The Product Rule

uvvuy

uvy

In words:

“Keep the first, differentiate the second” +

“Keep the second, differentiate the first”

Examples: 1. Differentiate xxy cos

xxy cosxu xv cos1u xv sin

uvvuy

)1(cos)sin( xxxy

xxxy cossin

xxxy sincos

Examples: 2. Differentiate xxy 3sin2

xxy 3sin22xu xv 3sinxu 2 xv 3cos3

uvvuy

)2(3sin)3cos3(2 xxxxy

xxxxy 3sin23cos3 2 Now watch this.

Examples: 3. Differentiate xxy 3112 2

Try this using “words” xxy 3112 2

21

3112 2 xxy

)2(12231)3(312

112 2

1212 xxxxy

12314312

12321

21

2

xxx

xy

2

121

312

12318

312

123 2

x

xx

x

xy

x

xxxxy

312

15681443 22

x

xxy

312

115260 2

12314312

12321

21

2

xxx

xy

The Quotient Rule

2v

vuuvy

v

uy

In words:

“Keep the denominator, differentiate the numerator”

“Keep the numerator, differentiate the denominator”

–

Denominator 2

Examples: 1. Differentiate1

12

2

x

xy

12 xu

12 xv

xu 2

xv 2

2v

vuuvy

22

22

1

)2(1)2(1

x

xxxxy

22

22

1

112

x

xxx

22 1

22

x

x

22 1

4

x

x

Examples: 2. Differentiate 1

22

x

xy

1

)2(121

2)2(1

2

22 21

21

x

xxxxy

Try this using “words” 21

1

22

x

xy

1

12122

222 21

21

x

xxxy

1

12122

222 21

21

x

xxxy

11

2

1

12

2

2

2

2

2

21

21

x

x

x

x

x

y

11

212 2

2

22

21

x

x

xxy

1

1

1

222 2

1

xxy

23

1

22

x

y

Add a denominator here

Derivatives of New Functions

Definitions:x

xcos

1sec

xx

sin

1cosec

xx

tan

1cot

Reminder:

x

xx

cos

sintan

continue

xy sec

xy cosec

xy cot

Derivative of xtan xxdx

d 2sectan xy tan

Use the Quotient Rule now

Proof:

x

xy

cos

sin

x

xxxxy

2cos

)sin(sin)(coscos

x

xxy

2

22

cos

sincos

xy

2cos

1

xy 2sec

Derivatives of xsec

xxxdx

dtansecsec

x cosec, xcot and

xxxdx

dcot cosec cosec

xxdx

d coseccot 2

Prove these and keep with your

notes.Use chain

rule or quotient rule

Example: Given that xxxf tansin)( 2 show that 24

f

xxxf tansin)( 2

xx 22 secsin )(xf ))(cos(sin2 tan xxx

xxxf tansin)( 2

xx

22

cos

1sin

1

))(cos(sin2

cos

sin xx

x

x

x2tan x2sin2

4f

4tan2

4

sin2 2

12

2

12

2

Exponential and Logarithmic Functions

Reminder: xexf )( xxf ln)( and are inverse to each other.

They are perhaps the most important functions in the applications of calculus in the real world.

Alternative notation: xex exp as written becan

xx elog as written becan ln

Two very useful results: xe x ln

xex )ln(

Also: Practise changing from exp to log and vice-versa.

Learn these!

xey xy ln

xy

x

y

Derivatives of the Exponential and Logarithmic Functionsxey xy lnxey

xy

1

Proof of (ii)

(i) (ii)

xy lnxy elog

yex ye

dy

dx

yedx

dy 1

xdx

dy 1

Examples: 1. Differentiatexey sin

xey sin

xey x cossinxexy sincos

Use the Chain Rule

2. Differentiatexexy 4

xexy 4 Use the Product Rule

34 4xeexy xx

)4(3 xexy x

3. Differentiate 1ln 2 xy 1ln 2 xy

)2(1

12

xx

y

1

22

x

xy

Use the Chain Rule

4. Differentiatex

xy

ln

x

xy

ln

Use the Quotient Rule

2ln

1)1(ln

x

xxx

y

2ln

1ln

x

xy

Note: In general

)()( )( xfxf exfedx

d •

)(

)()(ln

xf

xfxf

dx

d •

Useful for reverse i.e. INTEGRATION

Higher Derivatives

Given that f is differentiable, if is also differentiable then its derivative is denoted by .

f f

The two notations are:

function1st

derivative2nd

derivative……

nth derivative

f ……

……

f f )(nf

dx

df2

2

dx

fdn

n

dx

fd

Example:

If , write down is first second and third derivatives and hence make a conjecture about its nth derivative.

xxey

xxey xx exe

dx

dy

xxxxx exeeexedx

yd2

2

2

xxxxx exeeexedx

yd32

3

3

n

n

dx

ydxx nexe

Conjecture: The nth derivative is n

n

dx

yd xx nexe

Rectilinear Motion

If displacement from the origin is a function of time I.e.

then

dt

dxv

)(tfx

v - velocity

2

2

dt

xd

dt

dva 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 3239 tttx

(a) Find the initial dislacement, velocity and acceleration of the body.

(b) Find the time at which the body is instantaneously at rest.

0t m 0)0()0(3)0(9 32 x

2369 ttdt

dxv m/s 9v

tdt

dva 66 2m/s 6a

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

x

y

)(xfy

)(xfy

A

B

Consider maximum turning point A.

Notice, gradient of for x in the neighbourhood of A is negative.

)(xf

The Nature of Stationary Points

i.e. is negative)(xf

Similarly, gradient of for x in the neighbourhood of B is positive.

)(xf

i.e. is positive)(xf

Consider a curve and the corresponding gradient function

)(xfy )(xfy

The Nature of Stationary Points

Rule for Stationary Points

• and minimum turning point0)( xf 0)(xf

• and maximum turning point0)( xf 0)(xf

• and possibly a point of inflexion but must check using a table of signs

0)( xf 0)(xf

Example:

Consider 4)( xxf 34)( xxf

At S.P. 0)( xf

04 3 x0 x

212)( xxf

0)0( f

Now what does look like? 4)( xxf

x

y

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

.4)( xexf x

xexf x 4)(

4)( xexf

At S.P. 0)( xf

04 xe

4 xe

4ln x

4ln44ln ey4ln44 y

xexf )(

04)4(ln 4ln ef

)4ln44,4(ln is a

Minimum Turning Point

What does this curve look like?

xey x 4

x

y

x

y

Optimisation Problems

A sector of a circle with radius r cm has an area of 16 cm2.

Show that the perimeter P cm of the sector is given by

.16

2)(

rrrP

(a)

(b) Find the minimum value of P.

(a)2

sector area

2

length arc

rr

2

16

2 rr

l

2

)216

πr

πr(l

r rl

32

l

r

r

lrP 2now

rrP

322

rrP

162

1322)( rrrP(b)

At SP 0)( rP

2322)( rrP

2

322

r

032

22

r

2

322

r

322 2 r

162 r

4r

364)( rrP

3

64

r

34

64)4( P

01 r = 4 gives a minimum

stationary value of

4

32)4(2)4( P

cm 16