Analytical Toolbox
Differential calculus
By
Dr J.P.M. Whitty
Learning objectives
•
After the session you will be able to:
• Define the derivative in terms of a
imitating function
• Differentiate simple algebraic
functions from first principles
• Apply rules of differentiation
• Use math software to solve simple
differentiation problems
2
Differential Calculus
•
Calculus is the mathematics of change
A point in three dimensions needs six pieces of
information to be fully described.
• We use the Derivative Function , f , to
describe the rate of change of the
original function f.
y
f (a+h)
rise
f(a)
f ( a  h)  f ( a )
f (a )  lim
h
h0
y = f x()
run
a
a+h
x
3
Example
Find the derivative (using first principles) of
f (x) = x2 - 8x + 9. Here we use the
definition:
f ( a  h)  f ( a )
h
h0
f (a )  lim
f (a  h)  (a  h) 2  8(a  h)  9
Where:
f (a  h)  a 2  2ah  h 2  8a  8h  9
Giving:

 

 a 2  2ah  h 2  8a  8h  9  a 2  8a  9 
f ' (a)  lim 

h 0
h


4
Example Cont:
Simplifying the denominator:
 2ah  h 2  8h 
f ' (a)  lim 

h2
0
h


 a 2  2ah  h 2  8a  8h  9  a  8a  9 
f ' (a)  lim 

h 0
h


2
 2ah  h  8h 
f ' (a)  lim 
That is:

h 0
h


0
f ' (a)  lim 2a  h  8
Cancel out the h’s

h 0
letting h  0 :
let a  3 gives f ' (3) :

f ' ( a )  2a  8
f ' (3)  2  3  8  14
5
Differentiation of standard
functions
It is usual in calculus
textbooks that to see
tables of standard
functions and their
respective differential
coefficients.
• For this
introductory course
you will only require
the following
y
dy
dx
c
0
ax
n
anx
n 1
sin ax
acos ax
cos ax
 asin ax
e ax  exp ax ae ax  a exp ax
6
Class Examples Time
Copy and complete
the following table.
y
dy
dx
5c
0
3x
3
4x
2
8x
5x
4
20x 3
3x  2 x  1
3
9x2  2
7
Class examples
Differentiate the following:
a) y  3x  6 x  x  5
4
3
dy
 3  4 x 3  6  3x 2  1x 0
dx
dy
ans)
 12 x 3  18 x 2  1
dx
b) y  x  3x  2 x  7 x  9
4
3
2
dydy

4 4 x43x3 3 3 3x32x2 222x21 x1771x 01x 0
dxdx
dy
ans )
 4 x3  9 x 2  4 x  7
dx
8
Another example
Find the first and second derivatives of:
f ( x)  2 x 4  3sin 3x  5e 2 x  cos x
f ' ( x)  8x3  9 cos 3x  10e 2 x  sin x
f ' ' ( x)  24 x 2  18 sin 3x  20e 2 x  cos x
9
Class Examples Time
Find the first and
second derivative of:
f ( x)  2 exp( 9 x)  3x 2  cos 2 x  2 sin 3x  e x
f ' ( x)  18 exp( 9 x)  6 x  2 sin 2 x  6 cos 3x  e x
f ' ' ( x)  162 exp( 9 x)  6  4 cos 2 x  18 sin 3x  e x
10
Differentiation Formulae
Several ‘short-cut’ rules exist for differentiation, eliminating
the need for tedious and repetitive use of limit expressions.
Proofs of these can be found in most calculus books.
Description
Constant
General Power Rule
Function
f (x) = c
f (x) = xn
Addition/Subtraction
f (x) = g (x)  h (x)
f ( x )  nx n1
f ( x )  g ( x )  h ( x )
Product Rule
f (x) = g (x) h (x)
f  ( x )  g ( x ) h  ( x )  g  ( x ) h( x )
Quotient Rule
f (x) =
g( x)
h( x )
Derivative
f ( x )  0
f ( x ) 
g  ( x ) h( x )  g ( x ) h  ( x )
(h( x)) 2
11
Revision Questions
1.
Complete the
following
table:
y
dy
dx
c
x
x2
4x 3
x4
2x 5
2x3  x  2
12
More Revision Qs
1.
Differentiate the following:
a) y  2 x 4  5 x 3  2 x  4
b) y  3x 4  2 x 3  4 x 2  5 x  1
2.
Find the first and second derivatives
of:
f ( x)  3x 3  4 sin 2 x  3e x  cos 4 x
f ( x)  9 exp( 4 x)  2 x 3  2 cos 3x  3 sin 4 x  7e 4 x
13
Revision Qs (solutions)
1.
Complete the
following
table:
y
dy
dx
c
x
0
1
x2
2x
4x 3
12x 2
x4
4x 3
2x 5
10x 4
2x3  x  2
6x2 1
14
More Revision Qs (Solutions)
1.
Differentiate the following:
dy
 8 x 3  15 x 2  2
dx
dy
 12 x 3  6 x 2  8 x  5
dx
2.
The first and second derivatives
f ' ( x)  9 x 2  8 cos 2 x  3e x  4 sin 4 x
f ' ' ( x)  18x  16 sin 2 x  3e x  16 cos 4 x
f ' ( x)  36 exp( 4 x)  6 x 2  6 sin 3x  12 cos 4 x  28e 4 x
f ' ( x)  144 exp( 4 x)  12 x  18 cos 3x  48 sin 4 x  112e
15
4x
The Product rule:
This is used when we have two primitive
functions multiplied together.
If f ( x)  g ( x)h( x)  uv
then
f ( x)  g ( x)h ( x)  g ( x)h( x)  u
dv
du
v
dx
dx
16
The Product rule: Proof
Here we employ the definition of the
derivative thus:
 g ( x  x)h( x  x)  g ( x)h( x) 
f ' ( x)  lim 

x 0

x


Re-writing this gives
f ' ( x) 
 g ( x  x)h( x  x)  h( x) g ( x  x)  h( x) g ( x  x)  g ( x)h( x) 
lim 

x 0
x



 h( x  x)  h( x) 
 g ( x  x)  g ( x) 
f ' ( x)  lim  g ( x  x)
  h( x)

x 0
x
x



 17

The Product rule: Proof

 h( x  x)  h( x) 
 g ( x  x)  g ( x) 
f ' ( x)  lim  g ( x  x)
  h( x)

x 0
x
x





Now use the fact that the terms in parenthesis
are derivatives of the individual functions, to
complete the proof, i.e.:
 g ( x  x)  g ( x) 
 h( x  x)  h( x) 
g ' ( x)  lim 
 and h' ( x)  lim


x 0
x 0

x

x




Hence Substitution gives:
f ' ( x)  lim g ( x  x)h' ( x)  h( x) g ' ( x)
x 0
Evaluation of the limit renders the required result
18
The Product rule:
Proof (Alternative nomenclature)
In many calculus text books, the less formal functions
of u and v are used. This makes the preceding
proof a little less tedious, as follows
If
f ( x )  uv
then
 (u  u )(v  v)  uv 
f ' ( x)  lim 

x 0

x


 (uv  vu  uv  uv)  uv 
f ' ( x)  lim 

x 0

x


0
v uv 
 vu  uv  uv 
 u
f ' ( x)  lim 
  lim
v  u 

x 0
x

0

x

x

x

x




dy
du
u 
v 
f ' ( x)  v lim    u lim    u
v
x 0 x
x 0 x
dx
dx
 
 
19
Lemma 1:
When using the calculus we rarely resort to
using formulae, we tend to remember a
mechanism of a pattern, these leads us to
the lemma.
Lemma 1:The first times the diff. of
the second-plus-the second times the
diff. of the first
20
Example
If y  x e
First
dy
dx
find
3 5x
Second


dy
 x 3 5e5 x  5e5 x
dx
Diff of the second
Diff of the first
21
Another Example:
If
y  e cos(3x) find
2x
dy
dx

dy
 e 2 x (3 sin( 3 x))  cos(3x) 2e 2 x
dx
Ans:

dy
 3e 2 x sin( 3x)  2e 2 x cos(3x)
dx
22
The quotient rule:
This is used when we have two primitive functions
dividend by one another
if
g ( x) u
f ( x) 

h( x ) v
h( x ) g ' ( x )  g ( x ) h' ( x )
Then f ' ( x) 
2
g ( x)
or
du
dv
v
u
dy
 dx 2 dx
dx
v
23
Lemma 2:
Again we tend to remember a mechanism of a
pattern, this leads us to the lemma.
Lemma 1:The bottom times the diff. of
the top-minus-the top times the diff. of
the bottom – ALL OVER THE
BOTTOM SQUARED
In a sense this is simply the reverse of the
product rule and then dividing by the square
of the bottom!
24
Example
If y  x / e
3
5x
dy
dx
find
bottom
 
top
bottom squared

dy
3
5x
 e5 x 3x 2  x 5e
dx
Diff of the top
 e 
5x 2
Diff of the bottom
dy 3e5 x x 2  5e5 x x 3

dx
e10 x
x 2 3  5 x 

e5 x
25
Alternative approach
An alternative approach is to use
logarithmic differentiation as follows:
y  x 3 / e5 x
Take logs
ln y  3 ln x   5x
Differentiate using the chain rule as appropriate:
3
dy
x
ln y  3 ln x   5x
 5x
dx e
2
dy x 3  5 x 

5x
dx
e
3 
  5
x 
26
Alternative approach
The logarithmic approach is especially
useful when the primitive functions are
themselves products (or quotients),
the details of the method are left to P7
of the recommended reading. But here
we shall consider another quick
example, i.e.:


x 1  x3
dy
if y 
find
2
1 x
dx
27
Logarithmic differentiation
The approach is always the same take
natural logs, remembering ever
primitive above the line is positive and
everything below is negative. Then
differentiate the resulting primitive
functions, remembering that the RHS is
equal to the 1/y multiplied by the
differential coefficient. The rest is just
algebra!
28
Logarithmic differentiation
Example
Taking logs of both sides gives:

 
ln y  ln x  ln 1  x  ln 1  x
3
2

Differentiate:
3
1 dy 1 3x
2x
 

3
y dx x 1  x 1  x 2
Substitute:
dy x 1  x 3  1 3x 3
2x 



2 
3
2
dx
1 x  x 1 x 1 x 


29
Use of mathematic Software
As the expressions get more and more
complicated the methods stay the
same but the algebra gets more and
more tedious. Thankfully these days
MATLAB has the answer that is the
symbolic toolbox we have been using
in previous lessons can be used to
differentiate complicated functions
such as that in the last example.
30
MATLAB: Differentiation
The process is the same as
usual. i.e. easy as ABC!
A.
B.
C.
Set up your symbolics in
MATLAB using the syms
command
Type in the expression
remembering the rules of
BIDMAS
Use the appropriate MATLAB
function in this case diff( ),
making pretty if required.


x 1  x3
dy
if y 
find
2
1 x
dx
MATLAB Commands :
syms x y
y  ( x * (1  x ^3)) /(1  x ^ 2)
diff(y)
pretty(ans )
31
MATLAB Solution
The commands are
very straight
forward but you
may have to do a
little bit of algebra
at the end to get
the result in the
same form
alternatively use
the MATLAB
commands to do it
for you!!
32
MATLAB Solution cont…
For instance try
factorizing the
resulting
expression
using the factor
function we
have used in
class
previously, thus
33
Summary
Have we met our learning objectives?
Specifically: are you able to:
• Define the derivative in terms of a imitating
function
• Differentiate simple algebraic functions
from first principles
• Apply rules of differentiation
• Use math software to solve simple
differentiation problems
34
Homework
Find the derivative of
3
2
x  2 x  5x  6
2. Find the derivative of
1.
( x 5  6x 2  2)( x 3  x  1)
3. Find the derivative of
x2  1
x2  1
35
Examination type questions
1. Differentiate the following with
respect to x.
a.
e cos 2 x
b.
ln(cos x)
e 3 x / sin 2 x
c.
d.
4x
Explain how you would use math
software to verify your results
36
Examination type questions
2. Evaluate the tangent to the curve
y  x  2 x  3x
3
2
at a point P(2,6) and find the equation of the
tangent of that line. Given that the
product of the normal and tangential line
gradients is minus unity find the equation
of the normal.
37
More Calculus
WELL DONE!!
You have now completed the first part of
this learning pack. Have a break and
then you can try the next part where
you will be introduced to the inverse
process to differentiation, namely:
Integration
38