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