Transcript Differential Calculus

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  9
2
x 3
x 3
The idea of limits
Concept of Continuity
E.g. f ( x )  x 2 is 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
x 3
as
lim f ( x )  9  f ( 3 )
x 3
x 3
lim  f ( x )  lim  f ( x )  9
x 3
x 3
=> f(x) is continuous at 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
dx



df ( x )
dx
 lim
Dx  0
Dy
Dx
 lim
Dx  0
f ( x  Dx )  f ( x )
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  3 x with respect to (w.r.t.) x
2
dy
dx
 lim
Dx 0
 lim
Dy
Dx
f ( x  Dx )  f ( x )
Dx
Dx 0
( x  D x )  3( x  D x )  ( x  3 x )
2
 lim
2
Dx
Dx 0
x  2 x( D x )  ( D x )  3 x  3 D x  x  3 x )
2
 lim
2
2
Dx
Dx 0
2 x( D x )  ( D x )  3D x
2
 lim
Dx 0
Dx
 lim ( 2 x  D x  3 )
Dx 0
 2x  3
To obtain the derivative of a function by its
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 )
dx

d ( cf ( x ))
dx
c
dx
df ( x )
dx


dg ( x )
dx
n
d( x )
dx
 nx
n 1
for any real number n
Examples
3 x  4 x  1 and
Differentiate
w.r.t. x

y  3x  4x  1
2

d( 3x )
dx

3
d( 4x )
dx
2
3

d ( 1 )
dy
dx
dx
dx
d( x )
4
d( x )
dx
 3( 2 x )  4  0
 6x  4
3
2
y  5x  8x  6x  9
2
dy
5x  8x  6x  9
2
dx
0
2
3
5
d (x )
2
 (8)
dx
d (x )
dx
 5 (3 x )  8 ( 2 x )  6  0
2
 15 x  16 x  6
2
6
d ( x)
dx

d (9 )
dx
Table of derivative (1)
df ( x )
Function f ( x )
Constant k
x
kx
kx
n
ln x
ln kx
Derivative
0
1
k
knx
1
x
1
x
n 1
dx
Table of Derivatives (2)
df ( x )
Function
e
f (x)
kx
Derivative
ke
kx
sin kx
k cos kx
sin( kx   )
cos kx
k co s( kx   )
cos( kx   )
dx
 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
Form of quotient function

e.g.1
y( x )  u ( x )v ( x )
y( x ) 
u( x )
v( x )
y ( x )  x cos x
2
This is a _________ function with u ( x ) 
and v ( x ) 
e x
x

e.g.2
y( x ) 
ln x
This is a _________ function with
and
t 1
2

e.g.3
y(t ) 
cos t

e.g.4
f ( t )  ( t  6 ) cos 2 t

e.g.5
f ( x )  (3 x  7 )e
2
2 x
u( x ) 
v( x ) 
The product rule
Consider the function
y( x )  u ( x )v ( x )
dy
Using the product rule,
dx

du
vu
dx
 u ' v  uv '
dy
e.g.1 Find
Solution:
dx
where y  x 2 cos x
dv
dx




e.g.2 Find y ' where y ( x ) 
Solution:
df
e.g.3 Find f ' 
dt
Solution:
where
xe
2x
f ( t )  t ln t
3

The quotient rule
Consider the function
u( x )
y( x )  .
v( x )
Applying the quotient rule,
dy
dx
e.g.1 Find y ' where y 
cos x
x
Solution:
v

du
u
dx
v
2
dv
dx  vu ' uv '
2
v
e x
x




e.g.2 Find y ' where y ( x ) 
ln x
Solution:
e.g.3 Find
Solution:
dy
y '
dt
t 1
2
where y ( t ) 
cos 3 t
More Example (1)
y
Differentiate
f ( x )  2  3x
dy
( 2  3x )
d( 2  3x )
 ( 2  3x )
dx
2
( 2  3 x )(  3 )  ( 2  3 x )( 3 )
( 2  3x )
 12
( 2  3x )
2
d( 2  3x )
dx
( 2  3x )
dx

2  3x
w.r.t. x
g( x )  2  3x


2  3x
2
More Example (2)
Differentiate
dy
y  ( 2 x  4 x  1 )( 3 x  2 x  5 )
2
d( 3x  2x  5 )
2
 ( 2x  4x 1)
2
dx
d( 2x  4x 1)
2
 ( 3x  2x  5 )
2
dx
 ( 2 x  4 x  1 )( 3
2
dx
2
2
dx
dx
dx

dx
d( 5 )
)  ( 3 x  2 x  5 )( 2
2
dx
 ( 2 x  4 x  1 )( 3( 2 x )  2 )  ( 3 x  2 x  5 )( 2 ( 2 x )  4 )
2
2
 ( 2 x  4 x  1 )( 6 x  2 )  ( 3 x  2 x  5 )( 4 x  4 )
2
2
 24 x  24 x  2 x  22
3
2
w.r.t. x
2
dx
2
dx
4
dx
dx

d ( 1 )
dx
)
Fundamental formulas for
differentiation II


Let f(x) and g(x) be differentiable functions
d ( f ( x )  g ( x ))
 f ( x )
dx

d ( f ( x ) / g ( x ))
dx
d ( g ( x ))
 g( x )
d ( f ( x ))
dx
g( x )

df ( x )
dx
 f(x)
dx
2
( g ( x ))
dg ( x )
dx
Fundamental formulas for
differentiation III
d cos( x )
  sin( x )
d sin( x )
dx
dx
d tan( x )
1

dx
(cos( x ))
d sec( x )
2
 (sec( x ))
d cot( x )
2
d csc( x )
 sec( x ) tan( x )
1

dx
dx
de
 cos( x )
(sin( x ))
2
  (csc( x ))
  csc( x ) cot( x )
dx
x
e
x
where
dx
d ln( x )
dx
e  1 x 
x
x
2
2!

1
x
where

x
3
3!

x
4
4!

x
5

5!
ln( e )  x
x
ln(x) is called natural logarithm (自然對數)
2
Differentiation of composite functions


y
x 1
2
To differentiate
w.r.t. x, we may have problems
as we don’t have a formula to do so.
The problem can be simplified by considering composite
function:
u  x 1
2
Let
so
y
and
u
y
x 1
2
we know derivative of y w.r.t. u (by formula):
1
dy
du

d
u
du

du
du
2

1
2
1
u2
1

1
2

u
1
2
but still don’t know
dy
dx
Chain Rule (鏈式法則)
Chain Rule states that :
given y=g(u), and u=f(x)
dy
dy

dx
So our problem
dy

x 1
d
dx
dx

1
2

u
du
y  g( u ) 
2

dy
du


dx
and
u
u  f ( x )  x 1
2
du
dy
dx
du

 ( 2 x )  x( x  1 )
2
2
du
dx
d
u

du
1
1
2
du


1

u
1
2
2
d x 1
2
dx
  2x
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
dy
dx


du
du
du
dy
du
du

dx
 3u
dx
du
2
3

du
3
yu
dx

d (cos( x ))
  sin( x )
dx
 3 u  (  sin( x ))   3 sin( x )(cos( x ))
2
2
Example 2
2
Differentiate y  e w.r.t. x
x
u
Simplify y by letting u  x so now y  e
By chain rule
dy
dy du
2

dx
dy
du

de

du
u
e
du
u
du
dy
dx
dx
dx

dy
du

du
dx

dx
2
 2x
dx
 e  2 x  2 xe
u
x
2
Example 3
Differentiate
y  ln( 2 x  3 )
2
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
du

d ln( u )

du
dx
1
du
du
u
dx
dy
dy
dx


du

du
dx

d( 2x  3 )
2

 4x
dx
1
u
 4x 
4x
2x  3
2
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
y or f ’’(x). That is
and is denoted by
dx
2
2
f ''( x)  f
(2)
( x) 
d y
dx
d

2
(
dy
)
dx dx
3

Similarly, the third derivative =
f
(3)
( x) 
d y
dx
n

the n-th derivative = f
(n)
(x)
d y
dx
n

d
dx
(
d
3

dx
n 1
dx
d
y
n 1
)
2
(
d y
dx
2
)
Example
2

Find
dy
4
dx
2
dx
dx
4

dx
dx
d y
2

3
dy d y d y
, 2 , 3
dx dx
dx
d
 12
dx
dx
(
dx
dy
3
3

4
2
8
dx
)  16
dx
dx
d y
dx
3
y  4 x  x  12 x  8 x  6
if
3
3
dx
dx
2
(
d y
dx
2

dx
dx
d
dx
)  48
3
d( 6 )
 16 x  3 x  24 x  8
3
2
dx
2
 24
dx
dx
2
dx

dx
2
dx
6
dx
dx
d(8 )
 48 x  6 x  24
2
dx

d ( 24 )
dx
 96 x  6
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
dx


d( 2x  3 )
2
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
dx




d( x 1)
2

 2x
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 1)
2

dx
when
dx
x  0,
dy
dx
 2 x , so
 2x  0
dy
0
dx
when
x  0,
when x=0
dy
 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
y  x  6x  9x  2
3
2
 3 x  12 x  9  3( x  1 )( x  3 )
2
dx
dy
 0,
x 1
when
x3
and
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

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
2
d y

2
dx
2
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  x  6x  9x  2
3
2
Solution
Find all stationary points first:

y  x  6x  9x  2
3
2
dy
 3 x  12 x  9  3( x  1 )( x  3 )
2
dx
dy
 0,
then
x 1
or
2
d y
x3
dx
dx
2
 6 x  12 x
2
d y
dx
2
 6 ( 1 )  12   6  0
2
x 1
d y
dx
 6 ( 3 )  12  6  0
2
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: