Transcript Document

Differentiation
3
• Basic Rules of Differentiation
• The Product and Quotient Rules
• The Chain Rule
• Marginal Functions in Economics
• Higher-Order Derivatives
• Implicit Differentiation and Related Rates
• Differentials
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Basic Differentiation Rules
d
1.
c  0
dx
 c is a constant 
Ex. f ( x)  5
f ( x)  0
 
d n
x  nx n1
2.
dx
 n is a real number 
Ex. f ( x)  x 7
f ( x)  7 x 6
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Basic Differentiation Rules
d
d
3.
 cf ( x)   c  f ( x) 
dx
dx
 c is a constant 
8
f
(
x
)

3
x
Ex.
 
f ( x)  3 8 x7  24 x7
d
d
d
4. dx  f  x   g  x    dx  f ( x)  dx  g ( x) 
Ex. f ( x)  7  x12
f ( x)  0 12x11  12x11
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
More Differentiation Rules
5. Product Rule
d
d
d
 f  x   g  x     f ( x) g ( x)   g ( x)  f ( x)
dx
dx
dx



 1   x  2 x  5  21x
Ex. f ( x)  x  2 x  5 3x  8x  1


3
f ( x)  3x 2  2 3x7  8 x 2
Derivative
of the first
function
7
3
2
6
 16 x

Derivative of
the second
function
f ( x)  30x9  48x7 105x6  40x4  45x2  80x  2
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
More Differentiation Rules
6. Quotient Rule
d
d
g ( x)  f ( x)   f ( x)  g ( x) 


f
x
  
d
dx
dx


2
dx  g ( x) 
g
(
x
)
 
Sometimes remembered as:
d  hi  lo d  hi   hi d lo



dx  lo 
lo lo
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
More Differentiation Rules
6. Quotient Rule (cont.)
3x  5
Ex. f ( x)  2
x 2
Derivative of
the numerator
f ( x) 


Derivative of
the denominator

3 x 2  2  2 x  3x  5
x
2
2

2
3x2  10 x  6
x
2
2

2
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
More Differentiation Rules
7. The Chain Rule
If h( x)  g  f ( x)  then
h( x)  g  f ( x)  f ( x)
Note: h(x) is a composite function.
Another Version:
If y  h( x)  g u  , where u  f ( x), then
dy dy du


dx du dx
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
More Differentiation Rules
The Chain Rule leads to
The General Power Rule:
If h( x)   f ( x)
n
 n, real  then
h( x)  n  f ( x) 
n 1
 f ( x)

Ex. f ( x)  3x  4 x  3x  4 x
2

1
2

f ( x)  3 x  4 x
2
3x  2

3x 2  4 x

2
1 2

12
6x  4
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Chain Rule Example
7
2
x

1


Ex. G( x) 


 3x  5 
6
 2 x  1    3x  5 2   2 x  1 3 
G( x)  7 
 
2

 3x  5  
 3x  5 

 2x 1 
G( x)  7 

 3x  5 
6
13
 3x  5
2

91 2 x  1
6
 3x  5 
8
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Chain Rule Example
Ex. y  u
, u  7 x  3x
dy dy du


dx du dx
52
8

2

5 32
 u  56 x 7  6 x
Sub in for u
2
32
5
8
2
 7 x  3x
 56 x 7  6 x
2


 140 x
7
 

 15x  7 x  3x 
8
2
32
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Marginal Functions
The Marginal Cost Function approximates the change in
the actual cost of producing an additional unit.
The Marginal Average Cost Function measures the rate
of change of the average cost function with respect to the
number of units produced.
The Marginal Revenue Function measures the rate of
change of the revenue function. It approximates the revenue
from the sale of an additional unit.
The Marginal Profit Function measures the rate of change
of the profit function. It approximates the profit from the sale
of an additional unit.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Cost Functions
Given a cost function, C(x),
the Marginal Cost Function is
C ( x )
the Average Cost Function is
C ( x)
C
x
the Marginal Average Cost Function is

C ( x)
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Revenue Functions
Given a revenue function, R(x),
the Marginal Revenue Function is
R( x )
Profit Functions
Given a profit function, P(x),
the Marginal Profit Function is
P( x )
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Elasticity of Demand
If f is a differentiable demand function defined by
x  f ( p)
Then the elasticity of demand at price p is given by
E  p  
Demand is:
pf   p 
f  p
Elastic if E(p) > 1
Unitary if E(p) = 1
Inelastic if E(p) < 1
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Elasticity of Demand
If the demand is elastic at p, then an increase in unit
price causes a decrease in revenue. A decrease in unit
price causes an increase in revenue.
If the demand is unitary at p, then with an increase
in unit price the revenue will stay about the same.
If the demand is inelastic at p, then an increase in
unit price causes an increase in revenue. A decrease
in unit price causes a decrease in revenue.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Consider the demand equation
p  0.02x  400  0  x  20,000
which describes the relationship between the unit
price p in dollars and the quantity demanded x of the
Acrosonic model F loudspeaker systems. Find the
elasticity of demand E ( p).
Solving the given demand equation for x in
terms of p, we find x  f ( p)  50 p  20, 000.
Therefore,
p  50 
pf ( p )
p
E ( p)  

=
f ( p)
50 p  20, 000 400  p
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
The monthly demand for T-shirts is given by
0  x  400
p  0.05x  25
where p denotes the wholesale unit price in dollars
and x denotes the quantity demanded. The monthly
cost function for these T-shirts is
C( x)  0.001x  2x  200
2
1. Find the revenue and profit functions.
2. Find the marginal cost, marginal revenue, and
marginal profit functions.
3. Find the marginal average cost function. ……
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Solution
1. Find the revenue and profit functions.
Revenue = xp
 x  0.05x  25  0.05x  25x
2
Profit = revenue – cost

 0.05 x  25 x  0.001x  2 x  200
2
2
 0.049 x2  23x  200

2. Find the marginal cost, marginal revenue, and
marginal profit functions.
Marginal Cost = C ( x )
 0.002 x  2
......
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Solution
2. (cont.) Find the marginal revenue and marginal
profit functions.
Marginal revenue = R( x )
 0.1x  25
Marginal profit = P( x )
 0.098 x  23
3. Find the marginal average cost function.
C( x)  0.001x  2  200x
1
2

C ( x)  0.001  200 x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Higher Derivatives
The second derivative of a function f is the derivative
of the derivative of f at a point x in the domain of the
first derivative.
Derivative
Second
Third
Fourth
nth
Notations
f 
d2y
dx 2
f 
d3y
dx3
(4)
d4y
dx 4
 n
dny
dx n
f
f
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example of Higher Derivatives
Given f ( x)  3x5  2 x3  14 find f ( x).
4
2

f ( x)  15x  6x
f ( x)  60x3 12x
2

f ( x)  180x 12
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example of Higher Derivatives
2x 1
Given f ( x) 
find f (2).
3x  2
f ( x) 
2  3 x  2   3  2 x  1
 3x  2 
f ( x)  14  3x  2
f (2) 
3
42
3(2)  2
3
2
3 

7
 3x  2 
2
 7  3 x  2 
2
42
 3x  2 
3
42 21
 3 
32
4
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Implicit Differentiation
y  3x3  4x  17
y is expressed explicitly as a function of x.
y3  xy  3x  1
y is expressed implicitly as a function of x.
To differentiate the implicit equation, we write f (x) in
place of y to get:
 f ( x)
3
 x  f ( x)  3x  1
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Implicit Differentiation (cont.)
Now differentiate  f ( x)  x  f ( x)  3x  1
3
using the chain rule:
3 f ( x) f ( x)  f ( x)  xf ( x)  3
2
which can be written in the form
3 y2 y  y  xy  3


2

y 3y  x  3  y
y 
subbing in y
Solve for y’
3 y
3 y2  x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Related Rates
Look at how the rate of change of one quantity is
related to the rate of change of another quantity.
Ex. Two cars leave an intersection at the same
time. One car travels north at 35 mi./hr.,
the other travels west at 60 mi./hr. How
fast is the distance between them changing
after 2 hours?
Note: The rate of change of the distance between them is
related to the rate at which the cars are traveling.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Related Rates
Steps to solve a related rate problem:
1. Assign a variable to each quantity.
Draw a diagram if appropriate.
2. Write down the known values/rates.
3. Relate variables with an equation.
4. Differentiate the equation implicitly.
5. Plug in values and solve.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Ex. Two cars leave an intersection at the same time. One
car travels north at 35 mi./hr., the other travels east at 60
mi./hr. How fast is the distance between them changing
after 2 hours?
dx
 60
dt
dy
 35
dt
x  120
y  70
Distance = z
y
x
From original
relationship
x2  y 2  z 2
dx
dy
dz
2x  2 y
 2z
dt
dt
dt
dz
2(120)(60)  2(70)(35)  2 10 193
dt
dz

dt

 69.5 mi./hr.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Increments
An increment in x represents a change from x1
to x2 and is defined by:
x  x2  x1
Read “delta x”
An increment in y represents a change in y and
is defined by:
y  f ( x  x)  f ( x)
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Differentials
Let y = f (x) be a differentiable function, then the
differential of x, denoted dx, is such that dx  x.
The differential of y, denoted dy, is
dy  f ( x)x  f ( x)dx
Note: y
measures actual change in y
dy measures approximate change in y
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Given
f ( x)  3x  x, find:
2
1. x as x changes from 3 to 3.02.
x  3.02  3  0.02
2. y and dy as x changes from 3 to 3.02.
y  f (3.02)  f (3)
 24.3412  24  0.3412
dy  f ( x)dx   6x 1 dx
  6(3) 1 (0.02)  0.34
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
The total cost incurred in operating a certain type of
truck on a 500-mile trip, traveling at an average speed
of v mph, is estimated to be
4500
C (v )  125  v 
dollars
v
Find the approximate change in the total operating cost when
the average speed is increased from 55 mph to 58 mph.
With v  55 and v  3, we find
4500 
4500 


v  dv  C (v)dv  1  2 
 3  1 
  3  1.46
v  v55

 3025 
so the total operating cost is found to decrease
by $1.46.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.