3
Differentiation
 Basic Rules of Differentiation
 The Product and Quotient Rules
 The Chain Rule
 Marginal Functions in Economics
 Higher Order Derivatives
3.1
Basic Rules of Differentiation
1.
2.
3.
4.
Derivative of a Constant
The Power Rule
Derivative of a Constant Multiple Function
The Sum Rule
Four Basic Rules
 We’ve learned that to find the rule for the derivative f ′of a
function f, we first find the difference quotient
lim
h0
f ( x  h)  f ( x)
h
 But this method is tedious and time consuming, even for
relatively simple functions.
 This chapter we will develop rules that will simplify the
process of finding the derivative of a function.
Rule 1: Derivative of a Constant
d
 We will use the notation
 f ( x )
dx
 To mean “the derivative of f with respect to x at x.”
Rule 1: Derivative of a constant
d
c  0
dx
 The derivative of a constant function is equal to zero.
Rule 1: Derivative of a Constant
 We can see geometrically why the derivative of a constant
must be zero.
 The graph of a constant function is a straight line parallel to
the x axis.
 Such a line has a slope that is constant with a value of zero.
 Thus, the derivative of a constant must be zero as well.
y
f(x) = c
x
Rule 1: Derivative of a Constant
 We can use the definition of the derivative to
demonstrate this:
f ( x  h)  f ( x)
h0
h
cc
 lim
h0 h
 lim 0
f ( x )  lim
h0
0
Rule 2: The Power Rule
Rule 2: The Power Rule
 If n is any real number, then
d n
n 1
x

nx


dx
Rule 2: The Power Rule
 Lets verify this rule for the special case of n = 2.
 If f(x) = x2, then
f ( x ) 
d 2
f ( x  h)  f ( x)
x

lim


h0
dx
h
( x  h)2  x 2
x 2  2 xh  h 2  x 2
 lim
 lim
h0
h0
h
h
2 xh  h 2
h (2 x  h )
 lim
 lim
h0
h0
h
h
 lim(2 x  h )  2 x
h0
Rule 2: The Power Rule
Practice Examples:
 If f(x) = x, then
d
f ( x )   x   1  x11  x 0  1
dx
 If f(x) = x8, then
f ( x ) 
d 8
x   8  x 81  8 x 7

dx
 If f(x) = x5/2, then
f ( x ) 
d 5/2
5 5/21 5 3/2
x

x
 x


dx
2
2
Example 2, page 159
Rule 2: The Power Rule
Practice Examples:
 Find the derivative of
f ( x)  x
d
f ( x ) 
dx
 
d 1/2
x  x 
dx
1 1/21
 x
2

Example 3, page 159
1
2 x
1 1/2
 x
2
Rule 2: The Power Rule
Practice Examples:
1
 Find the derivative of f ( x ) 
3
x
f ( x ) 
d  1  d 1/3
 x 


3
dx  x  dx
1
  x 1/31
3
1 4 / 3
1
 x
  4/3
3
3x
Example 3, page 159
Rule 3: Derivative of a Constant Multiple Function
Rule 3: Derivative of a Constant Multiple Function
 If c is any constant real number, then
d
d
cf
(
x
)

c


 f ( x)
dx
dx
Rule 3: Derivative of a Constant Multiple Function
Practice Examples:
3
 Find the derivative of f ( x )  5x
d
f ( x )   5 x 3 
dx
d 3
 5 x 
dx
 5  3x 2 
 15 x 2
Example 4, page 160
Rule 3: Derivative of a Constant Multiple Function
Practice Examples:
3
 Find the derivative of f ( x ) 
x
f ( x) 
d
3 x 1/ 2 

dx
 1 3/ 2 
 3  x 
 2


Example 4, page 160
3
2 x 3/ 2
Rule 4: The Sum Rule
Rule 4: The Sum Rule
d
d
d
 f ( x )  g ( x )   f ( x )   g ( x )
dx
dx
dx
Rule 4: The Sum Rule
Practice Examples:
 Find the derivative of
f ( x )  4 x 5  3x 4  8 x 2  x  3
d
5
4
2
4
x

3
x

8
x
 x  3

dx
d 5
d 4
d 2
d
d
 4  x   3  x   8  x    x    3
dx
dx
dx
dx
dx
f ( x ) 
 4 5x 4   3  4 x 3   8  2 x   1  0
 20 x 4  12 x 3  16 x  1
Example 5, page 161
Rule 4: The Sum Rule
Practice Examples:
 Find the derivative of
t2 5
g (t )   3
5 t
d  t2 5  d  1 2

g (t )    3    t  5t 3 
dt  5 t  dt  5

1 d 2
d 3
  t   5 t 
5 dt
dt
1
 2t   5  3t 4 
5
2t 15 2t 5  75
  4 
5 t
5t 4

Example 5, page 161
Applied Example: Conservation of a Species
 A group of marine biologists at the Neptune Institute of
Oceanography recommended that a series of conservation
measures be carried out over the next decade to save a
certain species of whale from extinction.
 After implementing the conservation measure, the
population of this species is expected to be
N (t )  3t 3  2t 2  10t  600
(0  t  10)
where N(t) denotes the population at the end of year t.
 Find the rate of growth of the whale population when
t = 2 and t = 6.
 How large will the whale population be 8 years after
implementing the conservation measures?
Applied Example 7, page 162
Applied Example: Conservation of a Species
Solution
 The rate of growth of the whale population at any time t is
given by
N (t )  9t 2  4t  10
 In particular, for t = 2, we have
N (2)  9  2   4  2   10  34
2
 And for t = 6, we have
N (6)  9  6   4  6   10  338
2
 Thus, the whale population’s rate of growth will be 34
whales per year after 2 years and 338 per year after 6 years.
Applied Example 7, page 162
Applied Example: Conservation of a Species
Solution
 The whale population at the end of the eighth year will be
N 8  3 8  2 8  10 8  600
3
2
 2184 whales
Applied Example 7, page 162
3.2
The Product and Quotient Rules
d
 f ( x ) g ( x )  f ( x ) g ( x )  g ( x ) f ( x )
dx
d  f ( x )  g ( x ) f ( x )  f ( x ) g ( x )

2


dx  g ( x ) 
g
(
x
)
 
Rule 5: The Product Rule
 The derivative of the product of two differentiable
functions is given by
d
 f ( x ) g ( x )  f ( x ) g ( x )  g ( x ) f ( x )
dx
Rule 5: The Product Rule
Practice Examples:
 Find the derivative of
f ( x )   2 x 2  1
f ( x )   2 x 2  1 x 3  3
d 3
d
3
x

3

x

3
2 x 2  1





dx
dx
  2 x 2  1 3x 2    x 3  3  4 x 
 6 x 4  3x 2  4 x 4  12 x
 x 10 x 3  3x  12 
Example 1, page 172
Rule 5: The Product Rule
Practice Examples:
3
 Find the derivative of f ( x )  x


x 1
d 1/2
d 3
1/2
f ( x )  x
x  1   x  1 x

dx
dx
3
1

 x 3  x 1/2    x1/2  1 3x 2
2

1 5/2
 x  3x 5/2  3x 2
2
7 5/2
 x  3x 2
2
Example 2, page 172
Rule 6: The Quotient Rule
 The derivative of the quotient of two differentiable
functions is given by
d  f ( x)  g ( x) f ( x)  f ( x) g ( x)



2
dx  g ( x) 
g
(
x
)
 
 g  x   0
Rule 6: The Quotient Rule
Practice Examples:
 Find the derivative of
x
f ( x) 
2x  4
d
d
 2 x  4 ( x)  x  2 x  4 
dx
dx
f ( x ) 
2
2
x

4


2 x  4 1  x  2 


2
2
x

4



Example 3, page 173
2x  4  2x
 2 x  4
2

4
 2 x  4
2
Rule 6: The Quotient Rule
Practice Examples:
 Find the derivative of
x2  1
f ( x)  2
x 1
d 2
d 2
2
 x  1 dx  x  1   x  1 dx  x  1
f ( x ) 
2
2
 x  1
2


Example 4, page 173
2
2
x

1
2
x

x


 
  1  2 x 
x
2
 1
2
2 x3  2 x  2 x3  2 x
x
2
 1
2

x
4x
2
 1
2
Applied Example: Rate of Change of DVD Sales
 The sales ( in millions of dollars) of DVDs of a hit movie
t years from the date of release is given by
5t
S (t )  2
t 1
 Find the rate at which the sales are changing at time t.
 How fast are the sales changing at:
✦ The time the DVDs are released (t = 0)?
✦ And two years from the date of release (t = 2)?
Applied Example 6, page 174
Applied Example: Rate of Change of DVD Sales
Solution
 The rate of change at which the sales are changing at
time t is given by
d  5t 
S (t )   2 
dt  t  1 
t



Applied Example 6, page 174
2
 1  5   5t  2t 
t
2
 1
5t  5  10t
2
t
2
 1
2
2
2

5 1  t 2 
t
2
 1
2
Applied Example: Rate of Change of DVD Sales
Solution
 The rate of change at which the sales are changing when
the DVDs are released (t = 0) is
2

5 1   0   5 1

S (0)  
5
2
2
2
1
 0   1


That is, sales are increasing by $5 million per year.
Applied Example 6, page 174
Applied Example: Rate of Change of DVD Sales
Solution
 The rate of change two years after the DVDs are
released (t = 2) is
2

5 1   2   5 1  4 
15
3



S (2) 

     0.6
2
2
2
25
5
 4  1
 2   1


That is, sales are decreasing by $600,000 per year.
Applied Example 6, page 174
3.3
The Chain Rule
h( x ) 
d
g  f ( x )   g   f ( x )  f ( x )
dx
dy dy du


dx du dx
Deriving Composite Functions


 Consider the function h( x )  x  x  1
2
2
 To compute h′(x), we can first expand h(x)
h( x )   x  x  1   x 2  x  1 x 2  x  1
2
2
 x 4  2 x 3  3x 2  2 x  1
and then derive the resulting polynomial
h( x )  4 x 3  6 x 2  6 x  2
 But how should we derive a function like H(x)?
H ( x )   x  x  1
2
100
Deriving Composite Functions
Note that H ( x )   x  x  1
2
100
is a composite function:
 H(x) is composed of two simpler functions
f ( x)  x 2  x  1
g ( x )  x100
and
 So that
H ( x )  g  f ( x )   f ( x )
100
  x  x  1
2
100
 We can use this to find the derivative of H(x).
Deriving Composite Functions
To find the derivative of the composite function H(x):
 We let u = f(x) = x2 + x + 1 and y = g(u) = u100.
 Then we find the derivatives of each of these functions
du
 f ( x )  2 x  1
dx
and
dy
 g (u )  100u 99
du
 The ratios of these derivatives suggest that
dy dy du


 100u 99  2 x  1
dx du dx
 Substituting x2 + x + 1 for u we get
99
dy
2
H ( x ) 
 100  x  x  1  2 x  1
dx
Rule 7: The Chain Rule
 If h(x) = g[f(x)], then
h( x) 
d
g  f ( x)   g   f ( x)  f ( x)
dx
 Equivalently, if we write y = h(x) = g(u),
where u = f(x), then
dy dy du


dx du dx
The Chain Rule for Power Functions
 Many composite functions have the special form
h(x) = g[f(x)]
where g is defined by the rule
g(x) = xn
(n, a real number)
so that
h(x) = [f(x)]n
 In other words, the function h is given by the power of a
function f.
 Examples:
h( x )   x  x  1
2
100
H ( x) 
1
5  x 
3 3
G( x)  2 x 2  3
The General Power Rule
 If the function f is differentiable and
h(x) = [f(x)]n
(n, a real number),
then
d
n
n 1
h( x )   f ( x )   n  f ( x )  f ( x )
dx
The General Power Rule
Practice Examples:
 Find the derivative of G( x )  x 2  1
Solution
1/2
2
 Rewrite as a power function: G( x )   x  1
 Apply the general power rule:
1/2 d
1 2
2
G( x )   x  1
x
 1

2
dx
1/2
1 2
  x  1  2 x 
2
x

x2  1
Example 2, page 184
The General Power Rule
Practice Examples:
5
2
 Find the derivative of f ( x )  x  2 x  3
Solution
 Apply the product rule and the general power rule:
f ( x )  x 2
d
5
5 d
 2 x  3   2 x  3 x 2
dx
dx
 x  5 2 x  3  2    2 x  3  2  x
4
2
 10 x 2  2 x  3  2 x  2 x  3
4
 2 x  2 x  3  5 x  2 x  3
4
 2 x  2 x  3  7 x  3
4
Example 3, page 185
5
5
The General Power Rule
Practice Examples:
 Find the derivative of f ( x ) 
4x
Solution
1
2
 7
2
f ( x)   4 x  7 
2
 Rewrite as a power function:
 Apply the general power rule:
f ( x )  2  4 x  7 
2

Example 5, page 186
16 x
4x
2
 7
3
3
8 x 
2
The General Power Rule
Practice Examples:
3
2
x

1


 Find the derivative of f ( x )  

3
x

2


Solution
 Apply the general power rule and the quotient rule:
 2x  1  d  2x  1 

f ( x)  3 



3
x

2
dx
3
x

2




2
 2x  1 
 3

3
x

2


2
  3x  2  2    2 x  1 3 


2
 3x  2 


2


 2 x  1  6 x  4  6 x  3 3  2 x  1
 3

 
2
4
 3x  2    3x  2 
3
x

2

 
2
Example 6, page 186
Applied Problem: Arteriosclerosis
 Arteriosclerosis begins during childhood when plaque
forms in the arterial walls, blocking the flow of blood
through the arteries and leading to heart attacks, stroke
and gangrene.
Applied Example 8, page 188
Applied Problem: Arteriosclerosis
 Suppose the idealized cross section of the aorta is circular
with radius a cm and by year t the thickness of the plaque is
h = g(t) cm
then the area of the opening is given by
A = p (a – h)2 cm2
 Further suppose the radius of an individual’s artery is 1 cm
(a = 1) and the thickness of the plaque in year t is given by
h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm
Applied Example 8, page 188
Applied Problem: Arteriosclerosis
 Then we can use these functions for h and A
h = g(t) = 1 – 0.01(10,000 – t2)1/2
A = f(h) = p (1 – h)2
to find a function that gives us the rate at which A is
changing with respect to time by applying the chain rule:
dA dA dh


 f ( h )  g (t )
dt dh dt


1
2 1/2
 2p (1  h )( 1)  0.01   10,000  t  ( 2t ) 
2




0.01
t

 2p (1  h ) 
1/2
 10,000  t 2  


0.02p (1  h )t

10,000  t 2
Applied Example 8, page 188
Applied Problem: Arteriosclerosis
 For example, at age 50 (t = 50),
h  g (50)  1  0.01(10,000  2500)1/2  0.134
 So that
dA 0.02p (1  0.134)50

 0.03
dt
10,000  2500
 That is, the area of the arterial opening is decreasing at the
rate of 0.03 cm2 per year for a typical 50 year old.
Applied Example 8, page 188
3.4
Marginal Functions in Economics
E ( p)  
Percentage
change in quantity
demanded
Percentage
change in price
 f ( p  h)  f ( p ) 
100 


f ( p)


h
 p  100 
 
Marginal Analysis
 Marginal analysis is the study of the rate of change of
economic quantities.
 These may have to do with the behavior of costs, revenues,
profit, output, demand, etc.
 In this section we will discuss the marginal analysis of
various functions related to:
✦ Cost
✦ Average Cost
✦ Revenue
✦ Profit
✦ Elasticity of Demand
Applied Example: Rate of Change of Cost Functions
 Suppose the total cost in dollars incurred each week by
Polaraire for manufacturing x refrigerators is given by the
total cost function
C(x) = 8000 + 200x – 0.2x2
(0  x  400)
a. What is the actual cost incurred for manufacturing the
251st refrigerator?
b. Find the rate of change of the total cost function with
respect to x when x = 250.
c. Compare the results obtained in parts (a) and (b).
Applied Example 1, page 194
Applied Example: Rate of Change of Cost Functions
Solution
a. The cost incurred in producing the 251st refrigerator is
C(251) – C(250) = [8000 + 200(251) – 0.2(251)2]
– [8000 + 200(250) – 0.2(250)2]
= 45,599.8 – 45,500
= 99.80
or $99.80.
Applied Example 1, page 194
Applied Example: Rate of Change of Cost Functions
Solution
b. The rate of change of the total cost function
C(x) = 8000 + 200x – 0.2x2
with respect to x is given by
C´(x) = 200 – 0.4x
So, when production is 250 refrigerators, the rate of
change of the total cost with respect to x is
C´(x) = 200 – 0.4(250)
= 100
or $100.
Applied Example 1, page 194
Applied Example: Rate of Change of Cost Functions
Solution
c. Comparing the results from (a) and (b) we can see they are
very similar: $99.80 versus $100.
✦ This is because (a) measures the average rate of change
over the interval [250, 251], while (b) measures the
instantaneous rate of change at exactly x = 250.
✦ The smaller the interval used, the closer the average rate
of change becomes to the instantaneous rate of change.
Applied Example 1, page 194
Applied Example: Rate of Change of Cost Functions
Solution
 The actual cost incurred in producing an additional unit
of a good is called the marginal cost.
 As we just saw, the marginal cost is approximated by the
rate of change of the total cost function.
 For this reason, economists define the marginal cost
function as the derivative of the total cost function.
Applied Example 1, page 194
Applied Example: Marginal Cost Functions
 A subsidiary of Elektra Electronics manufactures a
portable music player.
 Management determined that the daily total cost of
producing these players (in dollars) is
C(x) = 0.0001x3 – 0.08x2 + 40x + 5000
where x stands for the number of players produced.
a. Find the marginal cost function.
b. Find the marginal cost for x = 200, 300, 400, and 600.
c. Interpret your results.
Applied Example 2, page 195
Applied Example: Marginal Cost Functions
Solution
a. If the total cost function is:
C(x) = 0.0001x3 – 0.08x2 + 40x + 5000
then, its derivative is the marginal cost function:
C´(x) = 0.0003x2 – 0.16x + 40
Applied Example 2, page 195
Applied Example: Marginal Cost Functions
Solution
b. The marginal cost for x = 200, 300, 400, and 600 is:
C´(200) = 0.0003(200)2 – 0.16(200) + 40 = 20
C´(300) = 0.0003(300)2 – 0.16(300) + 40 = 19
C´(400) = 0.0003(400)2 – 0.16(400) + 40 = 24
C´(600) = 0.0003(600)2 – 0.16(600) + 40 = 52
or $20/unit, $19/unit, $24/unit, and $52/unit, respectively.
Applied Example 2, page 195
Applied Example: Marginal Cost Functions
Solution
c. From part (b) we learn that at first the marginal cost is
decreasing, but as output increases, the marginal cost
increases as well.
This is a common phenomenon that occurs because of
several factors, such as excessive costs due to overtime and
high maintenance costs for keeping the plant running at
such a fast rate.
Applied Example 2, page 195
Applied Example: Marginal Revenue Functions
 Suppose the relationship between the unit price p in
dollars and the quantity demanded x of the Acrosonic
model F loudspeaker system is given by the equation
p = – 0.02x + 400
(0  x  20,000)
a. Find the revenue function R.
b. Find the marginal revenue function R′.
c. Compute R′(2000) and interpret your result.
Applied Example 5, page 199
Applied Example: Marginal Revenue Functions
Solution
a. The revenue function is given by
R(x) = px
= (– 0.02x + 400)x
= – 0.02x2 + 400x
Applied Example 5, page 199
(0  x  20,000)
Applied Example: Marginal Revenue Functions
Solution
b. Given the revenue function
R(x) = – 0.02x2 + 400x
We find its derivative to obtain the marginal revenue
function:
R′(x) = – 0.04x + 400
Applied Example 5, page 199
Applied Example: Marginal Revenue Functions
Solution
c. When quantity demanded is 2000, the marginal revenue
will be:
R′(2000) = – 0.04(2000) + 400
= 320
Thus, the actual revenue realized from the sale of the
2001st loudspeaker system is approximately $320.
Applied Example 5, page 199
Applied Example: Marginal Profit Function
 Continuing with the last example, suppose the total cost
(in dollars) of producing x units of the Acrosonic model F
loudspeaker system is
C(x) = 100x + 200,000
a. Find the profit function P.
b. Find the marginal profit function P′.
c. Compute P′ (2000) and interpret the result.
Applied Example 6, page 199
Applied Example: Marginal Profit Function
Solution
a. From last example we know that the revenue function is
R(x) = – 0.02x2 + 400x
✦ Profit is the difference between total revenue and total
cost, so the profit function is
P(x) = R(x) – C(x)
= (– 0.02x2 + 400x) – (100x + 200,000)
= – 0.02x2 + 300x – 200,000
Applied Example 6, page 199
Applied Example: Marginal Profit Function
Solution
b. Given the profit function
P(x) = – 0.02x2 + 300x – 200,000
we find its derivative to obtain the marginal profit
function:
P′(x) = – 0.04x + 300
Applied Example 6, page 199
Applied Example: Marginal Profit Function
Solution
c. When producing x = 2000, the marginal profit is
P′(2000) = – 0.04(2000) + 300
= 220
Thus, the profit to be made from producing the 2001st
loudspeaker is $220.
Applied Example 6, page 199
Elasticity of Demand
 Economists are frequently concerned with how strongly do
changes in prices cause quantity demanded to change.
 The measure of the strength of this reaction is called the
elasticity of demand, which is given by
percentage change in quantity demanded
E ( p)  
percentage change in price
Note: Since the ratio is negative, economists use the negative
of the ratio, to make the elasticity be a positive number.
Elasticity of Demand
 Suppose the price of a good increases by h dollars from p to
(p + h) dollars.
 The percentage change of the price is
Percentage
change in price
=
Change in price
Price
100  
h
100 
p
 The percentage change in quantity demanded is
Percentage
change in quantity
demanded

Change in quantity
demanded
Quantity demanded
at price p
100
 f ( p  h)  f ( p) 

100

f ( p)


Elasticity of Demand
 One good way to measure the effect that a percentage
change in price has on the percentage change in the
quantity demanded is to look at the ratio of the latter to the
former. We find
E ( p) 
Percentage
change in quantity
demanded
Percentage
change in price
f ( p  h)  f ( p )
f ( p)

h
p
 f ( p  h)  f ( p ) 
100 


f ( p)



h
 p  100 
 
f ( p  h)  f ( p )
h

f ( p)
p
Elasticity of Demand
 We have
f ( p  h)  f ( p )
h
E ( p) 
f ( p)
p
 If f is differentiable at p, then, when h is small,
f ( p  h)  f ( p)
 f ( p )
h
 Therefore, if h is small, the ratio is approximately equal to
f ( p ) pf ( p )
E ( p) 

f ( p)
f ( p)
p
 Economists call the negative of this quantity the elasticity of demand.
Elasticity of Demand
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
pf ( p )
E ( p)  
f ( p)
Note: Since the ratio is negative, economists use the negative
of the ratio, to make the elasticity be a positive number.
Applied Example: Elasticity of Demand
 Consider the demand equation for the Acrosonic model F
loudspeaker system:
p = – 0.02x + 400
(0  x  20,000)
a. Find the elasticity of demand E(p).
b. Compute E(100) and interpret your result.
c. Compute E(300) and interpret your result.
Applied Example 7, page 201
Applied Example: Elasticity of Demand
Solution
a. Solving the demand equation for x in terms of p, we get
x = f(p) = – 50p + 20,000
From which we see that
f ′(p) = – 50
Therefore,
pf ( p )
p( 50)
E ( p)  

f ( p)
50 p  20,000
p

400  p
Applied Example 7, page 201
Applied Example: Elasticity of Demand
Solution
b. When p = 100 the elasticity of demand is
100 

1
E (100) 

400  100  3
✦ This means that for every 1% increase in price we can
expect to see a 1/3% decrease in quantity demanded.
✦ Because the response (change in quantity demanded) is
less than the action (change in price), we say demand is
inelastic.
✦ Demand is said to be inelastic whenever E(p) < 1.
Applied Example 7, page 201
Applied Example: Elasticity of Demand
Solution
c. When p = 300 the elasticity of demand is
300 

E (300) 
3
400   300 
✦ This means that for every 1% increase in price we can
expect to see a 3% decrease in quantity demanded.
✦ Because the response (change in quantity demanded) is
greater than the action (change in price), we say demand
is elastic.
✦ Demand is said to be elastic whenever E(p) > 1.
✦ Finally, demand is said to be unitary whenever E(p) = 1.
Applied Example 7, page 201
3.5
Higher Order Derivatives
2 4
8 7/3
8
f ( x )      x 7/3 
x 
9 3
27
27 x 2 3 x
dv d  ds  d 2 s d
a
    2  8t   8
dt dt  dt  dt
dt
Higher-Order Derivatives
 The derivative f ′ of a function f is also a function.
 As such, f ′ may also be differentiated.
 Thus, the function f ′ has a derivative f ″ at a point x in the
domain of f if the limit of the quotient
f ( x  h )  f ( x )
h
exists as h approaches zero.
 The function f ″ obtained in this manner is called the
second derivative of the function f, just as the derivative f ′
of f is often called the first derivative of f.
 By the same token, you may consider the third, fourth,
fifth, etc. derivatives of a function f.
Higher-Order Derivatives
Practice Examples:
 Find the third derivative of the function f(x) = x2/3 and
determine its domain.
Solution
2  1  4/3
2 4/3
2 1/3
 We have f ( x )  x
and f ( x )     x   x
3
3  3
9
 So the required derivative is
2  4  7/3 8 7/3
8

f ( x)      x 
x 
9 3
27
27 x 7/3
 The domain of the third derivative is the set of all real
numbers except x = 0.
Example 1, page 208
Higher-Order Derivatives
Practice Examples:
 Find the second derivative of the function f(x) = (2x2 +3)3/2
Solution
 Using the general power rule we get the first derivative:
1/2
1/2
3
2
2
f ( x )   2 x  3  4 x   6 x  2 x  3
2
Example 2, page 209
Higher-Order Derivatives
Practice Examples:
 Find the second derivative of the function f(x) = (2x2 +3)3/2
Solution
 Using the product rule we get the second derivative:
1/2
1/2
d
d
2
2
f ( x )  6 x   2 x  3   2 x  3   6 x 
dx
dx
1/2
1/2
1
2
2
 6 x     2 x  3  4 x    2 x  3   6
2
 12 x  2 x  3
2
2
 6  2 x  3
2

Example 2, page 209
1/2
6  4 x 2  3
2 x2  3
1/2
 6  2 x  3
2
1/2
 2 x 2   2 x 2  3 


Applied Example: Acceleration of a Maglev
 The distance s (in feet) covered by a maglev moving along
a straight track t seconds after starting from rest is given
by the function
s = 4t2
(0  t  10)
 What is the maglev’s acceleration after 30 seconds?
Solution
 The velocity of the maglev t seconds from rest is given by
v
ds d
  4t 2   8t
dt dt
 The acceleration of the maglev t seconds from rest is given
by the rate of change of the velocity of t, given by
d
d  ds  d 2 s d
a  v     2  8t   8
dt
dt  dt  dt
dt
or 8 feet per second per second (ft/sec2).
Applied Example 4, page 209
3.6
Implicit Differentiation and Related Rates
Rocket
x
Spectator
Launch Pad
4000 ft
y
Differentiating Implicitly
 Up to now we have dealt with functions in the form





y = f(x)
That is, the dependent variable y has been expressed
explicitly in terms of the independent variable x.
However, not all functions are expressed explicitly.
For example, consider
x2 y + y – x2 + 1 = 0
This equation expresses y implicitly as a function of x.
Solving for y in terms of x we get
( x 2  1) y  x 2  1
x2  1
y  f ( x)  2
x 1
which expresses y explicitly.
Differentiating Implicitly
 Now, consider the equation




y4 – y3 – y + 2x3 – x = 8
With certain restrictions placed on y and x, this equation
defines y as a function of x.
But in this case it is difficult to solve for y in order to
express the function explicitly.
How do we compute dy/dx in this case?
The chain rule gives us a way to do this.
Differentiating Implicitly
 Consider the equation y2 = x.
 To find dy/dx, we differentiate both sides of the equation:
d
d
2
y    x

dx
dx
 Since y is a function of x, we can rewrite y = f(x) and find:
d
d
2
2
y    f ( x )

dx
dx
 2 f ( x ) f ( x )
dy
 2y
dx
Example 1, page 216
Using the chain rule
Differentiating Implicitly
 Therefore the above equation is equivalent to:
dy
2y
1
dx
 Solving for dy/dx yields:
dy
1

dx 2 y
Example 1, page 216
Steps for Differentiating Implicitly
 To find dy/dx by implicit differentiation:
1. Differentiate both sides of the equation with
respect to x.
(Make sure that the derivative of any term
involving y includes the factor dy/dx)
2. Solve the resulting equation for dy/dx in
terms of x and y.
Differentiating Implicitly
Examples
 Find dy/dx for the equation y 3  y  2 x 3  x  8
Solution
 Differentiating both sides and solving for dy/dx we get
d
d
3
3
y

y

2
x

x

 8


dx
dx
d
d
d
d
d
3
3
y    y    2 x    x    8

dx
dx
dx
dx
dx
dy dy
3y

 6x2  1  0
dx dx
dy
2
2
3
y

1

1

6
x


dx
dy 1  6 x 2
 2
dx 3 y  1
2
Example 2, page 216
Differentiating Implicitly
Examples
 Find dy/dx for the equation x 2 y 3  6 x 2  y  12
 Then, find the value of dy/dx when y = 2 and x = 1.
Solution
d 2 3
d
d
d
2
x
y

6
x

y



12 




dx
dx
dx
dx
x2 
d
dy
3
3 d
2
y

y

x

12
x





dx
dx
dx
3x 2 y 2
dy
dy
 2 xy 3  12 x 
dx
dx
dy
 3x y  1 dx  2 xy 3  12 x
dy 2 xy 3  12 x

dx 1  3x 2 y 2
2
Example 4, page 217
2
Differentiating Implicitly
Examples
 Find dy/dx for the equation x 2 y 3  6 x 2  y  12
 Then, find the value of dy/dx when y = 2 and x = 1.
Solution
 Substituting y = 2 and x = 1 we find:
dy 2 xy 3  12 x

dx 1  3x 2 y 2
2(1)(2)3  12(1)

1  3(1)2 (2)2
16  12
1  12
28

11

Example 4, page 217
Differentiating Implicitly
Examples
 Find dy/dx for the equation
Solution
x2  y2  x2  5
d 2
d 2
d
2 1/2
x

y

x

 5




dx
dx
dx
1 2
dy 
2 1/2 
x

y
2
x

2
y



  2x  0
2
dx 

x  y
2

2 1/2
dy 

2
x

2
y

  4x
dx 

dy
2
2 1/2
x y
 2x  x  y 
dx
dy
2
2 1/2
y
 2x  x  y   x
dx
dy 2x  x  y

dx
y
2
Example 5, page 219

2 1/2
x
Related Rates
 Implicit differentiation is a useful technique for solving a
class of problems known as related-rate problems.
 Here are some guidelines to solve related-rate problems:
1. Assign a variable to each quantity.
2. Write the given values of the variables and their rate
of change with respect to t.
3. Find an equation giving the relationship between the
variables.
4. Differentiate both sides of the equation implicitly with
respect to t.
5. Replace the variables and their derivatives by the
numerical data found in step 2 and solve the equation
for the required rate of change.
Applied Example: Rate of Change of Housing Starts
 A study prepared for the National Association of Realtors
estimates that the number of housing starts in the
southwest, N(t) (in millions), over the next 5 years is related
to the mortgage rate r(t) (percent per year) by the equation
9n2 + r = 36
 What is the rate of change of the number of housing starts
with respect to time when the mortgage rate is 11% per
year and is increasing at the rate of 1.5% per year?
Applied Example 6, page 220
Applied Example: Rate of Change of Housing Starts
Solution
 We are given that r = 11% and dr/dt = 1.5 at a certain
instant in time, and we are required to find dN/dt.
1. Substitute r = 11 into the given equation:
9 N 2  11  36
25
N2 
9
5
N
3
Applied Example 6, page 220
(rejecting the
negative root)
Applied Example: Rate of Change of Housing Starts
Solution
 We are given that r = 11% and dr/dt = 1.5 at a certain
instant in time, and we are required to find dN/dt.
2. Differentiate the given equation implicitly on both
sides with respect to t:
d
d
d
2
9
N

r

 
 36


dt
dt
dt
dN dr
18 N
 0
dt dt
Applied Example 6, page 220
Applied Example: Rate of Change of Housing Starts
Solution
 We are given that r = 11% and dr/dt = 1.5 at a certain
instant in time, and we are required to find dN/dt.
3. Substitute N = 5/3 and dr/dt = 1.5 into this equation and
solve for dN/dt:
 5  dN
18  
 1.5  0
 3  dt
dN
30
 1.5
dt
dN
1.5

dt
30
dN
 0.05
dt
 Thus, at the time under consideration, the number of
housing starts is decreasing at rate of 50,000 units per year.
Applied Example 6, page 220
Applied Example: Watching a Rocket Launch
 At a distance of 4000 feet from the launch site, a spectator
is observing a rocket being launched.
 If the rocket lifts off vertically and is rising at a speed of
600 feet per second when it is at an altitude of 3000 feet,
how fast is the distance between the rocket and the
spectator changing at that instant?
Rocket
x
Spectator
Launch Pad
4000 ft
Applied Example 8, page 221
y
Applied Example: Watching a Rocket Launch
Solution
1. Let
y = altitude of the rocket
x = distance between the rocket and the spectator
at any time t.
2. We are told that at a certain instant in time
y  3000
and
dy
 600
dt
and are asked to find dx/dt at that instant.
Applied Example 8, page 221
Applied Example: Watching a Rocket Launch
Solution
3. Apply the Pythagorean theorem to the right triangle we
find that x 2  y 2  40002
Therefore, when y = 3000, x  30002  40002  5000
Rocket
x
Spectator
Launch Pad
4000 ft
Applied Example 8, page 221
y
Applied Example: Watching a Rocket Launch
Solution
4. Differentiate x 2  y 2  40002 with respect to t, obtaining
dx
dy
2x
 2y
dt
dt
5. Substitute x = 5000, y = 3000, and dy/dt = 600, to find
dx
2  5000   2  3000  600 
dt
dx
 360
dt
Therefore, the distance between the rocket and the
spectator is changing at a rate of 360 feet per second.
Applied Example 8, page 221
3.7
Differentials
y
T
f(x + x)
f(x)
y
dy
P
x + x
x
x
x
Increments
 Let x denote a variable quantity and suppose x changes
from x1 to x2.
 This change in x is called the increment in x and is denoted
by the symbol x (read “delta x”).
 Thus,
x = x2 – x1
Examples:
 Find the increment in x as x changes from 3 to 3.2.
Solution
 Here, x1 = 3 and x2 = 3.2, so
x = x2 – x1 = 3.2 – 3 = 0.2
Example 1, page 227
Increments
 Let x denote a variable quantity and suppose x changes
from x1 to x2.
 This change in x is called the increment in x and is denoted
by the symbol x (read “delta x”).
 Thus,
x = x2 – x1
Examples:
 Find the increment in x as x changes from 3 to 2.7.
Solution
 Here, x1 = 3 and x2 = 2.7, so
x = x2 – x1 = 2.7 – 3 = – 0.3
Example 1, page 227
Increments
 Now, suppose two quantities, x and y, are related by an
equation y = f(x), where f is a function.
 If x changes from x to x + x, then the corresponding
change in y is called the increment in y.
 It is denoted y and is defined by
y = f(x + x) – f(x)
y
f(x + x)
y
f(x)
x + x
x
x
Example 1, page 227
x
Example
 Let y = x3.
 Find x and y when x changes
a. from 2 to 2.01, and
b. from 2 to 1.98.
Solution
a. Here, x = 2.01 – 2 = 0.01
Next, y  f ( x  x )  f ( x )  f (2.01)  f (2)
 (2.01)3  23  8.120601  8  0.120601
b. Here, x = 1.98 – 2 = – 0.02
Next, y  f ( x  x )  f ( x )  f (1.98)  f (2)
 (1.98)3  23  7.762392  8  0.237608
Example 2, page 228
Differentials
 We can obtain a relatively quick and simple way of
approximating y, the change in y due to small change x.
 Observe below that near the point of tangency P, the
tangent line T is close to the graph of f.
 Thus, if x is small, then dy is a good approximation of y.
y
T
f(x + x)
f(x)
y
dy
P
x + x
x
x
x
Differentials
 Notice that the slope of T is given by dy/x (rise over run).
 But the slope of T is given by f ′(x), so we have
dy/x = f ′(x) or dy = f ′(x) x
 Thus, we have the approximation
y ≈ dy = f ′(x)x
 The quantity dy is called the differential of y.
y
T
f(x + x)
f(x)
y
dy
P
x + x
x
x
x
The Differential
 Let y = f(x) define a differentiable function x. Then
1. The differential dx of the independent variable x is
dx = x
2. The differential dy of the dependent variable y is
dy = f ′(x)x = f ′(x)dx
Example
 Approximate the value of
26.5 using differentials.
Solution
 Let’s consider the function y = f(x) = x .
 Since 25 is the number nearest 26.5 whose square root is
readily recognized, let’s take x = 25.
 We want to know the change in y, y, as x changes from
x = 25 to x = 26.5, an increase of x = 1.5.
 So we find
 1

1

y  dy  f ( x)x  
  1.5    1.5  0.15
 10 
 2 x x 25 
 Therefore,
26.5  25  y  0.15
26.5  25  0.15  5.15
Example 4, page 229
Applied Example: Effect of Speed on Vehicular Operating
 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 
v
dollars.
 Find the approximate change in the total operating cost
when the average speed is increased from 55 to 58 mph.
Applied Example 5, page 230
Applied Example: Effect of Speed on Vehicular Operating
Solution
 Total operating cost is given by
4500
C (v )  125  v 
v
 With v = 55 an v = dv = 3, we find
4500 

C  dC  C (v )dv   1  2 
 3
v  v 55

 4500 
 1 
  3  1.46
 3025 
so the total operating cost is found to decrease by $1.46.
 This might explain why so many independent truckers
often exceed the 55 mph speed limit.
Applied Example 5, page 230
End of
Chapter