Calculus Revision A2 Level Table of derivatives

NB x

xcos

1sec

xx

2

2

cos

1sec

Chain rule otherwise known as the ‘function of a function’ or ‘composite’ rule.

Example Obtain the derivative of

(i) 412 x (ii) xe 2 (iii) 14ln x (iv) x3sin (v) x2cos

(i) 412 xy Let 12 xu

4uy 2dx

du

3

3

124

4

x

udu

dy

3

3

128

2124

x

x

dx

du

du

dy

dx

dy

dx

du

du

dy

dx

dy

* sec tan

sin cos

cos sin

1 ln

AS From

2

1

xx

xx

xx

xx

ee

anxax

dx

dyy

xx

nn

There are those of you who will do these without introducing u.

2124

12

3

4

xdx

dy

xy

differentiate with derivative

respect to bracket of bracket

The remainder will be done in this way, but the substitution will also be given for those who

would prefer to use it.

(ii) xx eey 22 {essential to insert brackets!}

xx eedx

dy 22 22 xu 2

(iii) 14ln xy

14

44

14

1

xxdx

dy 14 xu

(iv) xxy 3sin3sin

xxdx

dy3cos333cos xu 3

(v) 22 coscos xxy

xxxxdx

dysincos2sincos2 xu cos

Product rule For uvy

Example Find the derivatives of (i) xxe (ii) xx ln2

(iii) xe x cos (iv) 53 12 xx

{A formal use of the formula isn’t necessary, unless you insist!}

dx

dvu

dx

duv

dx

dy

dx

duv

dx

dvu

dx

dy

uvy

(leave 1st differentiate 2nd) + (leave 2nd differentiate 1st)

(i) xxey

x

xx

ex

exedx

dy

1

.1

(ii) xxy ln2

xxx

xxx

xdx

dy

ln2

ln212

(iii) xey x cos

xexe

xexedx

dy

xx

xx

cossin

cossin

(iv) 53 12 xxy

31612

1231012

1231210.

42

42

5243

xxx

xxxx

xxxxdx

dy

Quotient rule For v

uy

1

,

dx

du

xu

x

x

edx

dv

ev

xdx

du

xu

2

,2

xdx

dv

xv

1

ln

x

x

edx

du

eu

,

x

dx

dv

xv

sin

cos

2

3

3

,

xdx

du

xu

4

4

5

1210

2125

12

x

xdx

dv

xv

2v

dx

dvu

dx

duv

dx

dy

Example Differentiate with respect to x

(i) 13

2

x

x (ii)

2

ln

x

x (iii)

1

1

x

x

e

e

{Again, use a formal approach if you wish, as in (i) below}

(i) 13

2

x

xy

2

2

2

13

2

13

32213

x

x

xx

v

dx

dvu

dx

duv

dx

dy

(ii) 2

ln

x

xy {without the u and v}

3

4

4

4

2

ln21

ln21

ln2

2ln1

x

x

x

xx

x

xxx

x

xxx

x

dx

dy

(iii) 1

1

x

x

e

ey

21

11

x

xxxx

e

eeee

dx

dy

2

2

22

1

2

1

x

x

x

xxxx

e

e

e

eeee

2

2

dx

du

xu

, 3

13

dx

dv

xv

Table of integrals

An important technique will be used with standard integrals from the given table.

If CxFdxxf )()( then CbaxFa

dxbaxf )(1

)(

Example Cx

dxx 4

43

CxCx

dxx

4

43

128

1

4

12

2

112

Example Cxdxx

ln1

Example cedxe xx

32

2132

Example Cxdxx

14ln

4

1

14

1

Example Cxxdx cossin

CxCxxdx 21

21

212

1 cos2cos1

sin

Example Cd sincos

xx

xx

ee

xx

x

nn

xx

cydxy

xx

nn

sin cos

cos sin

ln 1

fromP1 1 1

1

1

Example Cd

4

34

3 sin3

1cos

Example Find the integral of 16

1

x

Cx

Cx

xdxx

163

1

16.

6

1

1616

1

21

2

1

2

1

Example Evaluate

1

0 2

12

1dx

x

1

0

1

0

1

1

0

21

0 2

122

1

1

12.

2

1

1212

1

x

x

dxxdxx

3

1

2

1

6

1

Log integrals (1)

Example Cxdxx

12ln

12

2

Example Cedxe

e x

x

x

1ln

1

since

Cx

dxx21

2

1

2

1

since

Cx

dxx1

12

Cxfdxxf

xf

)(ln)(

)(

Example

dx

x

xdx

x

xdx

x

x

1

2

2

1

1

2

2

1

1 222

Cx 1ln2

1 2

Example Cxdxx

xxdx sinln

sin

coscot

Log integrals (2)

You will notice that log integrals in the data booklet include modulus signs. These are

not generally necessary, but …..

Example 231

0

1

0

ln2ln3ln2ln2

1

xdxx

Example 2ln1ln2ln2

1 1

0

1

0

xdx

x

Introduce here, and

2ln2ln1ln2

11

0

dx

x

{it has been explained why this procedure is acceptable}

Implicit Differentiation

So far we have met curves with Cartesian equation in the form )(xfy i.e. y is

expressed explicitly in terms of x.

Some curves can’t conveniently be expressed explicitly in this way when the relationship

between x and y is contained implicitly in an equation.

e.g. a circle 053222 yxyx

Using ‘dx

d’ as an operator

(1) )()( xfxfdx

d

Retrospectively

here.

dx

dyyfyf

dx

d)()( …… chain rule

Example 23 3xxdx

d ,

dx

dyyy

dx

d 23 3

(2) Applying to products

Example 232323 .. ydx

dxyx

dx

dyx

dx

d

dx

dyyxyx

dx

dyyxyx

322

322

23

23

(3) Tangents and normals

Example Find the equation of the tangent to xyyx 32 22 at the point (1, 2)

,42 2 xxdx

d ,22

dx

dyyy

dx

d ,333

dx

dyxyxy

dx

d

dx

dyxy

dx

dyyx 3324

At (1, 2) dx

dy

dx

dy3644

2dx

dy

Equation 122 xy

xy 2

Integration techniques

Integration as the reverse of differentiation

Example Find 52 1xdx

d. Hence evaluate

1

0

42 1 dxxx

42

4252

110

2151

xx

xxxdx

d

5

2

1

0

1

0

421110 xdxxx

1321101

0

42 dxxx

10

311

1

0

42 dxxx

Simpson’s rule for approximate integration

The definite integral b

a

dxxf

is given by the area

bounded by xfy , ax , bx and 0y

Divide the area into an even number of strips 2n, each of width h. There will be 12 n ordinates

12321 ... , , , nyyyy

The value of the integral is given approximately by

where n

abh

2

Example Evaluate 25

9 dxx using eight strips

Strip width = 2

x0 = 9 y0 = 3

x1 = 11 y1 = 3.316625

x2 = 13 y2 = 3.605551

x3 = 15 y3 = 3.872983

x4 = 17 y4 = 4.123106

x5 = 19 y5 = 4.358899

x6 = 21 y6 = 4.582576

x7 = 23 y7 = 4.795832

x8 = 25 y8 = 5

xfy

x

y

b a xfy

x

y

b a

1231224220

..4..2

3

1 nnn

b

ayyyyyyyyhdxxf

795.4358.487.331.34582.4123.4605.325323

1

25

9 dxx

333.65

Parametrics

1. Equations of the form )(xfy or 0),( yxf are called Cartesian equations.

Example xey , 05222 xyx

2. Equations of the form )( ),( tgytfx where t is a third variable are called

parametric equations; t is the parameter

They define a curve which has points with coordinates of the form )(),( tgtf . As t

varies the curve is defined.

Parametric differentiation Where t is a parameter

Example Given that 12 tx and 12 ty obtain an expression for

(i) dx

dy (ii)

2

2

dx

yd in terms of t.

(i) tdt

dx2 2

dt

dy

ttdx

dy 1

2

12

(ii)

32

2

2

2

1

2

11

ttt

dx

dt

dt

dx

dyd

dx

dx

dyd

dx

yd

Tangents and normal to a curve at a specific point Find the gradient through differentiation, and use xxmyy

Example Find the equations of the tangent and normal to the graph of xxy ln at

the point where 2 ex

dt

dxdt

dy

dx

dy

xxy ln x

xxdx

dy 1.ln

1ln x

When 2 ex , 222 2)ln( eeey and 112 dx

dy

Tangent 22 12 exey

02 exy

Normal Grad = 1 22 12 exey

23 exy

Integration techniques

Integration by substitution –the substitution will be given

Example Find (i) xdx3sin by substituting xu cos

(ii)

dxx

x

1

0 22 1 by substituting 12 xu

(i) xdxxxdx sinsinsin 23

dux2cos1 xdxdu sin

AxxAuu

duu

duu

coscos3

1

1

3

31

3

2

2

(ii) In a definite integral the limits are changed according to the substitution

4

1

1

1

1

1

21

21

2

1

21

2

1

1

21

2

1

2

21

2

1 2

211

0 22

u

u

duu

duu

dxx

x

xdxdu

xu

sin

cos

xdxdu

xu

2

12

xdxdu 21

2 ,1

1 ,0

ux

ux

Integration by parts

This method is used for some products such as for example:

,sin xx ,2cos2 xx xxe , xx ln3 , xex 212

It can also be used to integrate, for example:

,ln x x1sin , x1tan

Formula dxdx

duvuvdx

dx

dvu

Example xdxxsin

Cxxx

xdxxx

dxxxxxdxx

sincos

coscos

1.coscossin

In words, for the evaluation of dxsecondfirst

Example xdxx ln3

Whether you use the formula or “words” the order here has to be changed

dxx

xxxdxxx

1

44lnln

443

Ax

xx

Ax

xx

dxxxx

16ln

4

44

1ln

44

1ln

444

443

4

1

dx

du

xu

,

xv

xdx

dv

cos

sin

first of

derivative Times

foundalready

Integralof Integral minus

second of

integral timesFirst

Example xdx1tan Here, introduce 1 as the “second”

dxx

xxxdxx

2

11

1

1tan1.tan

Axxx

dxx

xxx

dxx

xxx

2

211

2211

2

1

1lntan

1

2tan

1tan

(Use same method for xdxln , xdx1sin , etc) Do them!

For a definite integral

Example

1

0

1

0

1

0

1. dxeexdxxe xxx

101

1

0

1

0

0

x

xx

ee

dxeex

1

11

11

21

1

1

e

ee

ee

Using partial fractions

Example Evaluate (i)

dx

xx

x

21

12

(ii)

2

1 2 12

2dx

xx

x

(i) 2

1

1

1

2121

12 33

33

xxxxxx

x

Cxxdxxx

dxxx

x

2ln1ln2

1

1

1

21

12

(ii) 12

62

12

2

12

22

412

3

22

xxx

A

xxx

A

xx

x

26122122 xxxAxx

2x 620 A 3A

2

2 12

623

12

2dx

xxxdx

xx

x

Cxx

x

xx

x

dxx

dxxdxx

12ln32

ln3

12ln31

2ln3

12

232

13

1

2

2

1

2

1 2

12ln32

ln312

2

xx

xdxxx

x

6

5ln31

32

5ln31

3ln321ln35ln312ln3

Integration of x2sin and x2cos etc

Here we need the rearranged double angle formulae for x2cos

i.e. xx 2cos1sin212 , xx 2cos1cos

212

Example dxxdxxxdx 2cos12cos1sin21

212

CxxCx

x

2sin

2

2sin41

21

21

Using the tables of standard derivatives and integrals in the formula booklet

Relevant to Core 4 are derivatives of inverse trig functions, and of xsec , xcot and xcosec

(Page 5)

The results can be reversed.

Example Cxdxx

1

2sin

1

1

Example Cxxdx cotcosec 2

Example Cxdxx

1

2tan

1

1

On page 6 the integrals of ,tan x ,cot x xcosec and xsec are relevant to Core 4

Volumes of revolution

When the shaded region is rotated through 2c about

0x the “volume of revolution” will be given by

….about 0y

b

a

dyxV

2

Example Find the volume of revolution of the graph of xy cos from 0x to 2

x ,

about 0x through 2c

4

2

0sin0

2

sin

22

2

2sin

2

2cos1

cos

2

0

0 21

0

2

2

2

2

xx

dxx

xdxV

a b x

y

0

a

b

x

y

0

2 x

y

0

b

a

dxyV

2

Differential equations

(1) Equations of the form )(xfdx

dy or )(

2

2

xgdx

yd integrate ‘at once’

Example xedx

dy Cey x *

Example xdx

ydsin

2

2

Axdx

dy cos

BAxxy sin *

* These are the general solutions of the differential equations where A, B, C are

arbitrary constants.

Particular solutions to differential equations can be found if boundary conditions are

given

Example Solve the equation 12 tdt

dx given that when 0t , 2x

12 tdt

dx Attx 2 {general solution}

0t , 2x A 02 2A

22 ttx

(2) Equations which reduce to dyygdxxf )()( are called variables separable.

Integrate both sides but include just one arbitrary constant.

(3)

Example xydx

dy2 xdxdy

y2

1

Cxy 2ln

2

222

x

xcxCx

Aey

Aeeeey

(4) An important application is to rate of growth and rate of decay.

It is important to recall that a rate of change (ROC) (with respect to time unless

otherwise specified) is a derivative with respect to time).

ROC positive there is growth (increase)

ROC negative there is decay (decrease)

Example The rate of decay of a certain radioactive element at any time is proportional

to the mass of the element remaining at that instant.

After 100 days, one third of a given mass 0m has disintegrated. How much is

left after a further 100 days?

Let m be the mass remaining at time t. The initial mass is 0m (when 0t )

‘Decay’ implies loss of mass, and “rate of decay” is given by a derivative

kmdt

dm where 0k

Separate variables kdtdmm

1

kdtdmm

1

Aktm ln

0mm when 0t Am 0ln 0

0lnln mktm

kt

kt

emm

em

m

ktm

m

ktmm

0

0

0

0

ln

lnln

032 mm when 100t kemm 100

0032

32100 ke

100

100

1

32

0

32

t

mm

e k

When 200t 100

200

32

0mm

094 mm