BY TR.

ALI DAMDEOK

Pure Core 3

Revision Notes
BY TR. ALI DAMDEOK
BY TR. ALI DAMDEOK

Pure Core 3

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

Cancelling common factors .................................................................................................................. 3
Multiplying and dividing fractions ....................................................................................................... 3
Adding and subtracting fractions .......................................................................................................... 3
Equations .............................................................................................................................................. 4

2 Functions ...............................................................................................................5
Notation ................................................................................................................................................ 5
Domain, range and graph ...................................................................................................................... 5
Defining functions ........................................................................................................................................................................... 6
Composite functions ............................................................................................................................. 7
Inverse functions and their graphs ........................................................................................................ 7
To find the inverse of a function x → y................................................................................................................................. 7
Domain and range of inverse functions ................................................................................................ 9
Modulus functions .............................................................................................................................. 10
Modulus functions y = |f (x)|.................................................................................................................................................10
Modulus functions y = f (|x|) ..................................................................................................................................................11
Standard graphs................................................................................................................................... 11
Combinations of transformations of graphs ........................................................................................ 12
Sketching curves ................................................................................................................................. 13

3 Trigonometry ........................................................................................................ 14
Sec, cosec and cot ............................................................................................................................... 14
Graphs ................................................................................................................................................................................................14
Inverse trigonometrical functions ....................................................................................................... 15
Graphs ................................................................................................................................................................................................15
Trigonometrical identities ................................................................................................................... 16
𝑃+𝑄 𝑃−𝑄
Proof of sin P + sin Q = 2 sin � � cos � � .................................................................................................................16
2 2
Finding exact values.....................................................................................................................................................................17
Proving identities. .........................................................................................................................................................................17
Eliminating a variable between two equations ...............................................................................................................18
Solving equations ..........................................................................................................................................................................18
R cos(x + α) ........................................................................................................................................ 19

4 Exponentials and logarithms ................................................................................ 21

Natural logarithms .............................................................................................................................. 21
Definition and graph ....................................................................................................................................................................21
Equations of the form eax + b = p ............................................................................................................................................22
BY TR. ALI DAMDEOK

5 Differentiation ...................................................................................................... 23
Chain rule ............................................................................................................................................ 23
Product rule ......................................................................................................................................... 23
Quotient rule ........................................................................................................................................ 24
Derivatives of ex and logex ≡ ln x. ..................................................................................................... 25
𝒅𝒚 𝟏
= 𝒅𝒙 ............................................................................................................................................. 26
𝒅𝒙
𝐝𝐲

Derivative of ax................................................................................................................................... 26
Trigonometric differentiation .............................................................................................................. 27
Chain rule – further examples ............................................................................................................. 27
Trigonometry and the product and quotient rules ............................................................................... 28

6 Numerical methods .............................................................................................. 30

Locating the roots of f(x) = 0 .............................................................................................................. 30
The iteration xn + 1 = g(xn) ................................................................................................................... 30
Conditions for convergence................................................................................................................. 31

7 Appendix.............................................................................................................. 32
Derivatives of sin x and cos x .............................................................................................................. 32
sinℎ
limℎ→0 = 1 ...............................................................................................................................................................................32
ℎ
Alternative formula for derivative ........................................................................................................................................32
Derivatives of sin x and cos x ...................................................................................................................................................33

Index ........................................................................................................................... 34
BY TR. ALI DAMDEOK

1 Algebraic fractions
Cancelling common factors
x 3 − 2 x 2 − 3x
Example: Simplify .
4 x 2 − 36
Solution: First factorise top and bottom fully –
x 3 − 2 x 2 − 3x x( x 2 − 2 x − 3) x( x − 3)( x + 1)
= =
4 x 2 − 36 4( x 2 − 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.
9 x 2 − 4 3x 2 − x − 2
Example: Simplify ÷
3x 2 − 2 x x3 + x2
Solution: First turn the second fraction upside down and multiply
9x 2 − 4 x3 + x2
= × factorise fully
3x 2 − 2 x 3x 2 − x − 2

(3 x − 2)(3 x + 2) x ( x + 1)
2
= × cancel all common factors
x(3 x − 2) (3x + 2)( x − 1)
x( x + 1)
Answer = .
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.
3x 5
Example: Simplify − 2 .
x − 7 x + 12 x − 4 x + 3
2

Solution: First factorise the denominators

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)

3x 2 − 3x − 5 x + 20 3x 2 − 8 x + 20
= = which cannot be simplified further.
( x − 1)( x − 3)( x − 4) ( x − 1)( x − 3)( x − 4)
BY TR. ALI DAMDEOK

Equations
x x −1 1
Example: Solve − =
x +1 x 2
Solution: First multiply both sides by the Lowest Common Denominator
x x −1 1
− = multiply both sides by 2x(x + 1)
x +1 x 2
⇒ x × 2x – (x – 1) × 2(x + 1) = x(x + 1)
⇒ 2x2 – 2x2 + 2 = x2 + x
⇒ x2 + x – 2 = 0,
⇒ (x + 2)(x – 1) = 0
⇒ x = –2 or 1
BY TR. ALI DAMDEOK

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

Notation
y = x2 – 3x + 7, f (x) = x2 – 3x + 7 and f : x → x2 – 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
sometimes is evident from the function: e.g. for √x, x can only take positive x or zero values.
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 ∈ ℜ: –2 < x ≤ 4.
y
6

Solution: First sketch the graph for values of x between – 2 4

range y=2x−3
and 4 (the domain), and we can see that we are only using 2

the y–axis from –7 to 5, domain x

−4 −2 2 4 6
not including y = –7 (since x ≠ –2), −2

but including y = 5 (since x can equal 4) −4

and so the range is
−6
y ∈ ℜ: –7 < y ≤ 5.
−8

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 possible domain is x ∈ ℜ: x ≥ 3.

To find the range we first sketch the graph 6

and we see that the graph will cover all of the 4 y=√(x−3)+1
range
y-axis from 1 upwards 2
domain x

and so the range is y ∈ ℜ: y ≥ 1. −2 2 4 6 8 10 12 14 16

BY TR. ALI DAMDEOK

Example: Find the largest possible domain and the range for the function
2x
f:x→ . Give the equations of its asymptotes.
x +1
Solution: The only problem occurs when the denominator is 0, and so x cannot be –1.
Thus the largest domain is x ∈ ℜ: x ≠ –1.

y
To find the range we sketch the graph 8

and we see that y can take any value 6

y=2x/(x+1)
except 2, range
4
asymptote y=2
2
so the range is y ∈ ℜ: y ≠ 2. domain domain x
0 −8 −6 −4 −2 2 4 6 8

−2
Asymptotes are x = –1 and y = 2 asymptote x=−1 range
−4

−6

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.
1
2) x → where x ∈ ℜ is not a function as the image of x = 3 is not defined,
x −3
1
but f : x → where x ∈ ℜ, x ≠ 3 is a function.
x −3
BY TR. ALI DAMDEOK

Composite functions
To find the composite function fg we must do g first.
Example: f : x → 3x – 2 and g : x → x2 + 1. Find fg and gf.

Solution: Think of f and g as ‘rules’

f is multiply by 3 subtract 2

g is square add 1

⇒ fg is square add 1 multiply by 3 subtract 2

giving (x2 + 1) × 3 – 2 = 3x2 + 1

⇒ fg : x → 3x2 + 1 or fg(x) = 3x2 + 1.

gf is multiply by 3 subtract 2 square add 1

giving (3x – 2)2 + 1 = 9x2 – 12x + 5

⇒ gf : x → 9x2 – 12x + 5 or gf (x) = 9x2 – 12x + 5.

Note that fg and gf are not the same.

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.
BY TR. ALI DAMDEOK

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

Solution: We have x → y = 3x – 2 4 y

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

x+2 2
(ii) solving for y ⇒ y=
3 y = f –1(x)
x
x+2
⇒ f –1
:x→ −4 −2 2 4

3
−2 y = f (x)

−4

x+3
Example: Find the inverse of g : x → .
2x − 5
y
x+3
Solution: We have x → y = 6
2x − 5 y=x

(i) interchanging x and y 4

y+3
⇒ x =
2y − 5 y = g –1(x)
2

(ii) solving for y y = g (x) x

⇒ x(2y – 5) = y + 3 −4 −2 2 4 6

⇒ 2xy – 5x = y + 3 −2

⇒ 2xy – y = 5x + 3
−4
⇒ y(2x – 1) = 5x + 3
5x + 3
⇒ y =
2x − 1
5x + 3
⇒ g –1 : x →
2x − 1

Note that f f –1 (x) = f –1

f (x) = x.
BY TR. ALI DAMDEOK

Domain and range of inverse functions

–1
Note that the domain of f (x) is the range of f (x),
–1
and that the range of f (x) is the domain of f (x).
–1
This is because the graph of y = f (x) is that of y = f (x) after a reflection in the line y = x.

Example: f (x) = (x – 3)2 + 4, x ∈ ℜ, x ≤ 3.

(a) Sketch the graph of y = f (x), and state its range.
(b) Find the inverse function, f –1(x) and sketch its graph on the same diagram.
Show the line y = x on your diagram.
(c) State the domain and range of f –1(x).

Solution:
(a) As the domain of f is x ≤ 3, we only y
have the ‘left’ part of the parabola.
y = f (x)
8
The range is f (x) ≥ 4, f (x) ∈ ℜ.

6
y=x
(b) To find the inverse, swap x and y, then
find y. 4

x = (y – 3)2 + 4
2 y = f –1(x)
⇒ y – 3 = ±√𝑥 − 4 .
x
From the reflection of y = f (x) in y = x, −2 2 4 6 8
we can see that we want the negative
sign −2

⇒ y = 3 – √𝑥 − 4.

(c) The domain of f –1(x) is x ≥ 4 (the range of f (x)).

The range of f –1(x) is f –1(x) ≤ 3 (the domain of f (x)).
BY TR. ALI DAMDEOK

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 = |x2 – 3x|.

Solution:
First sketch the graph of y = x2 – 3x Then reflect the portion below
the x-axis in the x–axis to give

3 y 3 y

y=x²−3x y=|x²−3x|
2 2

1 1

x x
−1 1 2 3 4 −1 1 2 3 4

−1 −1

−2 −2
BY TR. ALI DAMDEOK

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 reflection of 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.
3 y
y=x²−3x
2
Example: Sketch the graph of y = |x|2 – 3|x|.
1

x
2 −4 −3 −2 −1 1 2 3 4
Solution: f (x) = x – 3x
−1
⇒ f (x) = x2 – 3x
−2
First sketch the graph of y = x2 – 3x for positive
values of x only. 3 y
y=|x|²−3|x|
2

Then reflect your graph in the y–axis to 1

x
complete the sketch. −4 −3 −2 −1 1 2 3 4

−1

−2

Standard graphs

3 y y
4 y=x⁴
y=x³ y=x
2 y=x²
3

1
2
x
−3 −2 −1 1 2
1
−1
x
−2 −1 1 2
−2
−1

y
y
2 3
y=1/x

1 2
y=√x
x
1
−3 −2 −1 1 2 3

x
−1
−1 1 2 3 4 5 6 7

−2 −1
BY TR. ALI DAMDEOK

Combinations of transformations of graphs

We know the following transformations of graphs:

y = f (x)
 a
translated through   becomes y = f (x – a) + b
 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 = 3x2 + 6 is the image of y = x2 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
0
3x2 + 6 = 3(x2 + 2) which is the image of y = x2 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 = sin x under a reflection in the x-axis
−π 
followed by a translation of   , or the translation followed by the reflection.
 0 
BY TR. ALI DAMDEOK

Sketching curves
When sketching curves, show the coordinates of the intercepts with the axes, and the
equations of any asymptotes − show the asymptotes with dotted lines.

2
Example: Sketch the curve 𝑦 = 4 − 𝑥, x > 0
Solution:
5 y
asymptote y=4
4
3
y=4−2/x
2
asymptote
x=0
1
x
−2 −1 1 2 3 4 5
−1
−2
−3

Note that the domain is x > 0, so no graph to the left of the y-axis.
x ≠ 0 ⇒ curve does not meet the x-axis
y=0 ⇒ x=1
−2
Thinking of y = 𝑥
translated up 4,
the horizontal asymptote is y = 4.
Do not forget that the y-axis, x = 0, is also an asymptote.
BY TR. ALI DAMDEOK

3 Trigonometry
Sec, cosec and cot
1 1
Secant is written sec x = ; cosecant is written cosec x =
cos x sin x
1 cos x
cotangent is written cot x = =
tan x sin x
Graphs

y y
y=cosecx
y=secx
2 2

y=sinx
y=cosx x x
−π π 2π −π π 2π

−2 −2

y
y=tanx y=cotx
2
Notice that your calculator does not have sec,
cosec and cot buttons so to solve equations x
involving sec, cosec and cot, change them −π π 2π
into equations involving sin, cos and tan and
then use your calculator as usual.
−2

Example: Find cosec 35o.

1 1
Solution: cosec 35o = = = 1 ⋅ 743 to 4 S.F.
sin 35 o
0 ⋅ 53576...

Example: Solve sec x = 3⋅2 for 0 ≤ x ≤ 2π

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 to 3 S.F.
BY TR. ALI DAMDEOK

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 .
Note that the graph of y = arcsin x is the reflection of part of the graph of y = sin x in the
line y = x.
Similarly for the inverses of cos x and tan x, as shown below.

Graphs
y = arcsin x y = arccos x

y
y=x y
y=arcsinx
π/2 π
y=arccosx
y=x
y=sinx
x
π/2
−3 −2 −1 1 2 3 4
y=cosx
x
−π/2
−3 −2 −1 1 2 3 4

y = arctan x

y
y=tanx y=x

y=arctanx
π/2

x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−π/2
BY TR. ALI DAMDEOK

Trigonometrical identities
You should learn these

sin2 θ + cos2 θ = 1 tan A - tan B

tan( A - B) =
tan2θ + 1 = sec2θ 1 + tan A tan B

1 + cot2θ = cosec2θ tan 2 A =

2 tan A
1 - tan 2 A
sin (A + B) = sin A cos B + cos A sin B
sin (A – B) = sin A cos B – cos A sin B 3 tan A − tan 3 A
tan 3A =
1 − 3 tan 2 A
sin 2A = 2 sin A cos A
cos (A + B) = cos A cos B – sin A sin B P+Q P -Q
sin P + sin Q = 2 sin cos
cos (A – B) = cos A cos B + sin A sin B 2 2

cos 2A = cos2 A – sin2 A P+Q P -Q

sin P – sin Q = 2 cos sin
= 2 cos2 A – 1 2 2

= 1 – 2 sin2 A P+Q P -Q
cos P + cos Q = 2 cos cos
sin2 A = ½ (1 – cos 2A) 2 2

cos2 A = ½ (1 + cos 2A) P+Q P -Q

cos P – cos Q = - 2 sin sin
2 2
sin2 ½ θ = ½ (1 – cos θ)
cos2 ½ θ = ½ (1 + cos θ) 2 sin A cos B = sin(A + B) + sin(A – B)

sin 3A = 3 sin A – 4 sin3 A 2 cos A sin B = sin(A + B) – sin(A – B)

cos 3A = 4 cos3A – 3 cos A 2 cos A cos B = cos(A + B) + cos(A – B)

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

tan( A + B) =
1 - tan A tan B

The last four formulae, 2 sin A cos B = sin(A + B) + sin(A – B) etc., are not in the formula
booklet, and should be learnt.
𝑃+𝑄 𝑃−𝑄
You should know the proofs of the four formulae sin P + sin Q = 2 sin 2
cos 2
, etc.
𝑷+𝑸 𝑷−𝑸
Proof of sin P + sin Q = 2 sin 𝟐
cos 𝟐

We know that sin(A + B) = sin A cos B + cos A sin B

and sin(A − B) = sin A cos B − cos A sin B
⇒ sin(A + B) − sin(A − B) = 2 sin A cos B
Now put P = A + B, and Q = A − B
𝑃+𝑄 𝑃−𝑄
⇒ P + Q = 2A, and P − Q = 2B, ⇒ A= 2
, and B = 2
𝑃+𝑄 𝑃−𝑄
⇒ sin P + sin Q = 2 sin 2
cos 2
.
The other formulae can be proved in a similar way.
BY TR. ALI DAMDEOK

Finding exact values

When finding exact values you may not use calculators.

Example: Find the exact value of cos 15o

Solution: We know the exact values of sin 45o, cos 45o and sin 30o, cos 30o
so we consider cos 15 = cos (45 – 30) = cos 45 cos30 + sin 45 sin 30
3 + 1 6 +
= 1
2
× 2
3
+ 1
2
× 12 = 2 2
= 4
2
.

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 sin2 θ + cos2 θ = 1

⇒ cos2 A = 1 – 9/25 = 16
/25 and sin2 B = 1 – 25/169 = 144
/169
⇒ cos A = ± /5 and 4
sin B = ± /13 12

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 A + 1
Example: Prove that = cot 2 A .
1 - cos 2 A

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

Solution: L.H.S. = = 2 = 2 = cot 2 A . Q.E.D.
1 - cos 2 A 1 - (1 - 2sin A) 2 sin A
BY TR. ALI DAMDEOK

Eliminating a variable between two equations

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

Solution: We remember that tan2θ + 1 = sec2θ

⇒ sec2θ – tan2θ = 1.
secθ = x + 1 and tanθ = y
⇒ (x + 1)2 – y2 = 1
⇒ y2 = (x + 1)2 – 1 = x2 + 2x.

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

Example: Solve the equation sec2A = 3 – tan A, for 0 ≤ A ≤ 360o.

Solution: We know that tan2A + 1 = sec2A

⇒ tan2A + 1 = 3 – tan A
⇒ tan2A + 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.

Example: Solve sin 3x − sin 5x = 0 for 0o ≤ x ≤ 90o.

𝑃+𝑄 𝑃−𝑄
Solution: Using the formula sin P − sin Q = 2 cos 2
sin 2

⇒ 2 cos 4x sin(−x) = 0 ⇒ cos 4x sin x = 0

⇒ cos 4x = 0, or sin x = 0

⇒ 4x = 90, 270, (450), … or x = 0, (180), …

⇒ x = 0o, 22⋅5o or 67⋅5o.

BY TR. ALI DAMDEOK

R cos(x + α)

An alternative way of writing a cos x ± b sin x using one of the formulae listed below
(1) R cos (x + α) = R cos x cos α – R sin x sin α
(2) R cos (x – α) = R cos x cos α + R sin x sin α
(3) R sin (x + α) = R sin x cos α + R cos x sin α
(4) 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 0o ≤ x ≤ 3600.

Solution: First re–write in the above form:

notice that the sin x is positive and the cos x is negative so we need formula (4).

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 I

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

Squaring and adding I and II ⇒ R2 cos2 α + R2 sin2 α = 122 + 52

⇒ R2 (cos2 α + sin2 α) = 144 + 25 but cos2 α + sin2 α = 1
⇒ R2 = 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 337⋅379…o or ……
and choosing the correct formula means that α is acute
⇒ α = 22⋅620…o.
⇒ 12 sin x – 5 cos x = 13 sin(x – 22⋅620…)

To solve 12 sin x – 5 cos x = 6

⇒ 12 sin x – 5 cos x = 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⋅1o or 175⋅1o .
BY TR. ALI DAMDEOK

Example: Find the maximum value of 12 sin x – 5 cos x and the smallest positive value
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 + 360no,
⇒ smallest positive value of x is 112⋅6o.
BY TR. ALI DAMDEOK

4 Exponentials and logarithms

Natural logarithms
Definition and graph y = ex y=x
e ≈ 2⋅7183 and logs to base e 2

are called natural logarithms.

log e x is usually written ln x.
y = ln x
−2 2
x
Note that y = e and y = ln x
are inverse functions and that
the graph of one is the reflection −2

of the other in the line y = x.

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

y = e2x y = ex
3

The graph of y = e2x is the graph of

y = ex stretched by a factor of 1/2 in 2

the direction of the x-axis.

y = e2x is above y = ex for x > 0, 1

and below for x < 0.

−2

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

y = e2x
 − 32 
y = e translated through   since
2x 6
y = e(2x+3)+4
 0 
−3
2x + 3 = 2(x – 2 )
4
and the graph of y = e(2x +3) + 4 is that of
0
y = e(2x + 3) translated through  
 4 y = e(2x+3)
2

−4 −2 2
BY TR. ALI DAMDEOK

Equations of the form eax + b = p

Example: Solve e2x + 3 = 5.

Solution: Take the natural logarithm of each side, remember that ln x is the inverse of ex

⇒ ln (e2x + 3) = ln 5 ⇒ 2x + 3 = ln 5
ln 5 − 3
⇒ x = = –0⋅695 to 3 S.F.
2

Example: Solve ln (3x – 5) = 4.

Solution:
(i) From the definition of logs loga x = y ⇔ x = ay
ln (3x – 5) = 4
⇒ 3x − 5 = e4
e4 + 5
⇒ x = = 19⋅9 to 3 S.F.
3
OR
(ii) Raise both sides to the power of e, .
⇒ eln (3x – 5) = e4 ⇒ (3x – 5) = e4 remember that ex is the inverse of ln x

e4 + 5
⇒ x = = 19⋅9 to 3 S.F.
3
BY TR. ALI DAMDEOK

5 Differentiation
Chain rule
If y is a composite function like y = (5x2 – 7)9
think of y as y = u9, where u = 5x2 – 7
then the chain rule gives
dy dy du
= ×
dx du dx
dy du
⇒ = 9u 8 ×
dx dx
dy
⇒ = 9(5 x 2 − 7) 8 × (10 x) = 90 x(5 x 2 − 7) 8 .
dx
du
The rule is very simple, just differentiate the function of u and multiply by .
dx
dy
Example: y = √(x3 – 2x). Find .
dx
1
Solution: y = √(x3 – 2x) = ( x 3 − 2 x) 2 . Put u = x3 – 2x
1
⇒ y= u 2
dy dy du
⇒ = ×
dx du dx
dy −1 du −1
⇒ = 12 u 2 × = 1
2
( x 3 − 2 x) 2
× (3 x 2 − 2)
dx dx
dy 3x 2 − 2
⇒ = 1
.
dx 2( x − 2 x )
3 2

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 = x2 × √(x – 5).

1
Solution: y = x2 × √(x – 5) = x2 × ( x − 5) 2

1
so put u = x2 and v = ( x − 5) 2

dy dv du
⇒ = u + v
dx dx dx
−1 1
= x2 × 1
2
( x − 5) 2
+ ( x − 5) 2 × 2x
2
x
= + 2x x − 5 .
2 x−5
BY TR. ALI DAMDEOK

Quotient rule

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

du dv
v − u
u dy dx dx .
y = ⇒ = 2
v dx v

2x − 3
Example: Differentiate y =
x 2 + 5x
2x − 3
Solution: y = , so put u = 2x – 3 and v = x2 + 5x
x + 5x
2

du dv
v − u
dy dx dx
⇒ = 2
dx v

( x 2 + 5 x) × 2 − (2 x − 3) × (2 x + 5)
=
( x 2 + 5 x) 2

2 x 2 + 10 x − (4 x 2 + 4 x − 15) − 2 x 2 + 6 x + 15
= = .
( x 2 + 5 x) 2 ( x 2 + 5 x) 2

3𝑥−2 𝑑𝑦
Example: If y = , find , expressing your answer as a single algebraic fraction in
√𝑥−1 𝑑𝑥
its simplest form.

3𝑥 − 2
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝑦= 1
(𝑥 − 1)2
1 −1
𝑑𝑦 (𝑥 − 1)2 × 3 − (3𝑥 − 2) × 12(𝑥 − 1) 2
⇒ =
𝑑𝑥 𝑥−1
1 (3𝑥 − 2)
(𝑥 − 1)2 × 3 −
𝑑𝑦 1
2(𝑥 − 1) �2
⇒ =
𝑑𝑥 𝑥−1
1 (3𝑥 − 2)
(𝑥 − 1)2 × 3 − 1
𝑑𝑦 1
2(𝑥 − 1) �2 2(𝑥 − 1) �2
⇒ = × 1
𝑑𝑥 𝑥−1 2(𝑥 − 1) �2
𝑑𝑦 6(𝑥 − 1) − (3𝑥 − 2) 6𝑥 − 6 − 3𝑥 + 2
⇒ = 3� = 3
𝑑𝑥 2(𝑥 − 1) 2 2(𝑥 − 1) �2
𝑑𝑦 3𝑥 − 4
⇒ = 3
𝑑𝑥 2(𝑥 − 1) �2
BY TR. ALI DAMDEOK

Derivatives of ex and logex ≡ ln x.

𝑑𝑦
y = ex ⇒ 𝑑𝑥
= ex
𝑑𝑦 1
y = ln x ⇒ 𝑑𝑥
= 𝑥
𝑑𝑦 1 1
y = ln kx ⇒ y = ln k + ln x ⇒ 𝑑𝑥
= 0+ 𝑥
= 𝑥
𝑑𝑦 1 1
or y = ln kx ⇒ 𝑑𝑥
= 𝑘𝑥
×𝑘 = 𝑥
using the chain rule

dy 1 k
y = ln xk ⇒ y = k ln x ⇒ = k× =
dx x x

Example: Find the derivative of f (x) = x3 – 5ex at the point where x = 2.

Solution: f (x) = x3 – 5ex

⇒ f ′(x) = 3x2 – 5ex
⇒ f ′(2) = 12 – 5e2 = –24⋅9

Example: Differentiate the function f (x) = ln 3x – ln x5

Solution: f (x) = ln 3x – ln x5 = ln 3 + ln x – 5 ln x = ln 3 – 4 ln x
−4
⇒ f ′(x) =
x
or we can use the chain rule
1 1 −4
⇒ f ′(x) = × 3 − 5 × 5x 4 =
3x x x

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

ln 3x
Solution: f (x) = log 10 3x = using change of base formula
ln 10
ln 3 + ln x ln 3 ln x
= = +
ln 10 ln 10 ln 10
1
1
⇒ f ′(x) = 0 + x
= .
ln 10 x ln 10
BY TR. ALI DAMDEOK

2 dy
Example: y = e x . Find .
dx

Solution: y = eu, where u = x2

dy dy du
⇒ = ×
dx du dx
dy du
= eu ×
2
⇒ = eu × 2x = 2x e x .
dx dx

dy
Example: y = ln 7x3. Find .
dx

Solution: y = ln u, where u = 7x3

dy dy du
⇒ = ×
dx du dx
dy 1 du 1 3
⇒ = × = 3
× 21x 2 = .
dx u dx 7x x

𝒅𝒚 𝟏
= 𝒅𝒙
𝒅𝒙
𝒅𝒚
𝑑𝑦 𝑑𝑥 𝑑𝑦 1
Using the chain rule we can see that 𝑑𝑥
× 𝑑𝑦 = 1, ⇒ 𝑑𝑥
= 𝑑𝑥
𝑑𝑦

𝑑𝑦
Example: x = sin2 3y. Find 𝑑𝑥 .
𝑑𝑥
Solution: First find 𝑑𝑦
as this is easier.
𝑑𝑥
𝑑𝑦
= 2 sin 3y cos 3y × 3 = 6 sin 3y cos 3y

𝑑𝑦 1 1 1 1
⇒ 𝑑𝑥
= 𝑑𝑥 = = = 3
cosec 6y
𝑑𝑦 6 sin 3 y cos 3 y 3 sin 6 y

Derivative of ax
𝑑𝑦
y = ax ⇒ 𝑑𝑥
= 𝑎 𝑥 ln 𝑎
Proof:
y = ax
⇒ ln y = ln ax = x ln a
1 𝑑𝑦
⇒ 𝑦 𝑑𝑥
= ln 𝑎
𝑑𝑦
⇒ 𝑑𝑥
= y ln a = ax ln a

You should know this proof.

BY TR. ALI DAMDEOK

𝑑𝑦
Example: If y = 5x + 2, find 𝑑𝑥
.

Solution 1: y = 5x × 52 = 25 × 5x
𝑑𝑦
⇒ 𝑑𝑥
= 25 × 5x ln 5

Solution 2: y = 5x + 2 = 5u
𝑑𝑦 𝑑𝑢 𝑑𝑢
⇒ 𝑑𝑥
= 5u ln 5 × 𝑑𝑥 = 5x + 2 ln 5 using the chain rule and
𝑑𝑥
=1

2 𝑑𝑦
Example: 𝑦 = 7𝑥 , find 𝑑𝑥

𝑑𝑦 𝑑𝑢 2
Solution: y = 7u ⇒ 𝑑𝑥
= 7u ln 7 × 𝑑𝑥 = 7𝑥 ln 7 × 2x using the chain rule

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 sin2 x + cos2 x = 1
tan x sec2 x
sec x sec x tan x tan2 x + 1 = sec2 x
cot x – cosec2 x
cosec x – cosec x cot x 1 + cot2 x = cosec2 x

Chain rule – further examples

dy
Example: y = sin4 x. Find .
dx
Solution: y = sin4 x. Put u = sin x

⇒ y = u4
dy dy du
⇒ = ×
dx du dx
dy du
⇒ = 4u 3 × = 4 sin 3 x × cos x .
dx dx
BY TR. ALI DAMDEOK

dy
Example: y = e sin x . Find .
dx
Solution: y = eu, where u = sin x
dy dy du
⇒ = ×
dx du dx
dy du
⇒ = eu × = esin x × cos x = cos x × esin x .
dx dx

dy
Example: y = ln (sec x). Find .
dx
Solution: y = ln u, where u = sec x
dy dy du
⇒ = ×
dx du dx
dy 1 du 1
⇒ = × = × sec x tan x = tan x .
dx u dx sec x

Trigonometry and the product and quotient rules

Example: Differentiate y = x2 × cosec 3x.

dy dv du
Solution: = u + v
dx dx dx

y = x2 × cosec 3x

Put u = x2 and v = cosec 3x

𝑑𝑦
= x2 × (– 3cosec 3x cot 3x) + cosec 3x × 2x
𝑑𝑥

= – 3x2 cosec 3x cot 3x + 2x cosec 3x.

BY TR. ALI DAMDEOK

tan 2 x
Example: Differentiate y =
7x3
du dv
v − u
u dy dx dx
Solution: y = ⇒ = 2
v dx v

tan 2 x
y =
7x3

Put u = tan 2x and v = 7x3

dy 7 x 3 × 2 sec 2 2 x − tan 2 x × 21x 2

⇒ =
dx (7 x 3 ) 2
dy 14 x 3 sec 2 2 x − 21x 2 tan 2 x
⇒ =
dx 49 x 6
dy 2 x sec 2 2 x − 3 tan 2 x
⇒ =
dx 7x 4
BY TR. ALI DAMDEOK

6 Numerical methods
Locating the roots of f(x) = 0

A quick sketch of the graph of y = f (x) can give a rough

idea of the roots of f (x) = 0.
y=f(x)
If y = f (x) changes sign between x = a and x = b and if
f (x) is continuous in this region then a root of f (x) = 0
lies between x = a and x = b. x
a b

The iteration xn + 1 = g(xn)

Example:
(a) Show that a root, α, of the equation f (x) = x3 – 8x – 7 = 0 lies between 3 and 4.
(b) Show that the equation x3 – 8x – 7 = 0 can be re–arranged as x = 3
8x + 7 .

(c) Starting with x1 = 3, use the iteration xn + 1 = 3 8 x n + 7 to find the first four
iterations for x.
(d) Show that your value of x4 is correct to 3 S.F.

Solution:
(a) 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.
(b) x3 – 8x – 7 = 0 ⇒ x3 = 8x + 7 ⇒ x= 3
8x + 7 .
(c) x1 = 3
⇒ x2 = 3
8× 3 + 7 ⇒ = 3⋅14138065239
⇒ x3 = 3⋅17912997899
⇒ x4 = 3⋅18905898325
(d) x4 = 3⋅19 to 3 S.F.
f (3⋅185) = 3⋅1853 – 8 × 3⋅185 – 7 = – 0⋅17 …
f (3⋅195) = 3⋅1953 – 8 × 3⋅195 – 7 = + 0⋅05 …
f (x) changes sign and f (x) is continuous
⇒ there is a root in the interval [3⋅185, 3⋅195]
⇒ α = 3⋅19 to 3 S.F.
BY TR. ALI DAMDEOK

Conditions for convergence

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

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

(i) will converge if –1 < g ′(α) < 1

y y
y=x y=x

y=g(x)

y=g(x)

x x
X1 X2 X3 X4 X1 X3 X4 X2

(a) will converge without oscillating (b) will oscillate and converge
if 0 < g ′(α) < 1, if –1 < g ′(α) < 0,

(ii) will diverge if y

g ′(α) < –1 or g ′(α) > 1.

y=g(x)
y=x

x
X1 X 2 X3 X4
BY TR. ALI DAMDEOK

7 Appendix
Derivatives of sin x and cos x
𝐬𝐢𝐧 𝒉
𝐥𝐢𝐦𝐡→𝟎 � 𝒉
� = 𝟏 C
A
OAB is a sector of a circle with centre O and radius r.
1 2
The area of the triangle OAB, 2
r sin h, r r tan h
1 2
is less than the area of the sector OAB, 2
rh
1 2 1 2 h
⇒ 2
r sin h < 2
rh O
r B
sin ℎ
⇒ ℎ
<1 …….I
1 2
Also the area of the sector OAB, 2
rh r tan h
1 2
is less than the area of the triangle OBC, 2
r tan h,
1 2 1 2 sin ℎ
⇒ 2
rh < 2
r tan h ⇒ cos h < ℎ
……..II
sin ℎ
I and II ⇒ cos h < ℎ
<1
and as ℎ → 0, lim cos ℎ = 1
ℎ→0
sin ℎ sin ℎ
⇒ 1 < lim � �<1 ⇒ lim � � =1
ℎ→0 ℎ ℎ→0 ℎ

h must be in RADIANS, as the formula for the area of sector is only true if h is in radians.

Alternative formula for derivative

y
R y = f (x)
The gradient of the curve at P will be nearly P
equal to the gradient of the line QR. Q S
QM = f (x – h) and RN = f (x + h)
⇒ RS = f (x + h) – f (x – h)
QS = MN = 2h
𝑓 (𝑥 + ℎ) – 𝑓 (𝑥 – ℎ) M N x
⇒ gradient of QR = 2ℎ x–h x x+h
and as h → 0, the gradient of QR → f ′(x) =
gradient of the curve at P
limit 𝑓 (𝑥 + ℎ) – 𝑓 (𝑥 – ℎ)
⇒ f ′(x) = ℎ→0 2ℎ

limit 𝑓(𝑥+ℎ)−𝑓(𝑥)
In C2 we used the formula f ′(x) = ℎ→0 ℎ
BY TR. ALI DAMDEOK

Derivatives of sin x and cos x

limit 𝑓 (𝑥 + ℎ) – 𝑓 (𝑥 – ℎ)
f (x) = sin x and f ′(x) = ℎ→0 2ℎ

limit sin (𝑥 + ℎ) – sin (𝑥 – ℎ)

⇒ f ′(x) = ℎ→0 2ℎ
limit (sin 𝑥 cos ℎ +cos 𝑥 sin ℎ ) − (sin 𝑥 cos ℎ−cos 𝑥 sin ℎ )
= ℎ→0 2ℎ
limit 2cos 𝑥 sin ℎ
= ℎ→0 2ℎ
limit sin ℎ
= ℎ→0
cos 𝑥 ℎ
limit sin ℎ
but ℎ→0
� ℎ
� = 1
𝑑
⇒ f ′(x) = 𝑑𝑥
(sin 𝑥) = cos x

𝑑
Similarly, we can show that 𝑑𝑥
(cos 𝑥) = – sin x

x must be in RADIANS.
BY TR. ALI DAMDEOK

Index
algebraic fractions graphs
adding and subtracting, 3 combining transformations, 12
equations, 4 standard functions, 11
multiplying and dividing, 3 inverse functions
chain rule, 23 domain and range, 9
further examples, 27 finding inverse, 7
cosecant, 14 graphs, 7
cotangent, 14 iteration
curves conditions for convergence, 31
sketching, 13 iteration equation, 30
derivative justification of accuracy, 30
ax, 26 locating the roots, 30
ex, 25 logarithm
ln x, 25 graph of ln x, 21
differentiation natural logarithm, 21
𝑑𝑦𝑑𝑥 = 1 𝑑𝑥𝑑𝑦, 26 product rule, 23
proof of results for sin x and cos x, 32 further examples, 28
trigonometric functions, 27 quotient rule, 24
equations further examples, 28
graphical solutions, 30 R cos(x + α), 19
exponential secant, 14
eax + b = p, 22 trigonometrical identities, 16
graph of y = e(ax + b) + c, 21 trigonometry
graph of y = ex, 21 finding exact values, 17
functions, 5 graphs of sec, cosec and cot, 14
combining functions, 7 harder equations, 18
defining functions, 6 inverse functions, 15
domain, 5 proving identities, 17
modulus functions, 10
range, 5