Tangents and Normals
• The equation of a tangent and normal takes
the form of a straight line i.e.
y  mx  c
• To find the equation you need to find a
value for x, y and m and then substitute to
find the value of c.

Find the equation of the tangent to the curve
y = x2 – 3x + 18 at the point (1, 16).
Equation has form
y  mx  c
x=1
y = 16
dy
m
 2x  3  2 1 3  1
dx

Substituting

y  mx  c  16 11 c  c  17

The equation of the tangent is

y 1x  17
To find the equation of the normal, use the
perpendicular gradient i.e.
mnormal 

1
mtangent
Topic 9

find equa tions of tang ents
Reference
Page
Exercise
Delt a
163
13.1
Sidebotham
249
13.4
NuLake
43
Theta (red)
82, 83
Senior Mathematics 165
1.12
8.1, 8.2
4A, 4B
Worksheet 2
Rules of Differentiation
Differentiating Trig Functions
A list of the trigonometry
differentials is given in your
formula sheet.
f (x)  sin x  f (x)  cos x


f (x)  cos x  f (x) sin x
2

f (x)  tan x  f (x)  sec x
f (x)  sec x  f (x)  sec x tan x


f (x)  cosecx  f (x) cosecx cot x
f (x)  cot x  f (x) cosec x

2
Topic 8

dif ferentiate trigono me tric func tions
(including reciprocal func tions )
Reference
Page
Delta
110
Sidebotham
113
Exercise
8.9, 8.10 , 8.11,
8.12
6.1
NuLake
33
1.9

Exponential
f (x)  e  f (x)  e
x
x
Topic 6

Reference
Sidebotham
Page
101,
102
117
Exercise
8.1, 8.2 ,
8.3
6.2
NuLake
25
1.7
Delta
dif ferentiate exponen tial fun ctions (base e
only)
Chain Rule applies when we have a function
of a function e.g.
Take two functions:
yu
5
and u  3x  4
Now combine them into one function by eliminating u
Function 1

y  (3x  4)
5
Function 2
Chain Rule applies when we have a function
of a function e.g.
Take two functions:
Note:
y  u and u  3x  4
dy
du
4
 5u and
3
du
dx
5
y  (3x  4)
Function 1
5
Function 2
dy dy du


dx du dx
Think of it like this:
Differentiate the first function as a whole

and then differentiate what is inside of it.
dy dy du
4
4


 5u  3  5(3x  4)  3
dx du dx
Think of it like this: Differentiate the first
function as a whole...
Differentiate function 1
dy
4
 5(3x  4)  3
dx
dy dy du
4
4


 5u  3  5(3x  4)  3
dx du dx
Think of it like this: Differentiate the first
function as a whole and then differentiate
what is inside of it.
dy
4
 5(3x  4)  3
dx
Then function 2
Example:
Function 1
Function 2
y  sin( 2x  4)
dy
 cos(2x  4)  2
dx
Differential of sin
Differential of 2x + 4
Topic 4

Reference
Page
Delta
78
dif ferentiate using the chain rule
Sidebotham 63, 64
NuLake
22
Exercise
6.7
3.4, 3.5
1.6
Differentiating logs
Note: You can only differentiate
natural log so any other base
needs to be converted first.
y  log e x  ln x
dy 1

dx x
Topic 7

Reference
Exercise
Sidebotham
Page
106,
107
121
NuLake
28
1.8
Delta
dif ferentiate loga rit hmic func tions (base e
only)
8.6, 8.7
6.3

Examples
y e
y  sin 5x
3x
y  ln( 3x 2  2x  4)

dy
 3e 3x
dx
dy
 5cos5x
dx
dy
1
 2
 6x  2
dx 3x  2x  4
Hard Example
1
y  ln(sec( 3x  4 x))
4
3
4
3
2
dy
1
3
3
2
 4(ln(sec( 3x 3  4 x)) 3 

sec(3x

4
x)tan(
3x

4
x)

9x
 4

3
dx
sec(3x  4 x)
1
2
3
4
Product Rule
( f .g) f .g g. f 


dy
dv
du
If y  uv then
u v
dx
dx
dx
Product Rule
f (x)  e sin 2x
5x
f
g
Product Rule
f (x)  e sin 2x
5x
( f .g) f .g g. f 
f (x)  e (2cos2x)  5e sin 2x
5x
5x
Product Rule
f (x)  e sin 2x
5x
( f .g) f .g g. f 
f (x)  e (2cos2x)  5e sin 2x
5x
5x
Topic 5a

Reference
2
3
dif ferentiate products, such as (3x Π7) (4x + 8)
or x2sinx
Page Exercise
Delta
79
6.8
Sidebotham
126
6.4
NuLake
14
1.4
Quotient Rule
 f  g. f  f .g
  
2
g
g 
du
dv
v
u
u
dy
dx
dx
If
y

then

2

v
dx
v
Quotient Rule
f
ln 3x
f (x) 
cos 4x
g

Quotient Rule
ln 3x
f (x) 
cos 4x
 f  g. f  f .g
  
2
g
g
 
1
cos 4 x  ln 3x4 sin 4 x

x
f (x) 
2
cos 4 x 
Quotient Rule
ln 3x
f (x) 
cos 4x
 f  g. f  f .g
  
2
g
g 
1
cos 4 x  ln 3x4 sin 4 x
x
f (x) 
2
cos 4 x 

Quotient Rule
ln 3x
f (x) 
cos 4x
 f  g. f  f .g
  
2
g
g 
1
cos 4 x  ln 3x4 sin 4 x
x
f (x) 
2
cos 4 x 

Quotient Rule
ln 3x
f (x) 
cos 4x
 f  g. f  f .g
  
2
g
g 
1
cos 4 x  ln 3x4 sin 4 x
x
f (x) 
2
cos 4 x 

Quotient Rule
ln 3x
f (x) 
cos 4x
 f  g. f  f .g
  
2
g
g 
1
cos 4 x  ln 3x4 sin 4 x
x
f (x) 
2
cos 4 x 

Topic 5b

dif ferentiate quotients, such as
Reference
x
1 x2
Page Exercise
Delt a
80
6.9
Sidebotham
126
6.4
NuLake
18
1.5
When a curve is written in the form y  f (x)
it is said to be defined explicitly.
When a curve is written in the form

it is said to be defined implicitly.
Example:

x 2  3y  y2  0
f (x, y)  0
Implicit differentiation
x  3y  y  0
2
2
Differentiating with respect to x

d x 2  d 3y  d y 2 


0
dx
dx
dx
d
d dy
Note :
 .
dx dy dx

dy
dy
2x  3  2y
0
dx
dx
Implicit differentiation
x  3y  y  0
2
2
Differentiating with respect to x

d x 2  d 3y  d y 2 


0
dx
dx
dx
d
d dy
Note :
 .
dx dy dx

dy
dy
2x  3  2y
0
dx
dx
Implicit differentiation
x  3y  y  0
2
2
Differentiating with respect to x

d x 2  d 3y  d y 2 


0
dx
dx
dx
d
d dy
Note :
 .
dx dy dx

dy
dy
2x  3  2y
0
dx
dx
Implicit differentiation
x  y  4 x  5y  8  0
2


2
dy
dy
2x  2y  4  5  0
dx
dx
dy
2y  5  2x  4
dx
Implicit differentiation
x  y  4 x  5y  8  0
2


2
dy
dy
2x  2y  4  5  0
dx
dx
dy
2y  5  2x  4
dx
dy 2x  4

dx 2y  5
Parametric Equations
dy dy dt


dx dt dx

Parametric Equations
dy dy dt dy dx




dx dt dx dt dt


Parametric Equations
dy
dy dy dt dy dx dt y 






dx dt dx dt dt dx x 
dt
Parametric Equations
x  3t  4, y  3t  2
2
dy
3
dt
dy 3 1
 
dx 6t 2t


dx
 6t
dt

Example 2
x  3cos2t, y  4 cos2t
dy 8sin 2t 4 tan2t


dx 6cos2t
3


Second derivative
dy 
d 
2
d y
dx  dt


2
dx
dt
dx

Second derivative
dy 
dy 
d 
d 
2
d y
dx  dt
dx  dx




2
dx
dt
dx
dt
dt

Second derivative
dy 
d 
dy 
dy 
dx 
d
d




d2y
dx  dt
dx  dx
dt





2
dx
dx
dt
dx
dt
dt
dt
Example 2
x  3cos2t, y  4 cos2t


dy 
d 
dx 
dt

dx
dt
dy 8sin 2t 4 tan2t


dx 6cos2t
3
8 2
sec 2t
2
d y
4
3


2
dx
6cos2t
9cos 3 2t
Topic 10

perform implicit differentiation

parametric dif ferentiation
Reference
Page
Delta
164, 225, 228
Sidebotham
129, 235
NuLake
61, 66
Exercise
13.2, 17.1 , 17.2
6.5, 13.1
1.16, 1.17