Core 3 Revision Notes - Mr Barton Maths

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Core 3 Revision Notes - Mr Barton Maths

Core Maths C3

Revision Notes

October 2012


Core Maths C3

1 Algebraic fractions .................................................................................................. 3

Cancelling common factors .................................................................................................................. 3

Multiplying and dividing fractions ....................................................................................................... 3

Adding and subtracting fractions .......................................................................................................... 3

Equations .............................................................................................................................................. 4

2 Functions ................................................................................................................. 4

Notation ................................................................................................................................................ 4

Domain, range and graph ...................................................................................................................... 4

Defining functions ............................................................................................................................................... 5

Composite functions ............................................................................................................................. 6

Inverse functions and their graphs ........................................................................................................ 6

Modulus functions ................................................................................................................................ 7

Modulus functions y = |f (x)| .............................................................................................................................. 7

Modulus functions y = f (|x|) ............................................................................................................................... 8

Standard graphs ..................................................................................................................................... 8

Combinations of transformations of graphs .......................................................................................... 9

3 Trigonometry ......................................................................................................... 10

Sec, cosec and cot ............................................................................................................................... 10

Graphs ................................................................................................................................................................ 10

Inverse trigonometrical functions ....................................................................................................... 11

Graphs ................................................................................................................................................................ 11

Trigonometrical identities ................................................................................................................... 11

Finding exact values .......................................................................................................................................... 12

Proving identities. .............................................................................................................................................. 12

Eliminating a variable between two equations ................................................................................................. 12

Solving equations .............................................................................................................................................. 12

R cos(x + α) ........................................................................................................................................ 13

4 Exponentials and logarithms ................................................................................. 14

Natural logarithms .............................................................................................................................. 14

Definition and graph .......................................................................................................................................... 14

Equations of the form e ax + b = p ....................................................................................................................... 15

5 Differentiation ........................................................................................................ 15

Chain rule ............................................................................................................................................ 15

Product rule ......................................................................................................................................... 16

Quotient rule ....................................................................................................................................... 16

Derivatives of e x and log e x ≡ ln x. ................................................................................................... 17

dy 1

= ............................................................................................................................................ 18

dx

dx

dy

Trigonometric differentiation .............................................................................................................. 18

Chain rule – further examples ............................................................................................................. 19

Trigonometry and the product and quotient rules ............................................................................... 19

C3 14/04/13 SDB 1


6 Numerical methods ................................................................................................ 20

Locating the roots of f(x) = 0 ............................................................................................................. 20

The iteration x n + 1 = g(x n ) ................................................................................................................... 20

Conditions for convergence ................................................................................................................. 21

Index ............................................................................................................................. 22

2

C3 14/04/13 SDB


1 Algebraic fractions

Cancelling common factors

3 2

x − 2x

− 3x

Example: Simplify

.

2

4x

− 36

Solution: First factorise top and bottom fully –

3 2

2

x − 2x

− 3x

x(

x − 2x

− 3) x(

x − 3)( x + 1)

=

=

2

2

4x

− 36 4( x − 9) 4( x − 3)( x + 3)

and now cancel all common factors, in this case (x – 3) to give

x(

x + 1)

Answer = .

4( x + 3)

Multiplying and dividing fractions

This is just like multiplying and dividing fractions with numbers and then cancelling common

factors as above.

Example: Simplify

2

2

9x

− 4 3x

− x − 2

÷

2

3 2

3x

− 2x

x + x

Solution: First turn the second fraction upside down and multiply

=

2

3 2

9x

− 4 x + x

×

, and then factorise fully

2

2

3x

− 2x

3x

− x − 2

2

(3x

− 2)(3x

+ 2) x ( x + 1)

= ×

, and cancel all common factors

x(3x

− 2) (3x

+ 2)( x −1)

Answer =

x(

x + 1)

x −1

.

Adding and subtracting fractions

Again this is like adding and subtracting fractions with numbers; but finding the Lowest

Common Denominator can save a lot of trouble later.

Example:

Solution:

=

=

Simplify

x

2

3x


− 7x

+ 12 x

First factorise the denominators

2

5

.

− 4x

+ 3

3x

5


; we see that the L.C.D. is (x – 3)(x – 4)(x – 1)

( x − 3)( x − 4) ( x − 3)( x −1)

3x(

x −1)

5( x − 4)


( x − 3)( x − 4)( x −1)

( x − 3)( x −1)(

x − 4)

=

2

2

3x

− 3x

− 5x

+ 20 3x

− 8x

+ 20

=

( x −1)(

x − 3)( x − 4) ( x −1)(

x − 3)( x − 4)

which cannot be simplified further.

C3 14/04/13 SDB 3


Equations

Example:

Solution:

Solve

x −1

− =

+ 1 x

x

x

1

2

First multiply both sides by the Lowest Common Denominator which in this case

is 2x(x + 1)

⇒ x × 2x – (x – 1) × 2(x + 1) = x(x + 1)


2x 2 – 2x 2 + 2 = x 2 + x

⇒ x 2 + x – 2 = 0, ⇒ (x + 2)(x – 1) = 0

⇒ x = –2 or 1

2 Functions

A function is an expression (often in terms of x) which has only one value for each value of x.

Notation

y = x 2 – 3x + 7, f (x) = x 2 – 3x + 7 and f : x → x 2 – 3x + 7

are all ways of writing the same function.

Domain, range and graph

The domain is the set of values which x can take:

this is sometimes specified in the definition

and sometimes is evident from the function: e.g. √x can only take positive x or zero.

The range is that part of the y–axis which is used.

Example:

Find the range of the function

f : x → 2x – 3 with domain x is a real number: –2 < x ≤ 3.

Solution: First sketch the graph for values of x

between – 2 and 3

and we can see that we are

only using the y–axis from –7 to 3,

and so the range is

y is a real number: –7 < y ≤ 3.

y

x

-6 -4 -2 2 4 6 8

-5

4

C3 14/04/13 SDB


Example:

Find the largest possible domain and the range for the function

f : x → x − 3 + 1.

Solution: First notice that we cannot have the square root of a negative number and so

x – 3 cannot be negative

⇒ x – 3 ≥ 0

⇒ largest domain is x real: x ≥ 3.

y

To find the range we first sketch the graph

and we see that the graph will cover all of the

y-axis from 1 upwards

and so the range is y real: y ≥ 1.

2 4 6 8

x

Example: Find the largest possible domain and the range for the function

2x f : x → .

x + 1

Solution: The only problem occurs when the denominator is 0, and so x cannot be –1.

Thus the largest domain is x real: x ≠ –1.

y

To find the range we sketch the graph

and we see that y can take any value

except 2,

so the range is y real: y ≠ 2.

5

x

-8 -6 -4 -2 2 4 6 8 1

Defining functions

Some mappings can be made into functions by restricting the domain.

Examples:

1) The mapping x → √x where x ∈ R is not a function as √–9 is not defined, but

if we restrict the domain to positive or zero real numbers then f : x → √ x where x ∈

R, x ≥ 0 is a function.

2) x →

but f : x →

1

x−3

1

x−3

where x ∈ R is not a function as the image of x = 3 is not defined,

where x ∈ R, x ≠ 3 is a function.

C3 14/04/13 SDB 5


Composite functions

To find the composite function fg we must do g first.

Example: f : x → 3x – 2 and g : x → x 2 + 1. Find fg and gf.

Solution:

f is

Think of f and g as ‘rules’

times by 3 subtract 2

g is

square add 1


fg is

square add 1

times by 3 subtract 2

giving (x 2 + 1) × 3 – 2 = 3x 2 + 1

⇒ fg : x → 3x 2 + 1 or fg(x) = 3x 2 + 1.

gf is

times by 3 subtract 2 square add 1

giving (3x – 2) 2 + 1 = 9x 2 – 12x + 5

⇒ gf : x → 9x 2 – 12x + 5 or gf (x) = 9x 2 – 12x + 5.

Inverse functions and their graphs

The inverse of f is the ‘opposite’ of f:

thus the inverse of ‘multiply by 3’ is ‘divide by 3’

and the inverse of ‘square’ is ‘square root’.

The inverse of f is written as f –1 : note that this does not mean ‘1 over f ’.

The graph of y = f –1 (x) is the reflection in y = x of the graph of y = f (x)

To find the inverse of a function x → y

(i) interchange x and y

(ii) find y in terms of x.

Example: Find the inverse of f : x → 3x – 2.

Solution: We have x → y = 3x – 2

(i) interchanging x and y ⇒ x = 3y – 2

6

(ii) solving for y ⇒ y =


f –1 : x →

x + 2

3

x + 2

3

y = x

y

2

x

-4 -2 2 4

y = f –1 (x)

-2

y = f (x)

C3 14/04/13 SDB


Example:

Find the inverse of g : x →

x + 3

.

2x

− 5

Solution: We have x → y =

x + 3

2x

− 5

(i)

(ii)

interchanging x and y

y + 3

⇒ x =

2y

− 5

solving for y

⇒ x(2y – 5) = y + 3

y

5

y = x

⇒ 2xy – 5x = y + 3

⇒ 2xy – y = 5x + 3

y = g –1 (x)

⇒ y(2x – 1) = 5x + 3

5x

+ 3

⇒ y =

2x

−1

x

5 1

y = g (x)


g –1 : x →

5x

+ 3

2x

−1

Note that f f –1 (x) = f –1 f (x) = x.

Modulus functions

Modulus functions y = |f (x)|

|f (x)| is the ‘positive value of f (x)’,

so to sketch the graph of y = |f (x)| first sketch the graph of y = f (x) and then reflect the

part(s) below the x–axis to above the x–axis.

Example: Sketch the graph of y = |x 2 – 3x|.

Solution: First sketch the graph of y = x 2 – 3x.

y

5

Then reflect the portion between x = 0 and x = 3 in

the x–axis

5

x

to give

y

5

5

x

C3 14/04/13 SDB 7


Modulus functions y = f (|x|)

In this case f (–3) = f (3), f (–5) = f (5), f (–8.7) = f (8.7) etc. and so the graph on the

left of the y–axis must be the same as the graph on the right of the y–axis,

so to sketch the graph, first sketch the graph for positive values of x only, then reflect

the graph sketched in the y–axis.

Example: Sketch the graph of y = |x| 2 – 3|x|.

Solution: First sketch the graph of y = x 2 – 3x for only

positive values of x.

y

5

x

-5 5

Then reflect the graph sketched in the y–axis to

complete the sketch.

y

5

x

-5 5

Standard graphs

3y

y = x 3

y = x

2

1

x

-4 -2 2 4

-1

-2

-3

y = x and y = x 3

y=x 4

y=x 2

y

4

3

2

1

x

-4 -3 -2 -1 1 2 3 4

1

y = x 2 and y = x 4

y

2

1

x

-4 -3 -2 -1 1 2 3 4

-1

-2

3

1

y =

x

4y

3

2

1

x

-1 1 2 3 4 5 6 7

-1

y = √x

8

C3 14/04/13 SDB


Combinations of transformations of graphs

We know the following transformations of graphs:

y = f (x)

⎛ a

translated through ⎜ ⎞ ⎝ b⎠ ⎟ becomes y = f (x - a) + b

stretched factor a in the y-direction becomes y = a × f (x)

⎛ x⎞

stretched factor a in the x-direction becomes y = f ⎜ ⎟

⎝ a⎠

reflected in the x-axis becomes y = – f (x)

reflected in the y-axis becomes y = f (–x)

We can combine these transformations:

Examples:

1) y = 2f (x – 3) is the image of y = f (x) under a stretch in the y-axis of factor 2

⎛3 ⎞

followed by a translation ⎜ ⎟ , or the translation followed by the stretch.

⎝0⎠

2) y = 3x 2 + 6 is the image of y = f (x) under a stretch in the y-axis of factor 3

⎛0 ⎞

followed by a translation ⎜ ⎟ ,

⎝6⎠

BUT these transformations cannot be done in the reverse order.

To do a translation before a stretch we have to notice that

3x 2 + 6 = 3(x 2 ⎛0⎞

+ 2) which is the image of y = f (x) under a translation of ⎜ ⎟

⎝2⎠

followed by a stretch in the y-axis of factor 3.

3) y = – sin(x + π) is the image of y = f (x) under a reflection in the x-axis followed

⎛−π


by a translation of ⎜ ⎟ , or the translation followed by the reflection.

⎝ 0 ⎠

C3 14/04/13 SDB 9


3 Trigonometry

Sec, cosec and cot

1

Secant is written sec x = ; cosecant is written cosec x =

cos x

1 cos x

cotangent is written cot x = =

tan x sin x

Graphs

1

sin x

and

y

1

y

1

y

1

x

90 180 270 360

x

90 180 270 360

x

90 180 270 360

-1

-1

-1

y = sec x y = cosec x y = cot x

Notice that your calculator does not have sec, cosec and cot buttons so to solve equations

involving sec, cosec and cot, change them into equations involving sin, cos and tan and then use

your calculator as usual.

Example: Find cosec 35 o .

Solution: cosec 35 o 1 1

=

o

= = 1. 743 to 4 S.F.

sin 35 0.53576...

Example: Solve sec x = 3.2 for 0 ≤ x ≤ 2π c

1

1

Solution: sec x = 3.2 ⇒ = 3.

2 ⇒ cos x = = 0.

3125

cos x

3.

2

⇒ x = 1.25 or 2π – 1.25 = 5.03 radians

10

C3 14/04/13 SDB


Inverse trigonometrical functions

The inverse of sin x is written as arcsin x or sin –1 x and in order that there should only be

one value of the function for one value of x we restrict the domain to – π / 2 ≤ x ≤ π / 2 .

Similarly for the inverses of cos x and tan x, as shown below.

Graphs

1

y

y

3

y

2

1

-2 -1 1 2

x

1

-2 -1 1 2

x

-1

-2 -1 1 2

x

-1

y = arcsin x

y = arccos x

y = arctan x

–1 ≤ x ≤ 1

–1 ≤ x ≤ 1

x ∈ R

– π / 2 ≤ arcsin x ≤ π / 2

0 ≤ arccos x ≤ π

– π / 2 < arctan x < π / 2

Trigonometrical identities

You should learn these

sin 2 θ + cos 2 θ = 1

tan 2 θ + 1 = sec 2 θ

1 + cot 2 θ = cosec 2 θ

sin (A + B) = sin A cos B + cos A sin B

sin (A – B) = sin A cos B – cos A sin B

sin 2A = 2 sin A cos A

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

cos 2A = cos 2 A – sin 2 A

= 2 cos 2 A – 1

= 1 – 2 sin 2 A

sin 2 A = ½ (1 – cos 2A)

cos 2 A = ½ (1 + cos 2A)

sin 2 ½ θ = ½ (1 – cos θ)

cos 2 ½ θ = ½ (1 + cos θ)

sin 3A = 3 sin A – 4 sin 3 A

cos 3A = 4 cos 3 A – 3 cos A

tan A + tan B

tan( A + B)

=

1- tan A tan B

tan A - tan B

tan( A - B)

=

1+ tan A tan B

tan 2

tan 3A =

2 tan

=

1- tan

A

2

sin P + sin Q =

sin P – sin Q =

cos P + cos Q =

A

A

3

3tan A − tan A

2

1 − 3tan A

2sin

2cos

2 cos

P + Q

2

P + Q

2

P + Q

2

P - Q

cos

2

P - Q

sin

2

P - Q

cos

2

P + Q P - Q

cos P – cos Q = – 2sin sin

2 2

2 sin A cos B = sin(A + B) + sin(A – B)

2 cos A sin B = sin(A + B) – sin(A – B)

2 cos A cos B = cos(A + B) + cos(A – B)

–2 sin A sin B = cos(A + B) – cos(A – B)

C3 14/04/13 SDB 11


Finding exact values

When finding exact values you may not use calculators.

Example:

Find the exact value of cos 15 o

Solution: We know the exact values of sin 45 o , cos 45 o and sin 30 o , cos 30 o

so we consider cos 15 = cos (45 – 30) = cos 45 cos30 + sin 45 sin 30

1 6 2

= 3 1 1 3 + 1

+

× + × = = .

2 2

2 2 2 2

4

Example: Given that A is obtuse and that B is acute, and sin A = 3 / 5 and cos B = 5 / 13 find

the exact value of sin (A + B).

Solution: We know that sin (A + B) = sin A cos B + cos A sin B so we must first find

cos A and sin B.

Using sin 2 θ + cos 2 θ = 1

⇒ cos 2 A = 1 – 9 / 25 = 16 / 25 and sin 2 B = 1 – 25 / 169 = 144 / 169

⇒ cos A = ± 4 / 5 and sin B = ± 12 / 13

But A is obtuse so cos A is negative and B is acute so sin B is positive

⇒ cos A = – 4 / 5 and sin B =

12 / 13

⇒ sin (A + B) = 3 / 5 × 5 / 13 + – 4 / 5 × 12 / 13 = –33 / 65

Proving identities.

Start with one side, usually the L.H.S., and fiddle with it until it equals the other side. Do not

fiddle with both sides at the same time.

cos 2

Example: Prove that A +1

= cot

2 A .

1- cos 2A

2

2

cos 2A +1 2 cos A -1 +1 2 cos A

2

Solution: L.H.S. = =

2

=

2

= cot A . Q.E.D.

1- cos 2A

1- (1- 2sin A)

2sin A

Eliminating a variable between two equations

Example: Eliminate θ from x = sec θ – 1, y = tan θ.

Solution: We remember that tan 2 θ + 1 = sec 2 θ

⇒ sec 2 θ – tan 2 θ = 1.

secθ = x + 1 and tanθ = y

⇒ (x + 1) 2 – y 2 = 1

⇒ y 2 = (x + 1) 2 – 1 = x 2 + 2x.

Solving equations

Here you have to select the ‘best’ identity to help you solve the equation.

Example: Solve the equation sec 2 A = 3 – tan A, for 0 ≤ A ≤ 360 o.

Solution: We know that tan 2 θ + 1 = sec 2 θ

⇒ tan 2 A + 1 = 3 – tan A

12

C3 14/04/13 SDB


⇒ tan 2 A + tan A – 2 = 0, factorising gives

⇒ (tan A – 1)(tan A + 2) = 0

⇒ tan A = 1 or tan A = –2

⇒ A = 45, 225, or 116.6, 296.6.

R cos(x + α)

An alternative way of writing a cos x ± b sin x using one of the formulae listed below

(I) R cos (x + α) = R cos x cos α – R sin x sin α

(II) R cos (x – α) = R cos x cos α + R sin x sin α

(III) R sin (x + α) = R sin x cos α + R cos x sin α

(IV) R sin (x – α) = R sin x cos α – R cos x sin α

To keep R positive and α acute we select the formula with corresponding + and – signs. The

technique is the same which ever formula we choose.

Example: Solve the equation 12 sin x – 5 cos x = 6 for 0 o ≤ x ≤ 360 0 .

Solution: First re–write in the above form:

notice that the sin x is +ve and the cos x is –ve so we need formula (IV).

R sin (x – α) = R sin x cos α – R cos x sin α = 12 sin x – 5 cos x

Equating coefficients of sin x, ⇒ R cos α = 12

Equating coefficients of cos x, ⇒ R sin α = 5

[i]

[ii]

Squaring and adding [i] and [ii] ⇒ R 2 cos 2 α + R 2 sin 2 α = 12 2 + 5 2

⇒ R 2 (cos 2 α + sin 2 α) = 144 + 25 but cos 2 α + sin 2 α = 1

⇒ R 2 = 169 ⇒ R = ±13

But choosing the correct formula means that R is positive ⇒ R = +13

Substitute in [i] ⇒ cos α = 12 / 13

⇒ α = 22.620 o or ....

But choosing the correct formula means that α is acute ⇒ α = 22.620 o .

⇒ 12 sin x – 5 cos x = 13 sin(x – 22.620)

and so to solve 12 sin x – 5 cos x = 6

we need to solve 13 sin(x – 22.620) = 6

⇒ sin(x – 22.620) = 6 / 13

⇒ x – 22.620 = 27.486 or 180 – 27.486 = 152.514

⇒ x = 50.1 o or 175.1 o .

C3 14/04/13 SDB 13


Example: Find the maximum value of 12 sin x – 5 cos x and the value(s) of x for which

it occurs.

Solution: From the above example 12 sin x – 5 cos x = 13 sin(x – 22.6).

The maximum value of sin(anything) is 1 and occurs when the angle is 90, 450, 810

etc. i.e. 90 + 360n

⇒ the max value of 13 sin(x – 22.6) is 13

when x – 22.6 = 90 + 360n ⇒ x = 112.6 + 360n o .

4 Exponentials and logarithms

Natural logarithms

Definition and graph

e ≈ 2.7183 and logs to base e

are called natural logarithms.

log e x is usually written ln x.

y = e x

y

y = x

2

y = ln x

x

-4 -2 2 4 6

-2

Note that y = e x and y = ln x

are inverse functions and that

the graph of one is the reflection

of the other in the line y = x.

-4

Graph of y = e (ax + b) + c.

y

y = e 2x

y = e x

The graph of y = e 2x is the graph of

6

y = e x stretched by a factor of 1 / 2 in

4

the direction of the x-axis.

2

-2 2

x

14

C3 14/04/13 SDB


The graph of y = e (2x + 3) is the graph of

y = e 2x 3

⎛−

2

translated through ⎟ ⎞

⎜ since

⎝ 0 ⎠

2x + 3 = 2(x – − 3

2

)

and the graph of y = e (2x +3) + 4 is that of

y = e (2x + 3) ⎛0⎞

translated through ⎜ ⎟

⎝4⎠

y = e 2x + 3 + 4

y

6

4

2

−8 −6 −4 −2 2

y = e 2x + 3

x

Equations of the form e ax + b = p

Example: Solve e 2x + 3 = 5.

Solution: Take the natural logarithm of each side, remembering that ln x is the

inverse of e x .

⇒ ln(e 2x + 3 ) = ln 5 ⇒ 2x + 3 = ln 5

⇒ x =

ln5

− 3

2

Example: Solve ln(3x – 5) = 4.

Solution:

ln x.

= –0.695 to 3 S.F.

Raise both sides to the power of e, remembering that e x is the inverse of

⇒ e ln(3x – 5) = e 4 ⇒ (3x – 5) = e 4

⇒ x =

e 4 + 5

3

= 19.9 to 3 S.F.

5 Differentiation

Chain rule

If y is a composite function like y = (5x 2 – 7) 9

think of y as y = u 9 , where u = 5x 2 – 7

then the chain rule gives



dy

dx

dy

dx

dy

dx

dy du

= ×

du dx

8 du

= 9 u ×

dx

= x

x x x

2 8

2

9(5

− 7) × (10 ) = 90 (5 −

The rule is very simple, just differentiate the function of u and multiply by

7)

8

.

du .

dx

C3 14/04/13 SDB 15


Example:

y = √(x 3 dy

– 2x). Find . dx

Solution: y = √(x 3 3

– 2x) = ( x − 2x)

2

. Put u = x 3 – 2x


1

2

y = u


dy dy du

= ×

dx du dx


dy

1 −1

du

−1

2

1 3

2

2

= u × = ( x − 2x)

× (3x

2)

2

2

dx dx


2

dy 3x

− 2

=

1

dx

3

2

2( x − 2x)

.

1

Product rule

If y is the product of two functions, u and v, then

dy dv du

y = uv ⇒ = u + v .

dx dx dx

Example: Differentiate y = x 2 × √(x – 5).

Solution: y = x 2 × √(x – 5) = x 2 1

2

× ( x − 5)

so put u = x 2 1

2

and v = ( x − 5)


dy

dx

dv du

= u + v

dx dx

= x 2 1

2

× 1 )


1

2

( x − 5 + ( x − 5) × 2x

2

2

x

= + 2x

x − 5 .

2 x − 5

Quotient rule

If y is the quotient of two functions, u and v, then

du dv

v − u

u

dy

y = ⇒ =

dx dx

.

2

v

dx v

2x

− 3

Example: Differentiate y =

x

2 + 5x

2x

− 3

Solution: y = , so put u = 2x – 3 and v = x 2 + 5x

x

2 + 5x

du dv

v − u

dy

⇒ =

dx dx

2

dx v

16

C3 14/04/13 SDB


( x

+ 5x)

× 2

2

=

2 2

( x


(2x

− 3) × (2x

+ 5)

+ 5x)

2x

+ 10x

− 2x

2

2

2

= =

2 2

2 2

( x


(4x

+ 5x)

+ 4x

−15)

( x

+ 6x

+ 15

+ 5x)

.

Derivatives of e x and log e x ≡ ln x.

y = e x

y = ln x



dy

dx

=

x

e

dy 1

=

dx x

y = ln kx ⇒ y = ln k + ln x ⇒

y = ln x k ⇒ y = k ln x ⇒

or use the chain rule.

dy 1 1

= 0 + =

dx x x

dy

dx

=

1

k ×

x

=

k

x

Example: Find the derivative of f (x) = x 3 – 5e x at the point where x = 2.

Solution: f (x) = x 3 – 5e x

⇒ f ′(x) = 3x 2 – 5e x

⇒ f ′(2) = 12 – 5e 2 = –24.9

Example: Differentiate the function f (x) = ln 3x – ln x 5

Solution: f (x) = ln 3x – ln x 5 = ln 3 + ln x – 5 ln x = ln 3 – 4 ln x

⇒ f ′(x) =

− 4

x

or we can use the chain rule

⇒ f ′(x) =

1

3x

1

x

4

× 3 − × 5x

=

5

− 4

x

Example: Find the derivative of f (x) = log 10 3x.

Solution: f (x) = log 10 3x =

=

ln 3

+

ln10

ln x

ln 3x

ln10

=

ln 3

ln 10

+

ln x

ln 10

(using change of base formula)

1

x 1

⇒ f ′(x) = 0 + = .

ln10 x ln10

C3 14/04/13 SDB 17


Example:

2

x dy

y = e . Find . dx

Solution: y = e u , where u = x 2


dy dy du

= ×

dx du dx


dy du

= e

dx

= e u

dx

× 2x = 2x ×

2

x

e .

Example: y = ln 7x 3 dy

. Find . dx

Solution: y = ln u, where u = 7x 3



dy

dx

dy dy du

= ×

dx du dx

du

= 1 × =

u dx 7x

1 2

=

3

3

× 21x

.

x

dy 1

=

dx

dx

dy

Example:

x = sin 2 dy

3y. Find . dx

Solution:

dx

First find dy

as this is easier.

dx

dy

= 2 sin 3 = 2 sin 3y cos 3y × 3 = 6 sin 3y cos 3y


dy 1 1

1

1

= = = =

3

cosec 6y

dx 6 sin3ycos3y

3sin 6y

dx

dy

Trigonometric differentiation

x must be in RADIANS when differentiating trigonometric functions.

f (x) f ′ (x) important formulae

sin x

cos x

cos x – sin x sin 2 x + cos 2 x = 1

tan x

sec 2 x

sec x sec x tan x tan 2 x + 1 = sec 2 x

cot x

– cosec 2 x

cosec x – cosec x cot x 1 + cot 2 x = cosec 2 x

18

C3 14/04/13 SDB


Chain rule – further examples

Example: y = sin 4 x.

dy

Find . dx

Solution: y = sin 4 x. Put u = sin x

⇒ y = u 4


dy dy du

= ×

dx du dx


dy 3 du

3

= 4u

× = 4sin x

dx dx

× cos x .

Example: y =

x dy

e sin . Find . dx

Solution: y = e u , where u = sin x


dy dy du

= ×

dx du dx


dy du

= e

dx

dx

= e sin x × cos x = cos x × e sin x .

Example:

dy

y = ln (sec x). Find . dx

Solution: y = ln u, where u = sec x


dy dy du

= ×

dx du dx


dy du

= 1 1

× = × sec x tan x

dx u dx sec x

= tan x .

Trigonometry and the product and quotient rules

Example: Differentiate y = x 2 × cosec 3x.

Solution: y = x 2 × cosec 3x

so put u = x 2 and v = cosec 3x

dy dv du

⇒ = u + v

dx dx dx

= x 2 × (– 3cosec 3x cot 3x) + cosec 3x × 2x

= – 3x 2 cosec 3x cot 3x + 2x cosec 3x.

C3 14/04/13 SDB 19


tan 2x

Example: Differentiate y =

3

7x

tan 2x Solution: y =

3 , so put u = tan 2x and v = 7x

3

7x

3

2

2

dy 7x

× 2sec 2x

− tan 2x

× 21x

⇒ =

3 2

dx

(7x

)

dy

dx

dy

dx

14x

sec

2x

3 2

⇒ =

6

2x

sec

2x

2

⇒ =

4

7x


21x

2

49x

− 3 tan 2x

tan 2x

6 Numerical methods

Locating the roots of f(x) = 0

A quick sketch of the graph of y = f (x) can help to give a rough idea of the roots of f (x) = 0.

If y = f (x) changes sign between x = a and x = b and f (x) is continuous in this region then

a root of f (x) = 0 lies between x = a and x = b.

Example: Show that a root of the equation f (x) = x 3 – 8x – 7 = 0 lies between 3 and 4.

Solution: f (3) = 27 – 24 – 7 = –4, and f (4) = 64 – 32 – 7 = +25

Thus f (x) changes sign and f (x) is continuous ⇒ there is a root between 3 and 4.

The iteration x n + 1 = g(x n )

Many equations of the form f (x) = 0 can be re–arranged as x = g (x) which leads to the

iteration x n + 1 = g (x n ).

Example: Show that the equation x 3 – 8x – 7 = 0

can be re–arranged as x = 3 8 x + 7 and find a solution correct to four decimal places.

Solution: x 3 – 8x – 7 = 0 ⇒ x 3 = 8x + 7 ⇒ x = 3 8 x + 7 .

From the previous example we know that there is a root between 3 and 4,

so starting with x 1 = 3 we use the iteration x n + 1 = 3 8 x + 7

to give x 1 = 3

x 2 = 3.14138065239

x 3 = 3.17912997899

x 4 = 3.18905898325

x 5 = 3.19166031127

x 6 = 3.19234114002

x 7 = 3.19251928098

x 8 = 3.19256588883

x 9 = 3.19257808284

n

-4 -2 2 4

-5

-10

-15

-20

y

x

20

C3 14/04/13 SDB


As x 8 and x 9 are both equal to 3.1926 to 4 D.P.

we conclude that a root of the equation is 3.1926 to 4 D.P.

Note that we have only found one root: the sketch shows that there are more roots.

Conditions for convergence

If an equation is rearranged as x = g(x) and if there is a root x = α

then the iteration x n+1 = g(x n ), starting with an approximation near x = α

(i) will converge if -1 < g ′(α) < 1

y

y

x

x 1 x 2

x 3 x3

x 4

x

x 5

x 2 x 1

10

(a) will converge without oscillating (b) will oscillate and converge

if 0 < g ′(α) < 1, if –1 < g ′(α) < 0,

(ii)

will diverge if

g ′(α) < -1 or g ′(α) > 1.

y

x 1 x 2 x 3

x

C3 14/04/13 SDB 21


Index

algebraic fractions

adding and subtracting, 3

equations, 4

multiplying and dividing, 3

chain rule, 15

further examples, 19

cosecant, 10

cotangent, 10

derivative

e x , 17

ln x, 17

differentiation

dy 1

= , 18

dx

dx

dy

trigonometric functions, 18

equations

graphical solutions, 20

exponential

e ax + b = p, 15

graph of y = e (ax + b) + c, 14

graph of y = e x , 14

functions, 4

combining functions, 6

defining functions, 5

domain, 4

inverse functions, 6

modulus functions, 7

range, 4

graphs

combining transformations, 9

standard functions, 8

iteration

conditions for convergence, 21

x n + 1 = g(x n ), 20

logarithm

graph of ln x, 14

natural logarithm, 14

product rule, 16

further examples, 19

quotient rule, 16

further examples, 19

R cos(x + α), 13

secant, 10

trigonometrical identities, 11

trigonometry

finding exact values, 12

harder equations, 12

inverse functions, 11

proving identities, 12

22

C3 14/04/13 SDB

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