Differential CalculusThe derivative, or derived function of f(x) denoted f`(x) is defined as

0

( ) ( )`( ) lim

h

f x h f xf x

h

( ) ( )f x h f x

h

x x + h

P

Q

x

y

( ) ( )PQ

f x h f xm

h

Leibniz Notation: `( ) limh o

y dyf x

x dx

Differentiation from first principles

Given that f(x) is differentiable, we can use the definition to prove that if 2( ) then `(5) 10f x x f

0

( ) ( )`( ) lim

h

f x h f xf x

h

2 2

0

( )limh

x h x

h

2 2 2

0

2limh

x xh h x

h

2

0

2limh

xh h

h

0

(2 )limh

h x h

h

0lim(2 )h

x h

As 0, `( ) 2h f x x

`(5) 10f

32. From first principles determine `( ) when ( ) 2f x f x x

0

( ) ( )`( ) lim

h

f x h f xf x

h

3 3

0

( ) 2 ( 2)limh

x h x

h

3 2 2 3 3

0

3 3 2 2limh

x x h xh h x

h

2 2 3

0

3 3limh

x h xh h

h

2 2

0

(3 3 )limh

h x xh h

h

2 2

0lim(3 3 )h

x xh h

2As 0, `( ) 3h f x x 2`( ) 3f x x

Further practice on page 29 Exercise 1AQuestions 1, 4, 5 and 7

TJ Exercise 1

But not just Yet……..

CAUTION Not all functions are differentiable.

y

x

y = tan(x). Here, tan(x) has ‘breaks’ in the graph where

the gradient is undefined

Although the graph is continuous, the derivative at zero is undefined as the left derivative is negative and the right derivative is positive.

x

y y x

For a function to be differentiable, it must be continuous.

Differentiation reminder:

Page 32 Exercise 3A Questions 1(a), (d), 2(a), (c), (d) 3(a), 4(a), 6(a)

TJ Exercise 2TJ Exercise 3

Further practice on page 29 Exercise 1AQuestions 1, 4, 5 and 7

TJ Exercise 1

The Product RuleIf ( ) ( ). ( ) , then:k x f x g x

`( ) `( ) ( ) `( ) ( )k x f x g x g x f x

Using Leibniz notation,

If ( ). ( ) , then:y f x g x

. ( ) . ( )dy df dg

g x f xdx dx dx

OR ` `dy

f g g fdx

21. Differentiate siny x x

2( )f x x ( ) sing x x

`( ) 2f x x `( ) cosg x x

` `dy

f g g fdx

22 sin cosx x x x

4 52. Differentiate ( 3) ( 3)y x x

4( ) ( 3)f x x 5( ) ( 3)g x x 3`( ) 4( 3)f x x 4`( ) 5( 3)g x x

` `dy

f g g fdx

3 5 4 44( 3) ( 3) 5( 3) ( 3)x x x x

3 4( 3) ( 3) (4( 3) 5( 3))x x x x

3 4( 3) ( 3) (4 12 5 15)x x x x 3 4( 3) ( 3) (9 3)x x x

3 43( 3) ( 3) (3 1)x x x

2 32. Differentiate ( 1) siny x x x

2( )f x x 3( ) ( 1)g x x

`( ) 2f x x 2`( ) 3( 1)g x x

`( . ) ( . )`dy

f g h g h fdx

`( . ) ( ` ` ) `. . . . ` . . `f g h f g h h g f g h f h g f g h

3 2 2 2 32 ( 1) sin 3 sin ( 1) ( 1) cosx x x x x x x x x

2( 1) (2( 1)sin 3 sin ( 1)cos )x x x x x x x x x

( ) sinh x x

`( ) cosh x x

Let .( . )y f g h

Page 35 exercise 4A Questions 1, 2(b) and 3Page 36 exercise 4B Questions 1(b), 3 and 4

TJ Exercise 4

The Quotient Rule( )

If ( ) , then:( )

f xk x

g x

2

`( ) ( ) `( ) ( )`( )

( ( ))

f x g x g x f xk x

g x

Using Leibniz notation,

( )If , then:

( )

f xy

g x

2

. ( ) . ( )

( ( ))

df dgg x f xdy dx dx

dx g x

OR

2

` `dy f g g f

dx g

3

1. Find sin

d x

dx x

3( )f x x ( ) sing x x

2`( ) 3f x x `( ) cosg x x

2

` `dy f g g f

dx g

2 3

2

3 sin cos

sin

x x x x

x

Page 37 exercise 5A Questions 1 to 4 and 7Page 38 Exercise 5B Questions 1 to 3

TJ Exercise 5

Sec, Cosec, cot and tan1

sec the secant of cos

x xx

1cos the cosecant of

sinec x x

x

1cot the cotangent of

tanx x

x

Unlike the sine and cosine functions, the graphs of sec and cosec functions have ‘breaks’ in them.

The functions are otherwise continuous but for certain values of x, are undefined.

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

y

x

1

1

2

2

3

3

4

4

5

5

6

6

1

1

2

2

3

3

4

4

5

5

6

6

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

– 6

– 6

secy x

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

y

x

1

1

2

2

3

3

4

4

5

5

6

6

1

1

2

2

3

3

4

4

5

5

6

6

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

– 6

– 6

cosec y x

1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6

y

x

1

1

2

2

3

3

4

4

5

5

6

6

1

1

2

2

3

3

4

4

5

5

6

6

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

– 6

– 6

coty x

In general, sec is undefined for 2

x x n

also cosec is undefined for x x n

In general, sec is undefined for 2

x x n

1sec

2 cos2

1

0 undefined

also cosec is undefined for x x n

1. Find the derivative of tan x

sintan

cos

d d xx

dx dx x

(quotient rule)

2

cos .cos ( sin ).sin

cos

x x x x

x

2

1

cos x

2sec x

Page 40 Exercise 7 questions 1, 2, 3(a), (c), (e), (g). 4(a)TJ Exercise 6 Questions 1 to 4

2. cosec d

xdx

( ) 1 ( ) sinf x g x x

`( ) 0 `( ) cosf x g x x

2

coscosec

sin

d xx

dx x

cos 1.

sin sin

x

x x

cosec cotx x

3. cot d

xdx

( ) cos ( ) sinf x x g x x

`( ) sin `( ) cosf x x g x x

2 2

2

sin coscot

sin

d x xx

dx x

2

1

sin x

2cosec x

cos

sin

x

x

Exponential and Logarithmic functions

If then x xdyy e e

dx

1If log then 0e

dyy x x

dx x

Proof 1.

0

( ) ( )`( ) lim

h

f x h f xf x

h

( ) xf x a

0lim

x h x

h

a a

h

0

1lim

x h

h

a a

h

0 0

1lim lim

h

x

h h

aa

h

0

1lim

h

x

h

aa

h

Let us examine this limit.

a 0.1 0.01 0.001 0.0001 0.00001 0.000001

1 0 0 0 0 0 0

1.5 0.413797 0.406288 0.405547 0.405473 0.405466 0.405465

2 0.717735 0.695555 0.693387 0.693171 0.69315 0.693147

2.5 0.959582 0.920502 0.916711 0.916333 0.916295 0.916291

2.7 1.044254 0.998201 0.993745 0.993301 0.993257 0.993252

2.718281 1.051709 1.005016 1.0005 1.00005 1.000005 1

3 1.161232 1.104669 1.099216 1.098673 1.098618 1.098613

4 1.486984 1.395948 1.387256 1.38639 1.386304 1.386295

h

When a approaches e, the limit approaches 1.

Proof 2. ( ) logef x x

logConsider e xe x Differentiating both sides with respect to x.

ln xde dx

dx dx

ln . ln1

xe d x

dx

Using the chain rule.

ln ln1x d x

edx

ln

1ln

x

dx

dx e

1ln

dx

dx x

31. Find xde

dx

3 33x xde e

dx

d2. Find

dxxxe

x x xdxe e xe

dx (1 )xe x

3. Find ln3d

xdx

1ln3 .3

3

dx

dx x

1

x

2

ln4. Find

d x

dx x

2

2 4

12 lnln x x xd x x

dx x x

4

2 lnx x x

x

3

1 2ln x

x

Higher derivativesFunction 1st derivative 2nd Derivative………..nth Derivative

( )f x '( )f x ''( )f x ( )nf x

dy

dx

2

2

d y

dx

n

n

d y

dxy

41. 3y x312

dyx

dx

22

236

d yx

dx

3

372

d yx

dx etc. etc. etc.

Page 43 Exercise 8A Questions 1(b), (d), 2(b), (c), (d), 3(a), (b),(c). 4(d), (e), 5(a), (c), (e), 6(b), (c),(e)

TJ Exercise 6

Page 46 Exercise 9A Qu. 1 to 6Review Chapter 2.1

Applications of Differential Calculus

Let displacement from an origin be a function of time.

( )x f t

Velocity is a rate of change of displacement.

dxv

dt

Acceleration is a rate of change of velocity.

dva

dt

2

2

d x

dt

Note: The use of units MUST be consistent.

A particle travels along the x axis such thatx(t) = 4t3 – 2t + 5,where x represents its displacement in metres from the origin ‘t’

seconds after observation began.

(a) How far from the origin is the particle at the start of observation?

(b) Calculate the velocity and acceleration of the particle after 3 seconds.

( ) When 0, (0) 5a t x

Hence the particle is 5m from the origin at the start of the observation.

2( ) 12 2dx

b v tdt

1(3) 12 9 2 106v ms

2

224

d xa t

dt 2(3) 24 3 72a ms

A body travels along a straight line such thatS = t3 – 6t2 + 9t + 1,where S represents its displacement in metres from the origin after

observation began.

(a) Find when (i) the velocity and (ii) the acceleration is zero.(b) When is the distance S increasing?(c) When is the velocity of the body decreasing?(d) Describe the motion of the particle during the first 4 seconds of

observation.

2( ) 3 12 9ds

a v t tdt

Let 0v

23 12 9 0t t 23( 4 3) 0t t

3( 1)( 3) 0t t 1 or 3t t

The velocity is zero when t = 1 or 3 seconds.

2

2( ) 6 12

d sii a t

dt Let 0a

6 12t 2t

The acceleration is zero when t = 2 seconds.

(b) S is increasing when 0ds

dt 3( 1)( 3) 0t t

y

x

1

1

2

2

3

3

4

4

5

5

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

10

10

– 1

– 1

– 2

– 2

– 3

– 3

– 4

– 4

– 5

– 5

This occurs when the graph of is above the x axis

ds

dt

1t 3t

Hence the distance S is increasing when t < 1 and when t > 3 seconds.

(c) V is decreasing when 2

20

d s

dt 6 12 0t 2t

Hence the velocity is decreasing when t < 2 seconds.

3 2

2

2

2

6 9 1

3 12 9

6 12

S t t t

dst t

dt

d st

dt

At t = 0, the particle is 1m from the origin with a velocity of 9ms-1 decelerating at a rate of 12ms-2.

At t = 1, the particle is 5m from the origin at rest decelerating at a rate of 6ms-2.At t = 2, the particle is 3m from the origin with a velocity of -3ms-1 with zero acceleration.At t = 3, the particle is 1m from the origin at rest accelerating at a rate of 3ms-2.

At t = 4, the particle is 5m from the origin with a velocity of 9ms-1 accelerating at a rate of 12ms-2.

Page 51 Exercise 1 Questions 1(a), (b), (d), (f), 2(a), (c), (e), 3, 4, 6, 7, 8, 10, 12.

TJ Exercise 7.

Extreme Values of a Function(Extrema)

Critical Points

A critical point of a function is any point (a, f(a)) where f `(a) = 0 or where f `(a) does not exist.

2 2 1

( ) 1 2

11 2 4

2

x x

f x x x

x x

Consider the function:y

x

1

1

2

2

3

3

4

4

– 1

– 1

– 2

– 2

1

1

2

2

3

3

4

4

5

5

– 1

– 1

– 2

– 2 Domain [ 2,4)

2 2 1

`( ) 1 1 2

12 4

2

x x

f x x

x

2 2 1

( ) 1 2

11 2 4

2

x x

f x x x

x x

2 2 1

`( ) 1 1 2

12 4

2

x x

f x x

x

Critical points are:y

x

1

1

2

2

3

3

4

4

– 1

– 1

– 2

– 2

1

1

2

2

3

3

4

4

5

5

– 1

– 1

– 2

– 2

A A (-2,4) f `(-2) does not exist. (Right differentiable at x = -2)

B

B (0,0) f `(0) = 0. (Turning point)

C C (1,1) f `(1) does not exist. Left derivative =2, right derivative = 1

D

D (2,2) f `(2) does not exist. Left derivative = 1, right derivative = 0.5

E

E is not a critical point as it is not in the domain of f.

Local extrema

Local extreme values occur either at the end points of the function, turning points or critical points within the interval of the domain.

Consider the function,

2

3

2

2 3 1

( ) 1 1

2 8 5 1 3

x x x

f x x x

x x x

y

x

1

1

2

2

3

3

– 1

– 1

– 2

– 2

– 3

– 3

1

1

2

2

3

3

4

4

5

5

– 1

– 1

– 2

– 2

Domain [ 3,3)

3 is the local maximum value

-1 is the local minimum value

If extrema occurs at end points then they are end point maximums or end point minimums.

In short,

(i) Local maximum / minimum turning points(ii) End point values(iii) Critical points

The Nature of Stationary Points.

If f `(a) = 0 then a table of values over a suitable interval centred at a provides evidence of the nature of the stationary point that must exist at a.

A simpler test does exist.

It is the second derivative test.

`( ) 0 and ``( ) 0 then minimum turning point.

`( ) 0 and ``( ) 0 then maximum turning point.

`( ) 0 and ``( ) 0 then draw a table of signs.

f x f x

f x f x

f x f x

If the second derivative test is easier to determine than making a table of signs then this provides an efficient technique to finding the nature of stationary points.

Page 56 Exercise 2 Questions 1, 3(a), (c), (e), (g), (i)Page 60 Exercise 3 Questions 1(a), (c), (e), 5(a) to (d)

TJ Exercise 8

Optimisation Problems

Optimisation problems appear in many guises – often in the context in which they are set can be somewhat misleading.

1. Read the question at least TWICE.2. Draw a sketch where appropriate. This should help you

introduce any variables you are likely to need. It may be that you come back to the diagram to add in an extra ‘x’ etc.

3. Try to translate any information in the question into a mathematical statement.

4. Identify the variables to be optimised and then express this variable as a function of one of the other variables.

5. Find the critical numbers for the function arrived at in step 4.6. Determine the local extrema and if necessary the global

extrema.

1. An open box with a rectangular base is to be constructed from a rectangular piece of card measuring 16cm by 10cm. A square is to be cut out from each corner and the sides folded up. Find the size of the cut out squares so that the resulting box has the largest possible volume.

xx

16 – 2x

10 – 2x

v lbh(16 2 )(10 2 )x x x

3 24 52 160x x x (max / min V occurs when = 0) dVdx

3 24 52 160V x x x

212 104 160dV

x xdx

0 for s.p.

212 104 160 0x x 24(3 26 40) 0x x

4(3 20)( 2) 0x x

22, 6 3x or

2

2 24 104d V

xdx

2

2When 2, 0d v

xdx

maximum volume at 2x

Hence the cut out squares should be 2cm in length.

Page 63 Exercise 4A. Note Question 8 can’t be done but try to prove me wrong!!

Page 64 Exercise 4B is VERY Difficult. (VERY)

TJ Exercise 9