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Engineering
Mathematics:
Differential Calculus
Contents
Concepts of Limits and Continuity
 Derivatives of functions
 Differentiation rules and Higher Derivatives
 Applications

Differential Calculus
Concepts of Limits and
Continuity
The idea of limits
f(x) x
2

Consider a function

The function is well-defined for all real values of x
The following table shows some of the values:

x
2.9
2.99
2.99
3
3.001
3.01
3.1
F(x)
8.41
8.94
8.994
9
9.006
9.06
9.61
lim f ( x )  lim x 2  9
x 3
x 3
The idea of limits
Concept of Continuity
E.g.
f ( x )  x 2is continuous at x=3?
The following table shows some of the values:
x
2.9
2.99
2.99
3
3.001
3.01
3.1
F(x)
8.41
8.94
8.994
9
9.006
9.06
9.61
lim f ( x )  9
lim f ( x )  9
x 3

and

lim f ( x ) exists as
x 3
x 3
lim f ( x )  lim f ( x )  9
x 3
x 3
lim f ( x )  9  f ( 3 ) => f(x) is continuous at x=3!
x 3
Differential Calculus
Derivatives of functions
Derivative (導數)
y



Given y=f(x), if
variable x is given
an increment Dx
from x=x0, then y
would change to
f(x0+Dx)
Dy= f(x0+Dx) – f(x)
Dy/Dx is the slope
(斜率) of triangular
ABC
Y=f(x)
f(x0+Dx)
B
Dy
f(x0)
A
C
Dx
x0
x0+Dx
x
Derivative





What happen with Dy/Dx as Dx tends to 0?
It seems that Dy/Dx will be close to the slope of the curve
y=f(x) at x0.
We defined a new quantity as follows
dy df ( x )
Dy
f ( x  Dx )  f ( x )

 lim
 lim
D
x

0
dx
dx
Dx Dx0
Dx
If the limit exists, we called this new quantity as the
derivative (導數) of f(x).
The process of finding derivative of f(x) is called
differentiation (微分法).
Derivative
y
Y=f(x)
f(x0+Dx)
B
Dy
f(x0)
A
C
Dx
x0
x0+Dx
Derivative of f(x) at Xo = slope of f(x) at Xo
x
Differentiation from first principle
Find the derivative of y  f ( x )  x  3x with respect to (w.r.t.) x
dy
Dy
 lim
dx Dx0 Dx
f ( x  Dx )  f ( x )
 lim
Dx 0
Dx
( x  Dx )2  3( x  Dx )  ( x 2  3 x )
 lim
Dx 0
Dx
x 2  2 x( Dx )  ( Dx )2  3x  3Dx  x 2  3 x )
 lim
Dx 0
Dx
2 x( Dx )  ( Dx )2  3Dx
 lim
Dx 0
Dx
To obtain the derivative of a function by its
 lim ( 2 x  Dx  3 )
2
Dx 0
 2x  3
definition is called
differentiation of the function from first principles
Differential Calculus
Differentiation rules and
Higher Derivatives
Fundamental formulas for
differentiation I



Let f(x) and g(x) be differentiable functions and c be a
constant.
d( c )
0
dx
d ( f ( x )  g( x )) df ( x ) dg( x )


dx
dx
dx

d ( cf ( x ))
df ( x )
c
dx
dx

d( xn )
 nx n1
dx
for any real number n
Examples

Differentiate 3x 2  4 x  1 and
w.r.t. x
5x3  8x 2  6 x  9
y  3x 2  4 x  1
y  5x3  8x 2  6 x  9
dy d ( 3 x 2 ) d ( 4 x ) d ( 1 )



dx
dx
dx
dx
d( x2 )
d( x )
3
4
0
dx
dx
 3( 2 x )  4  0
dy
d ( x3 )
d (x2 )
d ( x) d (9)
5
 (8)
6

dx
dx
dx
dx
dx
 5(3 x 2 )  8(2 x)  6  0
 6x  4
 15 x 2  16 x  6
Table of derivative (1)
Function f ( x )
Constant k
x
kx
kx n
ln x
ln kx
df ( x )
Derivative dx
0
1
k
knx n1
1
x
1
x
Table of Derivatives (2)
Function f ( x )
e kx
sin kx
sin(kx   )
cos kx
cos(kx   )
df ( x )
Derivative dx
ke kx
k cos kx
k cos(kx   )
 k sin kx
k sin(kx   )
Angles in
radians
Differential Calculus
Product Rule, Quotient Rule
and Chain Rule
- The product rule and the quotient rule
Form of product function
y( x )  u( x )v( x )
Form of quotient function
u( x )
y( x ) 
v( x )

e.g.1
y( x )  x 2 cos x
This is a _________ function with
and
u( x ) 
v( x ) 

e.g.2
ex  x
y( x ) 
ln x
This is a _________ function with
and

e.g.3
t2 1
y(t ) 
cos t

e.g.4
f (t )  (t 2  6)cos 2t

e.g.5
f ( x )  (3 x  7)e 2 x
u( x ) 
v( x ) 
The product rule
Consider the function
Using the product rule,
y( x )  u( x )v( x )
dy du
dv
 vu
dx dx
dx
 u ' v  uv '
dy
e.g.1 Find
where y  x 2 cos x
dx
Solution:




e.g.2 Find y ' where y( x ) 
Solution:
xe 2 x
df
3
e.g.3 Find f ' 
where f (t )  t ln t
dt
Solution:

The quotient rule
Consider the function
Applying the quotient rule,
u( x )
y( x )  .
v( x )
du
dv
v u
dy
vu 'uv '
dx
dx


2
dx
v
v2
cos x
y
'
e.g.1 Find
where y 
x
Solution:




ex  x
e.g.2 Find y ' where y( x ) 
ln x
Solution:
dy
e.g.3 Find y '  dt
Solution:
t2 1
where y(t ) 
cos 3t
More Example (1)
2  3x
Differentiate y 
w.r.t. x
2  3x
f ( x )  2  3x
g( x )  2  3x
d ( 2  3x )
d ( 2  3x )
( 2  3x )
 ( 2  3x )
dy
dx
dx

dx
( 2  3 x )2
( 2  3x )( 3 )  ( 2  3x )( 3 )

( 2  3 x )2
 12

( 2  3 x )2
More Example (2)
Differentiate y  ( 2 x 2  4 x  1 )( 3x 2  2 x  5 )
w.r.t. x
dy
d ( 3x 2  2 x  5 )
d( 2x2  4x 1)
2
2
 ( 2x  4x 1)
 ( 3x  2 x  5 )
dx
dx
dx
dx 2
dx d ( 5 )
dx 2
dx d ( 1 )
2
2
 ( 2 x  4 x  1 )( 3
2 
)  ( 3x  2 x  5 )( 2
4 
)
dx
dx dx
dx
dx
dx
 ( 2 x 2  4 x  1 )( 3( 2 x )  2 )  ( 3x 2  2 x  5 )( 2( 2 x )  4 )
 ( 2 x 2  4 x  1 )( 6 x  2 )  ( 3x 2  2 x  5 )( 4 x  4 )
 24 x 3  24 x 2  2 x  22
Fundamental formulas for
differentiation II



Let f(x) and g(x) be differentiable functions
d ( f ( x )  g( x ))
d ( g( x ))
d ( f ( x ))
 f ( x )
 g( x ) 
dx
dx
dx
df ( x )
dg( x )
g( x )
 f(x)
d ( f ( x ) / g( x ))
dx
dx

dx
( g( x ))2
Fundamental formulas for
differentiation III
d cos( x )
  sin( x )
dx
d sin( x )
 cos( x )
dx
d cot( x )
1
2



(csc(
x
))
dx
(sin( x ))2
d tan( x )
1

 (sec( x ))2
2
dx
(cos( x ))
d sec( x )
 sec( x ) tan( x )
dx
de x
 ex
dx
d ln( x ) 1

dx
x
d csc( x )
  csc( x ) cot( x )
dx
x 2 x3 x 4 x5
where e  1  x      
2! 3! 4! 5!
x
where ln( e x )  x
ln(x) is called natural logarithm (自然對數)
Differentiation of composite functions
2
y

x
 1 w.r.t. x, we may have problems
 To differentiate

as we don’t have a formula to do so.
The problem can be simplified by considering composite
function:
Let
u  x 2  1 so y  u and y  x 2  1
we know derivative of y w.r.t. u (by formula):
1
2
1
1
dy d u du
1 2 1 1  2


 u  u
du
du
du 2
2
dy
but still don’t know
dx
Chain Rule (鏈式法則)
Chain Rule states that :
given y=g(u), and u=f(x)
dy dy du


dx du dx
2
So our problem y  g( u )  u and u  f ( x )  x  1
dy d x  1 dy du



dx
dx
du dx
1
1  12
 u  ( 2 x )  x( x 2  1 ) 2
2
2
1
dy d u 1  2

 u
du
du
2


du d x 2  1

 2x
dx
dx
Example 1
Differentiate y  (cos( x ))3
w.r.t. x
Simplify y by letting u  cos( x ) so now
By chain rule
dy dy du
dx

dy du 3

 3u 2
du du
du

y  u3
dx
du d (cos( x ))

  sin( x )
dx
dx
dy dy du


 3u 2  (  sin( x ))  3 sin( x )(cos( x ))2
dx du dx
Example 2
Differentiate y  e w.r.t. x
x2
u
Simplify y by letting u  x so now y  e
By chain rule
dy dy du
2
dx
dy deu

 eu
du du

du

dx
du dx 2

 2x
dx dx
dy dy du
u
x2


 e  2 x  2 xe
dx du dx
Example 3
Differentiate y  ln( 2 x 2  3 ) w.r.t. x
Simplify y by letting u  2 x  3 so now y  ln( u )
By chain rule
dy dy du
2
dx
dy d ln( u ) 1


du
du
u

du

dx
du d ( 2 x 2  3 )

 4x
dx
dx
dy dy du 1
4x


  4x  2
dx du dx u
2x  3
Higher Derivatives (高階導數)

If the derivatives of y=f(x) is differentiable function of x, its
derivative is called the second derivative (二階導數) of y=f(x)
2
d
and is denoted by y or f ’’(x). That is
dx 2
2
d
y d dy
( 2)
f ' ' ( x)  f ( x)  2  ( )
dx
dx dx


3
2
d
y
d
d
y
( 3)
Similarly, the third derivative = f ( x)  3 
( 2)
dx
dx dx
n
n 1
d
y
d
d
y
(n)
the n-th derivative = f ( x )  n  ( n 1 )
dx
dx dx
Example

Find
dy d 2 y d 3 y
, 2 , 3 if
dx dx dx
y  4 x 4  x 3  12 x 2  8 x  6
dy
dx 4 dx 3
dx 2
dx d ( 6 )
4

 12
8 
 16 x 3  3x 2  24 x  8
dx
dx dx
dx
dx
dx
d 2 y d dy
dx 3
dx 2
dx d ( 8 )
2

(
)

16

3

24


48
x
 6 x  24
2
dx
dx dx
dx
dx
dx
dx
d3y d d2y
dx 2
dx d ( 24 )
 ( 2 )  48
6 
 96 x  6
3
dx
dx dx
dx
dx
dx
Differential Calculus
Applications
Slope of a curve

Recall that the derivative of a curve evaluate at a point is
the slope of the curve at that point.
Y=f(x)
f(x0+Dx)
B
Dy
f(x0)
A
Derivative of f(x) at Xo
= slope of f(x) at Xo
C
Dx
x0
x0+Dx
x
Slope of a curve


Find the slope of y=2x+3
at x=0
To find the slope of a
curve, we have to
compute the derivative of
y and then evaluate at a
point
dy d ( 2 x  3 )

2
dx
dx

The slope of y at x=0
equals 2
(y=mx+c now m=2)
Slope of a curve






Find the slope of y  x  1
at x=0, 2, -2
2
dy d ( x 2  1 )

 2x
dx
dx
The slope of y = 2x
The slope of y (at x=0)
= 2(0) = 0
The slope of y (at x=-2)
=2(-2) = -4
The slope of y (at x=2)
=2(2) = 4
X=-2
X=2
X=0
Local maximum and minimum point


dy
0
For a continuous function, the point at which
dx
is called a stationary point.
This gives the point local maximum or local minimum of
the curve
D
B
X1
C
A
X2
First derivative test (Max pt.)
Given a continuous function y=f(x)
If dy/dx = 0 at x=xo & dy/dx changes from +ve to –ve through x0,
x=x0 is a local maximum point
local maximum point
x=x0
First derivative test (Min pt.)
Given a continuous function y=f(x)
If dy/dx = 0 at x=xo & dy/dx changes from -ve to +ve through x0,
x=x0 is a local minimum point
x=x0
local minimum point
Example 1

Determine the position of any local maximum and
minimum of the function y  x 2  1
First, find all stationary point (i.e. find x such that dy/dx = 0)
dy d ( x 2  1 )

 2 x , so
dx
dx
when
dy
x  0,
 2x  0
dx
dy
0
dx
when
when x=0
dy
x  0,
 2x  0
dx
By first derivative test x=0 is a local minimum point
Example 2
Find the local maximum and minimum of y  x 3  6 x 2  9 x  2
Find all stationary points first:
dy
3
2
y  x  6x  9x  2
 3 x 2  12 x  9  3( x  1 )( x  3 )
dx
dy
 0, when
x  1 and
x3
dx
x
0
x<1
x=1
1<x<3
x=3
x>3
dy/dx
+
+
0
-
0
+
Example 2 (con’d)
By first derivative test,
 x=1 is the local maximum (+ve -> 0 -> -ve)
 x=3 is the local minimum (-ve -> 0 -> +ve)
Second derivative test
Second derivative test states:

There is a local maximum point in y=f(x) at x=x0,
2
dy
d
y < 0 at x=x0.
if
 0 at x=x0 and
dx

There is a local minimum point in y=f(x) at x=x0,
2
dy
d
y > 0 at x=x0.
if
 0 at x=x0 and
dx

dx 2
dx 2
d2y
If dy/dx = 0 and
2 =0 both at x=x0, the second derivative
dx
test fails and we must return to the first derivative test.
Example
Find the local maximum and minimum point of
y  x3  6 x 2  9 x  2
Solution
Find all stationary points first:

dy
y  x  6x  9x  2
 3 x 2  12 x  9  3( x  1 )( x  3 )
dx
2
dy
d
y
 0, then
x  1 or
x3
 6 x  12 x
2
dx
dx
3
2
d2y
d2y
 6( 1 )  12  6  0
 6( 3 )  12  6  0
2
2
dx x 1
dx x 3
By second derivative test, x=1 is max and x=3 is min.
So, (1,2) is max. point and (3,-2) in min. point.
Practical Examples 1
 e.g.1 A rectangular block, with square base of side x
mm, has a total surface area of 150 mm2. Show that
the volume of the block is given by V  1 (75 x  x 3 ) .
2
Hence find the maximum volume of the block.

Solution:
Practical Examples 2


A window frame is made in the shape of a rectangle with
a semicircle on top. Given that the area is to be 8 ,
8

show that the perimeter of the frame is P   r (  2).
r
2
Find the minimum cost of producing the frame if 1 metre
costs $75.
Solution: