiate the inverse function of exy , which is lny x :

ln e

de

d

d 1 1.

d e

y

y

y

y x x

x

y

y

x x

d 1ln

d

yy x

x x

You can look at the relationship between the gradients of the graphs y = e

x and y = ln x using the Geogebra resource Differentiating ln x.


© MEI, 25/03/09 4/5

Example 2 Differentiate:

(i) lnx x (ii) ln

1 ln

x

x (iii) 3ln(1 )x

Solution

(i) Using the product rule with u x and lnv x

11 22 1

2

d 1

d 2

d 1ln

d

uu x x x

x x

vv x

x x

d d d

d d d

1 1ln

2

1 ln

2

2 ln

2

y u vv u

x x x

x xx x

x

x x

x

x

(ii) Using the quotient rule with lnu x and 1 lnv x

d 1ln

d

d 11 ln

d

uu x

x x

vv x

x x

2

2

2

2

d d

d d d

d

1 1(1 ln ) ln

(1 ln )

1 ln ln

(1 ln )

1

(1 ln )

u vv u

y x x

x v

x xx x

x

x x

x x

x x

(iii) Using the chain rule with 31u x

31u x

2d3

d

ux

x

d 1ln

d

yy u

u u


© MEI, 25/03/09 5/5

2

2

3

d d d

d d d

13

3

(1 )

y y u

x u x

xu

x

x

You can also look at the Differentiating logs and exponentials video.

MEI Core 3

© MEI, 04/04/11 1/5


Section 4: Differentiating trigonometric functions


The derivative of sin x

The derivative of cos x

The derivative of tan x

Differentiating functions involving sine, cosine and tangent

The derivative of sin x You can trace out the derivatives of sin x and cos x using the Geogebra resource Differentiating sin x and cos x. What does the derivative of siny x look like?

It is clearly a periodic function, with period 2, as the values of the

gradient of the function must repeat itself every 2 just as y does;

Using this information, you can sketch the graph of the derivative of sin x:

-1

1

2

The gradient is zero here

The gradient takes its largest positive values here

The gradient takes its largest negative value here

-1

1

2

MEI C3 Differentiating Section 4 Notes and Examples

© MEI, 04/04/11 2/5

This suggests that cosy x might fit the picture, and this is in fact the case,

although you need to know more about trig functions to prove this. Note: It is important to remember that this result is only true if x is measured in radians! You can see why radians are the natural choice when differentiating trigonometric functions by using the Geogebra resource The derivative of sine x.

The derivative of cos x

What about the derivative of cos x? The graph of cos x is that of sin x

translated 2

to the left, so the gradient function will be that of sin x translated

2

to the left:

This looks like the reflection of y = sin x in the x-axis, which is y = sin x. So:

d d(sin ) cos , (cos ) sin

d dx x x x

x x

You can see the full justification for these results in the Differentiating sine and cosine from first principles video.

The derivative of tan x What about the derivative of tan x? You can do this by applying the quotient rule to trig functions.

-1

1

2

y = cos x

gradient of y = cos x


© MEI, 04/04/11 3/5

Example 1 Using the derivatives of sin x and cos x, prove that the derivative of tan x is sec

2x.

Solution

You know from C2 that sin

tancos

xx

x .

Using the quotient rule with u = sin x and v = cos x

dsin cos

d

dcos sin

d

uu x x

x

vv x x

x

2

2

2 2

2

2

2

d d

d d d

d

cos cos sin ( sin )

(cos )

cos sin

cos

1

cos

sec

u vv u

y x x

x v

x x x x

x

x x

x

x

x

This is another result which you should remember and may be quoted.

2d(tan ) sec

dx x

x

Differentiating functions involving sine, cosine and tangent You can now apply all the differentiation techniques you have learned to functions involving sines, cosines and tangents. Functions of the form siny kx , cosy kx and tany kx , where k is a

constant, can be differentiated using the chain rule. Example 2

Differentiate sin 2y x

Solution

Using the chain rule with 2u x :

since 2 2cos sin 1x x


© MEI, 04/04/11 4/5

d2 2

d

sin cos

uu x

x

dyy u u

du

d d d

d d d

cos 2

2cos 2

y y u

x u x

u

x

The result from Example 2 can be generalised:

2d d d(sin ) cos , (cos ) sin , (tan ) sec

d d d kx k kx kx k kx kx k kx

x x x

You can use these results directly, without having to write out the chain rule. Here are some more examples. Example 3 Differentiate:

(i) 2cos x

(ii) sin x

(iii) e tanx x

(iv) sin

1 cos

x

x,

Solution (i) Using the chain rule:

2

dcos sin

d

d2

d

uu x x

x

yy u u

u

d d d

d d d

2 ( sin )

2cos sin

y y u

x u x

u x

x x

(ii) sin sin180

xy x

dcos

d 180 180

cos180

y x

x

x

You can only differentiate trig functions when they are measured in radians, so convert to radians

by multiplying by 180

Using the standard result for

differentiating coskx


© MEI, 04/04/11 5/5

(iii) Using the product rule:

2

de e

d

dtan sec

d

x xuu

x

vv x x

x

2

d d d

d d d

e sec e tanx x

y v uu v

x x x

x x

(iv) Using the quotient rule:

d

sin cosd

d1 cos sin

d

uu x x

x

vv x x

x

2

2

2 2

2

2

d d

d d d

d

(1 cos )cos sin ( sin )

(1 cos )

cos cos sin

(1 cos )

cos 1

(1 cos )

1

1 cos

u vv u

y x x

x v

x x x x

x

x x x

x

x

x

x

The Chain Rule hexagonal puzzle involves differentiation of functions involving exponential, logarithmic and trigonometric functions using the chain rule.

Notice the use of trigonometric identities to

simplify results.

MEI Core 3

© MEI, 25/08/10 1/5


Section 5: Implicit differentiation Notes and Examples These notes contain subsections on

Differentiating functions of y with respect to x

Differentiating implicit functions

Differentiating products of x and y

Logarithmic differentiation

Differentiating functions of y with respect to x

Suppose that y is a function of x, say 2x . You can differentiate y to get 2x; but

what about differentiating 2y , or sin y or ln y ? You could convert them to

functions of x, i.e. 4 2 2, sin lnx x xor ; but can you find the derivative of these

functions without converting into functions of x?

For example, suppose 2y x as above and you want to find the derivative of

y² with respect to x. Let 2u y . You want to find d

d

u

x

2 d2

d

uu y y

y

2 d2

d

yy x x

x .


2

3

d d d

d d d

d 2

d

2 2

4

u u y

x y x

yy

x

x x

x

(which is indeed the derivative of x4) The other functions of y can be differentiated in the same way:

To differentiate sin y with respect to x, let sinu y . You want to find d

d

u

x.

dsin cos

d

uu y y

y

2 d2

d

yy x x

x


© MEI, 25/08/10 2/5

2d d[ ] 2

d d

yy y

x x , as before.


d d d

d d d

dcos

d

2 cos

u u y

x y x

yy

x

x y

To differentiate ln y with respect to x, let lnu y . You want to find d

d

u

x.

d 1ln

d

uu y

y y

2 d2

d

yy x x

x


2

d d d

d d d

1 d

d

1 2

2

u u y

x y x

y

y x

xx

x

This idea can be generalised:

If y is a function of x, then d df d

[f ( )]d d d

yy

x y x .

Differentiating implicit functions An implicit function is an equation between two variables, say x and y, in which y is not given explicitly as a function of x. The equation of a circle is an example of an implicit function.

For example, 2 2 25x y is the equation of a circle, centre O and radius 5.

You can differentiate an implicit function like this by differentiating each side of the equation with respect to x:

2 2 25x y

2 2d d d[ ] [ ] [25]

d d dx y

x x x

d

2 2 0d

yx y

x


© MEI, 25/08/10 3/5

You can now find d

d

y

x in terms of x and y by rearranging this equation:

d

2 2 0d

yx y

x

d

2 2d

yy x

x

d 2

d 2

y x x

x y y

This result enables you, for example, to find the gradient at the point (3, 4) of

the circle. Here, x = 3 and y = 4, so d 3

d 4

y x

x y .

Example 1

Show that the curve y + y3 = 2sin x has a turning point at the point 2

,1 .

Solution 3 2siny y x

Differentiating implicitly gives:

3d d[ ] [2sin ]

d dy y x

x x

2d d3 2cos

d d

y yy x

x x ,

2d(1 3 ) 2cos

d

yy x

x

2

d 2cos

d 1 3

y x

x y

At the point 2,1 x =

2 and y = 1

22cosd

0d 1 3

y

x

So 2,1 is a turning point.

Differentiating products of x and y

What is the derivative of 2xy with respect to x?

You can use the product rule to do this.

d1

d

uu x

x and 2 d d

2d d

v yv y y

x x .

So 2

2

d d d( )

d d d

d2

d

v uxy u v

x x x

yxy y

x


© MEI, 25/08/10 4/5

Example 2

Given that y2 + 3xy + x

3 = 25, find

d

d

y

x in terms of x and y.

Solution 2 33 25y xy x

Differentiating implicitly gives:

2 3d d[ 3 ] [25]

d dy xy x

x x

2d d2 3 3 3 0

d d

y yy x y x

x x

2d(2 3 ) 3 3

d

yy x y x

x

2d (3 3 )

d 2 3

y y x

x y x

You may also like to look at the Implicit differentiation video.

Logarithmic differentiation The technique of implicit differentiation can be used to simplify differentiating complicated product and quotient functions, as in the following example. Example 3

Find the derivative of 3

2

(1 )(1 2 )

1

x xy

x

Solution 3

2

(1 )(1 2 )

1

x xy

x

Take logarithms of each side: 3

2

3 2

2

(1 )(1 2 )ln ln

1

ln(1 ) ln(1 2 ) ln 1

1ln(1 ) 3ln(1 2 ) ln(1 )

2

x xy

x

x x x

x x x

Now differentiate implicitly:

2

1 d 1 3 1 2( 2)

d 1 1 2 2 (1 )

y x

y x x x x

The product rule is used here to

differentiate 3xy, with u = 3x and v = y.

It is not necessary to write out all the details of the product rule if you don’t

feel that you need to.


© MEI, 25/08/10 5/5

2

3

22

d 1 6

d 1 1 2 (1 )

(1 )(1 2 ) 1 6( )

1 1 2 (1 )1

y xy

x x x x

x x x

x x xx

This is quite a fearsome looking expression! But try

differentiating y directly and you will appreciate that

the logarithmic differentiation is considerably easier!

MEI Core 3

© MEI, 26/03/09 1/4

Techniques for integration

Section 1: Integration by substitution Notes and Examples These notes contain subsections on

Integration by substitution

Integration by inspection

Integration by substitution The chain rule allows you to differentiate a function of x by making a substitution of another variable u, say. What is the corresponding integration

method? Suppose you want to find the integral of (2x + 1)

5. If you wanted to differentiate

this function, you could use the substitution u = 2x + 1 and then apply the chain rule. However, you are trying to integrate with respect to x, and if you integrate

5u to give 616u , then you have integrated with respect to u. The key is the ‘dx’

in the integral – this needs to be changed to du, using the equation

d

d dd

xx u

u

which comes from letting x 0 in the equation

d

d

xx u

u .

Here is how the method is set out:

To integrate: 5(2 1) dx x

Let u = 2x + 1 d

2d

u

x du = 2dx dx = ½ du.

5 5

6

6

1(2 1) d d

2

1

12

12 1

12

x x u u

u c

x c

In the definite integral example below, you need to calculate new limits in terms of u for the limits which are for values of x.

MEI C3 Integration Section 1 Notes and Examples

© MEI, 26/03/09 2/4

Example 1

Evaluate 1

2

01 dx x x .

Solution

Let 21u x d

2d

d 2 d

1d d

2

ux

x

u x x

x ux

When x = 0, u = 1 + 02 = 1

when x = 1, u = 1 + 12 = 2.

12

32

3 32 2

1 22

0 1

2121

21 22 3

1

13

13

11 d d

2

d

2 1

(2 2 1)

x x x x u ux

u u

u

In Example 1, the x’s cancelled out giving an expression in terms of u only.

This is not always the case, so sometimes you need to express other parts of

the integral in terms of the new variable of integration. This is shown in Example 2.

Example 2

Find 1 2 dx x x .

Solution

Let u = 1 + 2x

12

d2

d

d 2d

d d

u

x

u x

x u

Also 12

1x u

5 32 2

1 12 2

3/ 2 1/ 214

1 2 24 5 3

5/ 2 3/ 21 110 6

1 2 d ( 1) d

( )d

(1 2 ) (1 2 )

x x x u u u

u u u

u u c

x x c

It’s a good idea to take out this fraction in front of the integral to simplify the integration.

Notice that the x’s

cancel out.


© MEI, 26/03/09 3/4

Notice that when the integration is indefinite, as in Example 2, you must express the final answer in terms of x, not the new variable u.

Integration by inspection

Look at 21 dx x x . You can integrate this without using a substitution, just as

you can use the chain rule to differentiate without writing the problem in terms

of a new variable. Notice that the function to be integrated (called the

integrand) consists of 1221 x multiplied by half the derivative of 21 x , which

is 2x. Now, try differentiating 3221 x :

3 12 2

12

2 232

2

d(1 ) (1 ) 2

d

3 (1 )

x x xx

x x

This is three times the integrand 1221x x ; so you must divide this by 3 to get

the result needed: 322 21

31 d (1 )x x x x c

You can spot integrals which may be integrated like this (this is called

integration by inspection or integration by recognition): just think about what you would use as u if you were to do a substitution. If the integrand is a multiple of a product of a function of u and the derivative of u, (i.e. it is of the

form df

g f ( )d

xx

) then you can integrate by inspection as shown in Example 3.

Example 3

Find 132 3(1 2 ) dx x x by inspection.

Solution

The integral will be of the form 4331 2x .

4 13 3

13

3 3 243

2 3

d(1 2 ) (1 2 ) 6

d

8 (1 2 )

x x xx

x x

This is 8 times the integrand, so 1 43 32 3 31

8(1 2 ) d (1 2 )x x x x c .

If you were to use a substitution, you

would use 31 2u x , which has

derivative 6x². The integrand is a

multiple of the product of a function of

u and the derivative of u.


© MEI, 26/03/09 4/4

Integration by inspection is not always possible. Look at Example 2 above and you will see that the integrand does not consist of a product of a function of u

and a multiple of the derivative of u. In such cases a substitution must be

used.

You may also like to look at the Integration by substitution video.

MEI Core 3

© MEI, 26/03/09 1/6


Section 2: Integrating other functions


Integrating exponential functions

Logarithmic integrals

Using the methods of substitution and inspection

Integrating trigonometric functions

Integrating exponential functions

Remember that the derivative of ex is ex. Therefore e .d ex xx c

Similarly, since the derivative of ekx is kekx:

1.e d ekx kxx c

k

The Geogebra resource Integrating ex shows graphically how the area under the graph of y = e

x is equal to the value of ex.

Example 1

Evaluate 2

0e dx x , giving your answer in terms of e.

Solution 12

12

2 2

0 0

2

0

1 0

e d e d

2e

2e 2e

2e 2

xx

x

x x

Logarithmic integrals

You can investigate the properties of the area under the graph of 1

yx

using

the Geogebra resource The area under the graph of y = 1/x (this was also included in the chapter on Natural logarithms and exponentials). You should

be able to see from this that the area has the same properties as a log function.


© MEI, 26/03/09 2/6

–5 –4 –3 –2 –1 1 2 3 4 5

–5

–4

–3

–2

–1

1

2

3

4

5

x

y

You can also use the Geogebra resource Integrating 1/x to see a graphical representation of the area under the graph.

The derivative of ln x is 1

x. It follows that

1d lnx x c

x , provided x > 0. Also, if

x < 0, the derivative of ln(x) is 1 1

.( 1)( )x x

, so 1

d ln( )x x cx

. These two

results are sometimes combined using a modulus sign:

1d lnx x c

x

When dealing with an indefinite integral, you need to include the modulus signs, unless a restricted domain is given so that only positive values of x are

involved. Care needs to be taken when using this result. The domain of x must be either

x > 0 or x < 0 – you cannot integrate across zero.

For example, suppose you wanted to find 3

1

1d

xx

. The graph below illustrates

the area represented by this integral.

From the graph, you can see that this area is not defined, as it includes the

value x = 0 for which the function 1

yx

is not defined.

Note: some integrals which involve points at which the function is not defined

can be evaluated – sometimes an area which appears to be finite does in fact have a finite value (just as the sum of an infinite geometric progression can be finite). However, it is not the case in the example above.

Example 2

Integrate

(i) 1

d2 1 x

x

(ii) 1

d3

xx


© MEI, 26/03/09 3/6

Solution

(i) 12

1d ln(2 1)

2 1

x x cx

(ii) 13

13

1 1d d

3

ln

x xx x

x c

The example above illustrates an important point. In the same way that the

integral of ekx is 1

ekx ck

, it is logical that the integral of 1

kx is

1ln kx c

k , and

in fact this is quite true. But in part (ii) of the example above, the integrand,

1

3x is treated as

1

xmultiplied by a constant 1

3, rather than as a function of a

function. The result is therefore given as 13ln x c rather than as 1

3ln 3x c .

The answer to this problem is that in fact these two expressions are the same.

Remember that by the laws of logarithms, ln 3 ln3 lnx x . So 13ln 3x c

may be written as 1 13 3ln ln3x c . But ln 3 is just a constant, and so it can be

considered as part of the arbitrary constant.

In general, it is easier to take any constant outside the integral, as in Example 2(ii), since this gives you a simpler expression to work with. However, for a

function like the one in Example 2(i) you need to use the standard result shown above.

Using the methods of substitution and inspection Example 3

Find 2

d1

xx

x.

Solution

Use the substitution 21u x

21u x d

2d

1d d

2

ux

x

x ux


© MEI, 26/03/09 4/6

2

12

212

1d d

1 2

1d

2

ln

ln 1

x xx x

x u x

xu

u c

x c

Notice that this integral can be done quicker by inspection, by noting that the

derivative of ln(1 + x2) is

2 2

1 22

1 1

xx

x x

. This is twice the integrand, so it

follows that 2122

d ln 11

xx x c

x

.

It is easy to spot integrals which can be done by substitution to give a logarithm. If the integrand is a fraction, the numerator of which is the derivative of the denominator, then the integral is the natural logarithm of the

denominator. This can be generalised:

f ( )d ln f ( )

f ( )

xx x c

x

As in Example 3, in some cases the numerator is a multiple of the derivative of the denominator, so you need to adjust things a little. In Examples 4 and 5 this approach is used.

Example 4

Find 2

2

ed

1 e

x

xx

Solution

2 2

122 2

212

e 2ed d

1 e 1 e

ln 1 e

x x

x x

x

x x

c

Example 5

Find 2

2

31

1d

3

xx

x x

, expressing the answer as a single logarithm.

Solution

The derivative of 3x + x3 is 3 + 3x2

= 3(1 + x2). This is three times the numerator of

the integrand. So

The derivative of the denominator is 22e x

, which is twice the numerator,

so you can integrate by inspection.

Notice that modulus signs are not

needed here, since 1 + x² is

always positive.

Again, since 21 e x is always positive,

no modulus signs are needed.


© MEI, 26/03/09 5/6

2 22 2

133 31 1

231

31

13

1 143 4

713 2

1 3 3d d

3 3

ln 3

(ln14 ln 4)

ln

ln

x xx x

x x x x

x x

You may also like to look at the Integration that leads to log functions

video.

Integrals of trigonometric functions

The derivative of sin x is cos x; the derivative of cos x is sin x. It follows that

sin d cos cos d sinx x x c x x x c

Another useful result to bear in mind is that the derivative of tan x is sec

2x, so

2sec d tanx x x c

Similarly, by looking at the derivatives of sin kx, cos kx and tan kx, you can see

that

21 1 1sin d cos cos d sin sec d tank k k

kx x kx c kx x kx c kx x kx c

Example 6

Find (i) sin3 dx x

(ii) / 6

2

0sec 2 dx x

(iii) 2sin cos dx x x

Solution

(i) 13

sin3 d cos3x x x c

(ii)

/ 6 / 62 12 00

12 3

12

sec 2 d tan 2

tan tan 0

3

x x x

remember that tan 60 3


© MEI, 26/03/09 6/6

(iii) 2sin cos dx x x can be done using the substitution u = sin x

d 1

sin cos d dd cos

uu x x x u

x x

2 2

2

313

313

1sin cos d cos d

cos

d

sin

x x x u x ux

u u

u c

x c

You could also do this by inspection: notice that the

integral is a product of sin²x

(a function of sin x) and cos x

(the derivative of sin x)

MEI Core 3

© MEI, 22/02/10 1/3


Section 3: Integration by Parts Notes and Examples These notes contain subsections on

Integration by parts

Definite integration by parts

Integration by parts Integration by parts is another technique which can sometimes be used to integrate the product of two simpler functions. It is useful in many cases where a substitution will not help, although it cannot be used for all functions. Activity 5.3 in the textbook (page 126), together with the text that follows on the general result for integration by parts, shows you how the following formula is derived from the product rule for differentiation:

d d

d dd d

v uu x uv v x

x x

This formula is called integration by parts. Example 1 shows how it works in practice. Example 1

Find cos dx x x

Solution

d

1d

uu x

x

dcos cos d sin

d

vx v x x x

x

Using the formula for integration by parts:

d d

d dd d

v uu x uv v x

x x

dcos d sin sin ( )d

d

sin sin d

sin cos

x x x x x x x xx

x x x x

x x x c

Split the integrand x cos x into

two parts u and d

d

v

x

There is no need for a ‘+c’ here, as it is added

to the final result


© MEI, 22/02/10 2/3

The choice of how to divide up the integrand between u and d

d

v

x is a matter of

experience. Usually, u is a simple function, such as a linear function of x, which becomes even simpler when differentiated. However, when the integrand involves a logarithm, this has to be ‘u’: ln x can’t be

integrated easily, so it can’t be d

d

v

x. This is shown in the following example:

Example 2

Find ln x xdx .

Solution

d 1ln

d

uu x

x x

212

d

d

vx v x

x


d dd d

d d

v uu x uv v x

x x

2 21 12 2

21 12 2

2 21 12 4

1ln d ln d

ln d

ln

x x x x x x xx

x x x x

x x x c

Definite integration by parts When using integration by parts on a definite integral, the formula for integration by parts becomes

d d

d dd d

b bb

aa a

v uu x uv v x

x x

Notice that the ‘uv’ part of the formula should be evaluated between the limits, as in this final example: Example 3

Find / 6

0sin 2 dx x x

.

Solution

d1

d

uu x

x


© MEI, 22/02/10 3/3

12

dsin 2 sin 2 d cos 2

d

vx v x x x

x


d dd d

d d

v uu x uv v x

x x

/ 6 / 61 12 200

1 1 12 6 3 2 2

/ 61

24 4 0

1 124 4 3 4

sin 2 d cos 2 ( cos 2 ).1d

cos 0 cos0 cos 2 d

sin 2

sin sin 0

3

24 8

3 3

24

x x x x x x x

x x

x

You may also like to look at the Integration by parts video.

Remember that 13 2

cos

and 13 2

sin 3

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