The Product Rule
y  uv
 y  uv  vu
In words:
“Keep the first,
differentiate the second”
+
“Keep the second,
differentiate the first”
Examples:
1. Differentiate
y  x cos x
 y  uv  vu
y  x cos x
ux
u  1
 y  x( sin x)  cos x(1)
 y   x sin x  cos x
 y  cos x  x sin x
v  cos x
v   sin x
Examples:
2. Differentiate
y  x sin 3x
2
y  x 2 sin 3x
u  x2
u  2 x
v  sin 3x
v  3 cos 3x
 y  uv  vu
2

 y  x (3 cos 3x)  sin 3x(2 x)
 y  3x 2 cos 3x  2 x sin 3x
Now
watch
this.
Examples:
y  2 x  1 1  3x
2
3. Differentiate
y  2 x  1 1  3x
2
Try this using “words”
y  2 x  1 1  3x 
1
2
2
1
1
 12
y  2 x  1 1  3x  (3)  1  3x 2 22 x  1(2)
2
2
y  
32 x  1
2
21  3x 
1
2
 41  3x  2 x  1
1
2
y  
y  
32 x  1
2
21  3x 
 41  3x  2 x  1
1
2
1
2
32 x  1
2
21  3x 
1
2

81  3x  2 x  1
21  3x 
1
2
 34 x 2  4 x  1  8 6 x 2  5 x  1
y 
2 1  3x 
 60 x 2  52 x  11
y 
2 1  3x 
The Quotient Rule
u
y
v
vu   uv
 y 
2
v
In words:
“Keep the denominator,
–
differentiate the numerator”
“Keep the numerator,
differentiate the denominator”
Denominator 2
Examples:
1. Differentiate
x2 1
y 2
x 1
vu  uv
 y 
v2
2
2

x  1(2 x)  x  1(2 x)
 y 
2
2
x  1




 
x  1

2x x 1  x 1
2
2
2 x2
x
2

x
2

1
4x
2
1
2
2
2
u  x 1
u  2 x
2
v  x2 1
v  2 x
Examples:
y
2. Differentiate
x
2x
2
2x
y
x

1
1
2
 1
2
Try this using “words”
 12
1 2
x  1 (2)  2 x 2 x  1 (2 x)



y 
2
x  1
1
2
2


1
2



2 x 1  2x x 1
 y 
2
x 1
2

2
2
 12

Add a
denominator
here



1
2

2 x 1  2x x 1
 y 
2
x 1
2x  1
2

x
 y 
 y 
2

2

2

2x
 12
2
 1 x  1
x 2  1
1
2

x
2
2

1
1
2
1
2
2
2 x2 1  2x2

 

 x2 1
1
 y 
1 
2
2
2
x 1
x 1

2

 y 
x
2
2

1
3
2
Derivatives of New Functions
Definitions:
1
sec x 
cos x
1
cosec x 
sin x
1
cot x 
tan x
Reminder:
sin x
tan x 
cos x
continue
y  sec x
10
5
p/2
5
10
p
3p/2
2p
y  cosec x
10
5
p/2
5
10
p
3p/2
2p
y  cot x
10
5
p/2
5
10
p
3p/2
2p
Derivative of
d
2
tan x  sec x
dx
tan x
y  tan x
sin x
y
cos x
cos x(cos x)  sin x( sin x)
 y 
cos 2 x
Proof:
cos x  sin x
 y 
2
cos x
1
 y 
2
cos x
2
2

 y  sec x
2
Use the
Quotient
Rule now
Derivatives of
sec x , cosec x and cot x
d
sec x  sec x tan x
dx
d
cosec x  cosec x cot x
dx
d
cot x  cosec 2 x
dx
Prove these
and keep
with your
notes.
Use chain
rule or
quotient rule
Example: Given that
f ( x)  sin x tan x show that
2
f ( x)  sin 2 x tan x
2
f ( x)  sin x  tan x
p

f    2
4
 f (x)  sin 2 x sec 2 x  tan x 2(sin x)(cos x)
1
sin x  2(sin x)(cos x) 
2
 sin x



2
cos x cos x 
1

 tan x  2 sin 2 x
p

2 p 
2 p 
 f    tan    2 sin  
4
4
4
2
 1 

 1  2
2
 2
2
Exponential and Logarithmic Functions
Reminder:
f ( x)  e x and f ( x)  ln x
are inverse to each other.
They are perhaps the most important functions in the
applications of calculus in the real world.
Alternative notation: e x can be written as exp x
ln x can be written as log e x
Two very useful results:
e
ln x
x
ln( e x )  x
Also:
Learn
these!
Practise changing from exp to log and vice-versa.
y  ln x
y  ex
y
2
1
6
4
2
2
1
2
yx
4
x
6
Derivatives of the Exponential and Logarithmic Functions
(i)
ye
x
x

 y e
y  ln x
 y  log e x
y
xe
dx

 ey
dy
dy 1

 y
dx e
Proof of (ii)
dy 1


dx x
y  ln x
1
 y 
x
(ii)
Examples:
1. Differentiate
ye
sin x
ye
sin x
Use the
Chain Rule
sin x

y  e cos x
y  cos x esin x
2. Differentiate
yx e
4 x
4 x
x
3

y  x e  e 4x
3 x

y  x e ( x  4)
y  x 4e x
Use the Product
Rule
3. Differentiate


y  ln x 2  1
1
y  2
(2 x)
x 1
2x
y  2
x 1


4. Differentiate
x
y
ln x
1
ln x(1)  x 
x

y
2
ln x 
ln x  1
y
2
ln x 


y  ln x  1
2
Use the
Chain Rule
x
y
ln x
Use the
Quotient Rule
Note:
In general


•
d f ( x)
e
 f ( x)e f ( x )
dx
•
d
f ( x)
ln f ( x) 
dx
f ( x)
Useful for reverse
i.e. INTEGRATION
Higher Derivatives
Given that f is differentiable, if f  is also differentiable
then its derivative is denoted by f  .
The two notations are:
function
1st
derivative
2nd
derivative
……
nth
derivative
f
f
f 
……
f (n )
df
dx
2
d f
dx 2
n
……
d f
dx n
Example:
If y  xex, write down is first second and third derivatives
and hence make a conjecture about its nth derivative.
y  xex
dy
 xex  e x
dx
2
d y
x
x
x
x
x

xe

e

e

xe

2
e
dx 2
d3y
x
x
x
x
x

xe

e

2
e

xe

3
e
dx 3
dny
dx n
Conjecture:
 xex  ne x
dny
x
x
th

xe

ne
The n derivative is dx n
Rectilinear Motion
If displacement from the origin is a function of time I.e. x  f (t )
then
dx
v
dt
v - velocity
dv d 2 x
a
 2
dt dt
a - acceleration
Example:
A body is moving in a straight line, so that after t
seconds its displacement x metres from a fixed
point O, is given by x  9t  3t 2  t 3
(a)
Find the initial dislacement, velocity and acceleration of
the body.
(b)
Find the time at which the body is instantaneously at rest.
2
3
x

9
(
0
)

3
(
0
)

(
0
)
0m

t 0
dx
2
v
 9  6t  3t  v  9 m/s
dt
dv
2
a

6
m/s

a
 6  6t
dt
Extreme Values of a Function
Understand the following terms:
• Critical Points
• Local Extreme Values
 Local maximum
 Local minimum
• End Point Extreme Values
 End Point maximum
 End Point minimum
See, MIA Mathematics 1, Pages 54 – 55
y
The Nature of Stationary Points
Consider a curve y  f (x) and the
corresponding gradient function y  f (x)
A
y  f (x)
Consider maximum turning point A.
Notice, gradient of f (x ) for x in
the neighbourhood of A is negative.
i.e. f (x) is negative
Similarly, gradient of f (x ) for x in
the neighbourhood of B is positive.
i.e. f (x) is positive
B
y  f (x)
The Nature of Stationary Points
Rule for Stationary Points
• f ( x)  0 and f (x)  0
 minimum turning point
• f ( x)  0 and f (x)  0
 maximum turning point
• f ( x)  0 and f (x)  0  possibly a point of inflexion
but must check using a table of signs
Now what does f ( x)  x 4 look like?
Example:
Consider
f ( x)  x 4
y
f ( x)  4 x 3
At S.P.
f ( x)  0
 4 x3  0
x0
f ( x)  12 x 2
 f (0)  0
Notice no Point of Inflexion.
Global Extreme Values
Understand the following terms:
• Global Extreme Values
 Global maximum
 Global minimum
See, MIA Mathematics 1, Pages 58 – 59
Example:
Find the coordinates and nature of the stationary point on the curve
f ( x)  e x  4 x.
f ( x)  e x  4 x
What does this
curve look like?
f ( x)  e x  4
At S.P.
f ( x)  0
 ex  4  0
 ex  4
 x  ln 4
 y  eln 4  4 ln 4
 y  4  4 ln 4
f ( x)  e x
f (ln 4)  eln 4  4  0
 (ln 4,4  4 ln 4)
is a
Minimum Turning Point
y
y  ex  4x
y
x
Optimisation Problems
A sector of a circle with radius r cm has an area of 16 cm2.
(a) Show that the perimeter P cm of the sector is given by
 16 
P(r )  2 r  .
r 

(b) Find the minimum value of P.
(a)
arc length area sector

2pr
pr 2
l
16

 2
2pr pr
16( 2πr )
32
l 
l 
2
πr r
r
l
r
r
P  2r  l
32
 P  2r 
r
 16 
 P  2 r  
r 

now
(b) P(r )  2r  32r 1
P(r )  2  32r 2
32
 2 2
r
At SP P(r )  0
32
 2 2  0
r
32
2 2
r
2r 2  32
r 2  16
r4
P(r )  64r 3
64
 3
r
64
 P(4)  3
4
1 0
 r = 4 gives a minimum
stationary value of
32
P(4)  2(4) 
4
 16 cm