Mathematics for Business
(Finance)
Instructor: Prof. Ken Tsang
Room E409-R11
Email: kentsang@uic.edu.hk
1
Chapter 2: Differentiation(微分): Basic Concepts
In this Chapter, we will encounter some
important concepts.
 The Derivative （导数）
 Product （乘法准则）and Quotient
Rules（除式准则）, Higher-Order
Derivatives（高次导数 ）
 The Chain Rule（链锁法则）
 Marginal Analysis（边际分析）,
Implicit Differentiation（隐函数求导）.
2
Section 2.1 The Derivative （导数）

Calculus is the mathematics of change, and the primary
tool for studying change is a procedure called
differentiation（微分）.

In this section, we will introduce this procedure and
examine some of its uses, especially in computing rates of
change.

Rate of changes, for example velocity, acceleration, the
rate of growth of a population, and many others, are
described mathematically by derivatives.
3
Example 1
If air resistance is neglected, an object dropped from a
great height will fall s(t )  16t 2 feet in t seconds. What
is the object’s instantaneous velocity after t=2 seconds?
Solution:
Average rate of change of s(t) over the time period [2,2+h]
by the ratio V  di s tan cetrave l e d s( 2  h)  s( 2)
ave
e l aps e dti m e
( 2  h)  2
16( 2  h) 2  16( 2) 2

 64  16h
h
Compute the instantaneous velocity by the limit
Vins  limVave  lim(64  16 h)  64
h 0
h 0
That is, after 2 seconds, the object is traveling at the rate
of 64 feet per second.
4
Rates of Change （变化率）
How to determine instantaneous rate of change or rate
of change of f(x) at x=c ?
Find the average rate of change of f(x) as x varies from
x=c to x=c+h
rateave
f ( x ) f (c  h)  f (c ) f (c  h)  f (c )



x
(c  h)  c
h
Compute the instantaneous rate of change of f(x) at x=c by
finding the limiting value of the average rate as h tends to 0
f (c  h)  f (c )
rateins  limrateave  lim
h 0
h 0
h
5
Rates of Change
(Linear function)
A linear function y(x)=mx+b changes at the constant rate m
with respect to the independent variable x. That is the rate of
change of y(x) is given by the slope or steepness of its graph
S l ope rateof ch an ge
ch an gei n y

ch an gei n x
y
y2  y1


x x2  x1
6
Rates of Change
(non-Linear function)
For the function that is
nonlinear, the rate of change
is not constant but varies
with x.
In particular, the rate of
change at x=c is given by the
steepness of the graph of f(x)
at the point (c,f(c)), which
can be measured by the slope
of the tangent line to the
graph at p.
7
The slope of secant line
The secant line（割线）: A line that intersects the curve at the
point x and point x+h
The average rate of change can be interpreted geometrically as the slope of the secant line from the point (x,f(x)) to
the point (x+h,f(x+h)).
8
The Derivative (导数)
A Difference Quotient for the function f(x): The express
f ( x  h)  f ( x )
h
The Derivative of a Function: The derivative of the
function f(x) with respect to x is the function f’(x) given
by
f ( x  h)  f ( x )
f ( x)  lim
h 0
h
The process of computing the derivative is called
differentiation. f(x) is differentiable at x=c if f’(x) exists
9
Example 2
Find the derivative of the function f ( x)  2x2 16x  35
Solution:
The difference quotient for f(x) is
f ( x  h)  f ( x) 2( x  h) 2  16( x  h)  35  2 x 2  16x  35

h
h
4 xh  2h 2  16h

 4 x  2h  16
h
Thus, the derivative of f(x) is the function
f ( x  h)  f ( x )
f ( x)  lim
 lim(4 x  2h  16)  4 x  16
h 0
h 0
h
10
Slope as a Derivative: The slope of the tangent
line to the curve y=f(x) at the point (c,f(c)) is
mtan  f (c)
Instantaneous Rate of Change as a Derivative:
The rate of change of f(x) with respect to x
when x=c is given by f’(c)
Remarks: Since the slope of the tangent line at (a,f(a))
is f’(a), the equation of the tangent line is
y  f (a)  f (a)(x  a)
The point-slope form
11
Example 3
What is the equation of the tangent line to the curve
y x
at the point where x=4 ?
Solution:
According to the definition of derivative, we have
f ( x  h)  f ( x )
xh  x
f ( x)  lim
 lim
h 0
h 0
h
h
( x  h  x )( x  h  x )
1
1
 lim
 lim

h 0
h 0
h( x  h  x )
xh  x 2 x
Thus, the slope of the tangent line to the curve at the
point where x=4 is f’(4)=1/4 . So the point-slope
form is y  2  1 ( x  4)  y  1 x  1
4
4
12
Example 4
Find the equation of the tangent line to the function
f ( x)  4x  8 x at x=16.
Solution:
We know that the equation of a tangent line is given by
So, we will need the derivative of the function
Now we need to evaluate the function and the derivative.
Thus, the point-slope form for a tangent line is:
y  32  3( x  16)  y  3x  16
13
Example 5
A manufacturer determines that when x thousand units of a
particular commodity are produced, the profit generated will be
p( x)  400x 2  6800x 12000
dollars. At what rate is profit changing with respect to the level of
production x when 9 thousand units are produced?
Solution:
We find that
p ( x  h)  p ( x )
p( x)  lim
h 0
h
 400( x  h) 2  6800( x  h)  12000  (400x 2  6800x  12000)
 lim
h 0
h
 400h 2  800hx  6800h
 lim
 800x  6800
h 0
h


14
Thus, when the level of production is x=9, the profit is changing
at the rate of p(9)  800(9)  6800 400dollars per
thousand units.
Which means that the tangent line to the profit curve y=p(x) is
sloped downward at point Q where x=9. Therefore, the profit
curve must be falling at Q and profit must be decreasing when
9 thousand units are being produced.
15
Significance of the Sign of the derivative f’(x): If the
function f is differentiable at x=c, then
f is increasing at x=c if f’(c)>0 and
f is decreasing at x=c if f’(c)<0
16
Alternative Derivative Notation
Given a function y=f(x) all of the following are equivalent
and represent the derivative of f(x) with respect to x.
If we want to evaluate the derivative at x=a all of the
following are equivalent
17
Differentiability（可微） and Continuity
（连续）
Continuity of a Differentiable Function: If the function f(x)
is differentiable at x=c, then it is also continuous at x=c.
Notice that a continuous function may not be differentiable
Graphs of four functions that are not differentiable at x=0.
18
Section 2.2 Techniques of
Differentiation
The Constant Rule: For any constant c, we have
d
c   0
dx
That is, the derivative of a constant is zero
Proof: Since f(x+h)=c for all x
f ( x)  lim
d
c  0
dx
h 0
f ( x  h)  f ( x )
cc
 lim
0
h

0
h
h
19
The Power Rule
For any real number n
d
[ x n ]  nx n 1
dx
In words, to find the derivative of xn, reduce the
exponent n of x by 1 and multiply your new power
of x by original exponent.
For Example
d
( x 3 )  3x 2
dx
1
1

d
d
1
( x) 
(x 2 ) 
x 2
dx
dx
2
20
The constant Multiple Rule If c is a constant and f(x)
is differentiable then so is cf(x) and
d
d
cf ( x)  c  f ( x)
dx
dx
That is, the derivative of a multiple is the multiple of the
derivative
For Example
d
d
4
(3 x )  3
( x 4 )  3(4 x 3 )  12 x 3
dx
dx
d 7 d
1 3 / 2 7  3 / 2
1/ 2
( )  (7 x )  7( x )  x
dx x dx
2
2
21
The Sum Rule: If f(x) and g(x) are differentiable then so is the sum
of s(x)=f(x)+g(x) and
d
d
d
[ f ( x)  g ( x)] 
[ f ( x)] 
[ g ( x)]
dx
dx
dx
That is, the derivative of a sum is the sum of the separate derivative
For Example
d
d
d
2
2
( x  7) 
(x ) 
(7)  2( x 3 )  0  2 x 3
dx
dx
dx
d
d
d
(2 x 5  3x 7 )  2
( x 5 )  3
( x 7 )  2(5 x 4 )  3( 7 x 8 )
dx
dx
dx
 10x 4  21x 8
22
Example 6
It is estimated that x months form now, the
population of a certain community will be
p( x)  x  20x  8000
2
a. At what rate will the population be changing with
respect to time 15 months from now?
b. By how much will the population actually change
during the 16th month?
23
Solution:
a. The rate of change of the population with respect to time
is the derivative of the population function. That is
Rate of change  p( x)  2 x  20
The rate of change of the population 15 months from now
will be p(15)  2(15)  20  50 people per month
b. The actual change in the population during the 16th month
is p(16)  p(15)  8576 8525 51 people
Note: Since the rate varies during the month, the actual
change in population during 16th month differs from the
monthly rate of the change at the beginning of the month.
24
The Derivative as a Rate of Change (Review)
25
Relative and Percentage Rates of Change
Relative and Percentage Rates of Change: The relative
rate of change of a quantity Q(x) with respect to x is
given by the ratio
Relativerateof
changeof Q(x)

 Q( x) dQ / dx
  Q( x)  Q

The corresponding percentage rate of change of Q(x)
with respect to x is
Percentagerate  100Q( x)
of changeof Q( x)  Q( x)


26
Example 7
The gross domestic product (GDP) of a certain country was
N (t )  t 2  5t  106 billion dollars t years after 1995.
a. At what rate was the GDP changing with respect to time in 2005?
b. At what percentage rate was the GDP changing with respect to
time in 2005?
Solution:
a. The rate of change of the GDP is the derivative N’(t)=2t+5.
The rate of change in 2005 was N’(10)=2(10)+5=25 billion
dollars per year.
b. The percentage rate of change of the GDP in 2005 was
100
N (10)
25
 100
 9.77% per year
N (10)
256
27
Example 8
3
2
Let p( x)  2x  3x 12x  5 , find all x where p’(x)>0, p’(x)=0 and
p’(x)<0.
Solution:
Based on the techniques of
differentiation, we have
p( x)  2(3 x 2 )  3(2 x)  12  0
 6 x 2  6 x  12
 6( x  1)(x  2)
So the solution is
p’(x)=0 at x=-1 and 2
p’(x)>0 at x<-1 and x>2
p’(x)<0 at -1<x<2
28
Rectilinear Motion: Motion of an object along a straightline in some way. If the position at time t of an object
moving along a straight line is given by s(t), then the object
has
ds
velocity
and
accelerati on
v(t )  s(t ) 
a(t )  v(t ) 
dt
dv
dt
The object is advancing when v(t)>0, retreating when v(t)<0,
and stationary when v(t)=0. It is accelerating when a(t)>0
and decelerating when a(t)<0
29
Example 9
The position at time t of an object moving along a line is given by
s(t )  t  6t  9t  5
3
2
a. Find the velocity of the object and discuss its motion between
times t=0 and t=4.
b. Find the total distance traveled by the object between times
t=0 and t=4.
c. Find the acceleration of the object and determine when the
object is accelerating and decelerating between times t=0 and
t=4.
30
Solution:
ds
a. The velocity is v(t )   3t 2  12t  9 . The object will be
dt
stationary when
v(t )  3t 2 12t  9  3(t 1)(t  3)  0
that is, at times t=1 and t=3. Otherwise, the object is either
advancing or retreating, as described in the following table.
Interval
0<t<1
1<t<3
Sign
of v(t)
Description
of Motion
+
Advancing
from s(0)=5 to s(1)=9
-
Retreating
from s(1)=9 to s(3)=5
+
Advancing
from s(3)=5 to s(4)=9
The motion of an object:
s(t )  t 3  6t 2  9t  5
3<t<4
31
b. The object travels from s(0)=5 to s(1)=9, then back to
s(3)=5, and finally to s(4)=9. Thus the total distance
traveled by the object is
D  | 9  5 |  | 5  9 |  | 9  5 |  12
  
0 t 1
1 t  3
3 t  4
c. The acceleration of the object is
dv
a( t ) 
 6t  12  6( t  2)
dt
The object will be accelerating [a(t)>0] when 2<t<4 and
decelerating [a(t)<0] when 0<t<2.
32
The Motion of a Projectile: An important example of
rectilinear motion .
Suppose an object is projected (e.g. thrown, fired, or
dropped) vertically in such a way that the only
acceleration acting on the object is the constant
downward acceleration g due to gravity (that means air
resistance is negligible). Thus, the height of the object
at time t is given by the formula
1 2
H (t )   gt  V0t  H 0
2
where H 0 andV0 are the initial height and velocity of the
object, respectively.
33
Example 10
Suppose a person standing at the top of a building 112 feet high
throws a ball vertically upward with an initial velocity of 96 ft/sec.
a. Find the ball’s height and velocity at time t.
b. When does the ball hit the ground and what is its impact velocity?
c. When is the velocity 0? What is the significance of this time?
d. How far does the ball travel during its flight?
34
Solution:
a. Since g  32, V0  96 and H0  112, the height of the ball above
the ground at time t is H (t )  16t 2  96t  112 feet. The
dH
velocity at time t is
v(t ) 
 32t  96 ft/sec
dt
b. The ball hits the ground when H=0. Solve the equation we find
t=7 and t=-1(discard). The impact velocity is v(7)=-128 ft/sec
(The negative sign means the direction of motion is different
with the direction of initial velocity. )
c. Set v(t)=0, solve t=3. Thus, the ball is at its highest point when
t=3 seconds.
d. The ball starts at H(0)=112 feet and rises to a maximum height
of H(3)=256. So
The total distance travelled=(256-112)+256=400 feet
35
Section 2.3 Product and Quotient Rules; Higher-Order
Derivative
The derivative of a product of functions is not the product of
separate derivative!! Similarly, the derivative of a quotient of
functions is not the quotient of separate derivative.
Suppose we have two function f(x)=x3 and g(x)=x6
36
The Product Rule（乘法准则） If the two functions f(x) and
g(x) are differentiable at x, then we have the derivative of the
product P(x)=f(x)g(x) is
d
 f ( x) g ( x)  f ( x) d [ g ( x)]  g ( x) d [ f ( x)]
dx
dx
dx
or equivalently,
( fg )  fg   gf 
Example 11
Using the product rule to find the derivative of the function
y  3 x 2 (2 x  x 2 )
Solution:
1
2 3
y( x)  x (2 x  x 2 )  3 x 2 (2  2 x)
3
2
5
2
5
2
5
4 3 2 3
10
8
 x  x  2x 3  2x 3 
x3  x3
3
3
3
3
37
Example 12
A manufacturer determines that t months after a new
product is introduced to the market, x(t )  t 2  3t hundred
units can be produced and then sold at a price of
3
2
p(t )  2t  30
dollars per unit .
a. Express the revenue R(t) for this product as a function
of time .
b. At what rate is revenue changing with respect to time
after 4 months? Is revenue increasing or decreasing at
this time?
38
Solution:
a. The revenue is given by
R(t )  x(t ) p(t )  (t 2  3t )(2t 3/ 2  30)
hundred dollars.
b. The rate of change of revenue R(t) with respect to time is given
by the derivative R’(t), which we find using the product rule:


d
d
 2t 3 / 2  30  (2t 3 / 2  30) [t 2  3t ]
dt
dt
3


 (t 2  3t )  2( t 1/ 2 )  (2t 3 / 2  30)[2t  3]
2


R(t )  (t 2  3t )
At time t=4, the revenue is changing at the rate R’(4)=-14
Thus, after 4 months, the revenue is changing at the rate of 14
hundred dollars per month. It is decreasing at that time since
R’(4) is negative.
39
The Quotient Rule（除式准则） If the two functions f(x) and g(x)
are differentiable at x, then the derivative of the quotient
Q(x)=f(x)/g(x) is given by d
d
d f ( x)
[
]
dx g ( x)
or equivalently,
g ( x)
dx
[ f ( x)]  f ( x)
dx
[ g ( x)]
2
g ( x)
if g ( x)  0
f '
gf '  fg '
( ) 
g
g2
Example 13
Using the quotient rule to find the derivative of the function
Solution:
W ( z ) 
3(2  z )  (3z  9)(1)
15

2
(2  z )
(2  z ) 2
40
Example 14
A biologist models the effect of introducing a toxin to a bacterial
colony by the function
t 1
P (t ) 
t2  t  4
where P is the population of the colony (in millions) t hours
after the toxin is introduced.
a. At what rate is the population changing when the toxin is
introduced? Is the population increasing or decreasing at this
time?
b. At what time does the population begin to decrease? By how
much does the population increase before it begins to decline?
41
Solution:
a. The rate of change of the population with respect to time is given
d
d
by the derivative, that is
(t 2  t  4) [t  1]  (t  1) [t 2  t  4]
P(t ) 
dt
(t 2  t  4) 2
dt
(t 2  t  4)(1)  (t  1)(2t  1)  t 2  2t  3

 2
2
2
(t  t  4)
(t  t  4) 2
That is the population is initially changing at the rate of
P’(0)=0.1875 million bacteria per hour, and it is increasing since
P’(0)>0.
b. Since the numerator of P’(t) can be factored as –(t-1)(t+3), the
denominator and the factor t+3 are both positive for all t≥0,
which means that for 0≤t<1 P(t) is increasing, and for t>1, P(t)
is decreasing. Therefore, before the population begins to decline,
it increases by P(1)-P(0)=1/3-1/4=1/12 million.
42
A Word of Advice: The quotient rule is somewhat cumbersome,
so don’t use it unnecessarily.
Example 15
Differentiate the function
y
2 x 4 x 1
  
3x 2 3 5 x
.
Solution:
Don’t use the quotient rule! Instead, rewrite the function as
y
2 2 1
4
x  x   1  x 1
3
3
5
and then apply the power rule term by term to get
dy 2
1
 ( 2 x 3 )   0  0  ( 1) x  2
dx 3
3
4
1
4
1 1
  x 3   x  2   3   2
3
3
3x
3 x
43
The Second Derivative （二次导数）

In applications, it may be necessary to compute the rate of change
of a function that is itself a rate of change. For example, the
acceleration of a car is the rate of change with respect to time of
its velocity.
The second derivative of a function is the derivative
of its derivative. If y=f(x), the second derivative is
denoted by
2
d y
or f ( x)
2
dx
The second derivative gives the rate of change of the
rate of change of the original function.
Note: The derivative f’(x) is called the first
derivative.
44
Alternate Notation: There is some alternate notation for higher
order derivatives as well.
df
d2 f
f ( x) 
dx
f ( x) 
dx2
Note: Before computing the second derivative of a function, always
take time to simplify the first derivative as much as possible.
Example 16
An efficiency study of the morning shift at a certain factory
indicates that an average worker who arrives on the job at 8:00
A.M. will have produced Q(t )  t 3  6t 2  24t units t hours later.
a. Compute the worker’s rate of production at 11:00 A.M.
b. At what rate is the worker’s rate of production changing with
respect to time at 11:00 A.M.?
45
Solution:
a. The worker’s rate of production is the first derivative
R(t )  Q(t )  3t 2  12t  24
of the output Q(t). The rate of production at 11:00 A.M. (t=3) is
R(3)  Q(3)  33 units per hour
b. The rate of change of the rate of production is the second
derivative R(t )  Q(t )  6t  12 of the output function. At 11:00 A.M.,
this rate is R(3)  Q(3)  6 units per hour per hour
The minus sign indicates that the worker’s rate of production is
decreasing; that is, the worker is slowing down. The rate of this
decrease in efficiency at 11:00 A.M. is 6 units per hour per hour.
46
The acceleration a(t) of an object moving along a straight line is
the derivative of the velocity v(t). Thus, the acceleration may be
d 2s
thought of as the second derivative of position; that is a (t )  2
dt
Example 17
If the position of an object moving along a straight line is given
3
2
by s(t )  t  3t  4 at time t, find its velocity and acceleration.
Solution:
ds
 3t 2  6t  4
The velocity of the object is v(t ) 
dt
and its acceleration is
dv d 2 s
a(t ) 

 6t  6
2
dt
dt
47
The nth Derivative: For any positive integer n, the nth derivative
of a function is obtained from the function by differentiating
successively n times. If the original function is y=f(x) the nth
derivative is denoted by
dny
dxn
or
f
(n)
( x)
Example 18
Find the fifth derivative of the function y=1/x
Solution:
dy

dx
d3y

3
dx
d5y

5
dx
d
1
d2y
d
2
1
2
2
3
(x )  x   2

(

x
)

2
x

dx
x
dx2
dx
x3
d
6
d4y
d
24
3
4
4
5
(2 x )  6 x   4

(

6
x
)

24
x

dx
x
dx4
dx
x5
d
120
(24x 5 )  120x 6   6
dx
x
48
Section 2.4 The Chain Rule 链锁法则
Suppose the total manufacturing cost at a certain factory is a function
of the number of units produced, which in turn is a function of the
number of hours the factory has been operating. If C, q, t, denote the
cost, units produced and time respectively, then
dC rateof changeof cost

dq with respect tooutput
(dollars per unit)
dq rat eof changeof output

(unit sper hour)
dt  with respect totime 
The product of these two rates is the rate of change of cost with
respect to time that is
dC dC dq

dt
dq dt
(dollars per hour)
49
The Chain Rule: If y=f(u) is a differentiable function of
u and u=g(x) is in turn a differentiable function of x,
then the composite function y=f(g(x)) is a differentiable
function of x whose derivative is given by the product
dy
dy du

dx du dx
or, equivalently, by
dy
 f ( g ( x )) g ( x )
dx
Note: One way to remember the chain rule is to pretend
the derivative dy/du and du/dx are quotients and to
“cancel” du.
50
Example 19
dy
Find
if y  ( x 2  2)3  3( x 2  2)2  1
dx
Solution:
We rewrite the function as y  u 3  3u 2  1,
where u  x 2  2 . Thus,
dy
 3u 2  6u
du
and
du
 2x
dx
and according to the chain rule,
dy dy du

 (3u 2  6u )(2 x)
dx du dx
 [3( x 2  2) 2  6( x 2  2)](2 x)
replace u with x 2  2
 3( x 2  2)[x 2 ](2 x)  6 x 3 ( x 2  2)
51
Example 20
The cost of producing x units of a particular commodity
is
1 2
C ( x )  x  4 x  53
3
dollars, and the production level t hours into a particular
production run is
x(t )  0.2t 2  0.03t units
At what rate is cost changing with respect to time after 4
hours?
52
Solution:
We find that
dC
2
 x4
dx
3
and
dx
 0.4t  0.03
dt
So according to the chain rule, we have
dC dC dx  2


  x  4 (0.4t  0.03)
dt
dx dt  3

When t=4, the level of production is x(4)=3.32 units,
and by substituting t=4 and x=3.32 into the formula for
dC
dC
dt , we get dt
t 4
2

  (3.32)  4[0.4(4)  0.03]  10.1277
3

Thus, after 4 hours, cost is increasing at the rate of
approximately $10.13 per hour.
53
Let’s look at the functions
then we can write the function as a composition.
differentiate a composition function using the Chain Rule.
The derivative is then
54
In general, we differentiate the outside function
leaving the inside function alone and multiple all
of this by the derivative of the inside function
F ( x) 
( x)
f
( g ( x))
g


 

 
derivativeof
outside function
leaveinside
function alone
derivativeof
inside function
Exercise
a. Based on the chain rule, find the derivative of
f ( x)  ( x 2  x)3 with respect to x
b. Find the derivative of f ( x) 
1
1 x2 1
55
The General Power Rule: For any real number n and
differentiable function h, we have
d
n
n 1 d
[h( x)]  n[h( x)]
[h( x)]
dx
dx
Think of [h(x)]n as the composite function
[h( x)]n  g[h( x)] where g  u n
By the chain rule, we have
d
d
n
n 1 d
[h( x)] 
g[h( x)]  g [h( x)]h( x)  n[h( x)]
[h( x)]
dx
dx
dx
56
Example 21
An environmental study of a certain suburban
community suggests that the average daily level
of carbon monoxide in the air will be
c ( p )  0.5 p  17 parts per million when the
population is p thousand. It is estimated that t
years from now, the population of the
community will be p(t )  3.1  0.1t 2 thousand.
At what rate will the carbon monoxide level be
changing with respect to time 3 years from now?
2
57
Solution:
The goal is to find dc/dt when t=3. Since
1
1


dc 1
1
 (0.5 p 2  17) 2 [0.5(2 p)]  p(0.5 p 2  17) 2
dp 2
2
and
dp
 0.2t
dt
it follows from the chain rule that
1

dc dc dp 1
0.1 pt
2

 p (0.5 p  17 ) 2 (0.2t ) 
dt dp dt 2
0.5 p 2  17
When t=3, p(3)=4 and substituting t=3 and p=4 into the formula
for dc/dt, we get dc
0.1(4)(3)
1.2
dt

0.5(4)  17
2

5
 0.24 partsper millionper year
58
Section 2.5 Marginal Analysis and Approximations Using
Increments
In economics, the use of the derivative to approximate
the change in a quantity that results from a 1-unit
increase in production is called marginal analysis 边际分析
For instance, suppose C(x) is the total cost of producing x units of
a particular commodity. If x0 units are currently being produced,
then the derivative
C ( x0  h)  C ( x0 )
C ( x0 )  lim
h 0
h
is called the marginal cost of producing x0 units. The limiting
value that defines this derivative is approximately equal to the
difference quotient of C(x) when h=1; that is
C ( x0  1)  C ( x0 )
C ( x0 ) 
 C ( x0  1)  C ( x0 )
1
59
Marginal Cost（边际成本）: If C(x) is the total cost of
producing x units of a commodity. Then the marginal cost
of producing x0 units is the derivative C( x0 ) , which
approximates the additional cost C( x0  1)  C( x0 ) incurred
when the level of production is increased by one unit,
from x0 to x0 +1, assuming x0 >>1.
60
Marginal Revenue （边际收入） and Marginal
Profit（边际利润）:
Suppose R(x) is the revenue function generated
when x units of a particular commodity are produced,
and P(x) is the corresponding profit function, when
x=x0 units are being produced, then
The marginal revenue is R( x0 ), it approximates
R( x0  1)  R( x0 ) , the additional revenue generated
by producing one more unit.
The marginal profit is P ( x0 ) , it approximates
P( x0  1)  P( x0 ) , the additional profit obtained by
producing one more unit, assuming x0 >>1.
61
Example 22
A manufacturer estimates that when x units of a particular commodity
are produced, the total cost will be C ( x)  1 x 2  3x  98 dollars, and
8
1
furthermore, that all x units will be sold when the price is p( x)  (75  x)
3
dollars per unit.
a. Find the marginal cost and the marginal revenue.
b. Use marginal cost to estimate the cost of producing the ninth unit.
c. What is the actual cost of producing the ninth unit?
d. Use marginal revenue to estimate the revenue derived from the
sale of the ninth unit.
e. What is the actual revenue derived from the sale of the ninth unit?
62
Solution:
a. The marginal cost is C’(x)=(1/4)x+3. The total revenue is
R( x)  xp( x)  x((75  x) / 3)  25x  x 2 / 3 , the marginal revenue is
R’(x)=25-2x/3.
b. The cost of producing the ninth unit is the change in cost as x
increases from 8 to 9 and can be estimated by the marginal cost
C’(8)=8/4+3=$5.
c. The actual cost of producing the ninth unit is C(9)-C(8)=$5.13
which is reasonably well approximated by the marginal cost C’(8)
d. The revenue obtained from the sale of the ninth unit is
approximated by the marginal revenue R’(8)=25-2(8)/3=$19.67
e. The actual revenue obtained from the sale of the ninth unit is
R(9)-R(8)=$19.33.
63
Marginal analysis is an important example
of a general Incremental approximation
procedure
64
Based on the fact that since
f ( x0 )  lim
h 0
f ( x0  h)  f ( x0 )
h
then for small h, the derivative f ( x0 ) is approximately
equal to the difference quotient f ( x  h)  f ( x ) . We indicate
h
this approximation by writing
0
f ( x0 ) 
f ( x0  h)  f ( x0 )
h
0
or, equivalent ly, f ( x0  h)  f ( x0 )  f ( x0 )h
Approximation by Increment If f(x) is differentiable
at x=x0 and △x is a small change in x, then
f ( x0  x)  f ( x0 )  f ( x0 )x
Or, equivalently, if
f  f ( x0  x )  f ( x0 )
, then
f  f ( x0 )x
65
Example 23
During a medical procedure, the size of a roughly tumor
is estimated by measuring its diameter and using the
formula V  4 R 3 to compute its volume. If the diameter
3
is measured as 2.5 cm with a maximum error of 2%, how
accurate is the volume measurement?
Solution:
A sphere of radius R and diameter x=2R has volume
4
4
x 3 1
1
3
3
V  R   ( )  x   ( 2.5) 3  8.181 cm 3
3
3
2
6
6
The error made in computing this volume using the diameter 2.5
when the actual diameter is 2.5+△x is V  V (2.5  x)  V (2.5)  V (2.5)x
To be continued
66
The corresponding maximum error in the calculation of
volume is
Maximumerrorin volume V  [V (2.5)]x
 [V (2.5)](0.02(2.5))
Since
V ( x) 
1
1
 (3x) 2  x 2
6
2
V (2.5) 
1
 (2.5) 2  9.817
2
it follows that
Maximumerrorin volume (9.817)(0.05)  0.491
3
cm
Thus, at worst, the calculation of the volume as 8.181
3
cm
is off by 0.491
, so the actual volume V must satisfy
7.690  V  8.672
67
The percentage change of a quantity expresses the change
in that quantity as a percentage of its size prior to the
change. In particular,
changein quant it y
Percent ageof change  100
size of quant it y
68
Example 24
The GDP of a certain country was N (t )  t 2  5t  200 billion dollars t
years after 1997. Use calculus to estimate the percentage change in
the GDP during the first quarter of 2005.
Solution:
Use the formula
N (t )t
Percentagechangein N  100
N (t )
with t=8, △t=0.25 and N’(t)=2t+5, to get
N (8)0.25
N (8)
[2(8)  5](0.25)
 100 2
 1.73%
(8)  5(8)  200
P ercent agechangein N  100
69
Differentials（微分）: The differential of x is dx=△x,
and if y=f(x) is a differentiable function of x, then
dy=f’(x)dx is the differential of y.
70
Example 25
In each case, find the differential of y=f(x).
a. f ( x)  x3  7 x 2  2
2
2
f
(
x
)

(
x

5
)(
3

x

2
x
)
b.
Solution:
a. dy  f ( x)dx  [3x2  7(2x)]dx  (3x2 14x)dx
b. By the product rule,
dy  f ( x)dx  [(x 2  5)(1  4 x)  (2 x)(3  x  2 x 2 )]dx
 (8 x 3  3x 2  14x  5)dx
71
Section 2.6 Implicit Differentiation（隐
函数求导） and Related Rates
So far the functions have all been given by equations of the form
y=f(x). A function in this form is said to be in explicit form. For
example, the functions
3
x
1
y  x 2  3x  1 y 
2x  3
and y  1  x 2
are all functions in explicit form（显式表达式）.
Sometimes practical problems will lead to equations in which the
function y is not written explicitly in terms of the independent
variable x. For example, the equations such as
x 2 y3  6  5 y3  xy and x2 y  2 y3  3x  2 y
are said to be in implicit form（隐式表达式）.
72
Example 26
Find dy/dx if x y  y  x
2
2
3
Solution:
We are going to differentiate both sides of the given equation with
respect to x. Firstly, we temporarily replace y by f(x) and rewrite the
x 2 f ( x)  ( f ( x))2. Secondly,
x3
equation as
we differentiate both
sides of this equation term by term with respect to x:
d 2
[ x f ( x)  ( f ( x))2 ] 
dx
d 2 
df
 2 df
 x dx  f ( x) dx ( x )  2 f ( x) dx 



 
d 2
[ x f ( x )]
dx
d
[( f ( x )) 2 ]
dx
d 3
[x ]
dx
3
x2
d 3
(x )
dx
To be continued
73
Thus, we have
df
df
df
2
x
 f ( x)(2 x)  2 f ( x)
 3x
gat her all
t erms
dx
dx
dx
df
2 df
x
 2 f ( x)
 3 x 2  2 xf ( x) on one side of t heequat ion
dx
dx
df
[ x 2  2 f ( x)]  3 x 2  2 xf ( x) combinet erms
dx
df 3 x 2  2 xf ( x)
df
 2
solve for
dx
x  2 f ( x)
dx
2
Finally, replace f(x) by y to get
dy
3 x 2  2 xy

dx
x2  2 y
74
Implicit Differentiation: Suppose an equation defines
y implicitly as a differentiable function of x. To find
df/dx
1. Differentiate both sides of equation with respect to x.
remember that y is really a function of x and use the
chain rule when differentiating terms containing y.
2. Solve the differentiated equation algebraically for
dy/dx.
Exercise
Find dy/dx by implicit differentiation where x3  y 3  xy
75
Computing the slope of a tangent line by implicit
differentiation
Example 27
Find the slope of the tangent line to the circle x 2  y 2  25 at
the point (3,4). What is the slope at the point (3,-4)?
Solution:
Differentiating both sides of the equation with respect to x, we have
2x  2 y
dy
dy
x
0

dx
dx
y
Thus, the slope at (3,4) is
The slope at (3,-4) is
dy
3

dx (3, 4)
4
dy
3 3


dx (3, 4)
4 4
76
Application to Economics
Example 28
Suppose the output at a certain factory is Q  2x3  x2 y  y3
units, where x is the number of hours of skilled labor used
and y is the number of hours of unskilled labor. The
current labor force consists of 30 hours of skill labor and
20 hours of unskilled labor.
Question: Use calculus to estimate the change in unskilled
labor y that should be made to offset a 1-hour increase in
skilled labor x so that output will be maintained at its
current level.
77
Solution:
If output is to be maintained at the current level, which is the value
of Q when x=30 and y=20, the relationship between skilled labor x
and unskilled labor y is given by the equation
80,000  Q(30,20)  2x3  x 2 y  y 3
The goal is to estimate the change in y that corresponds to a 1-unit
increase in x when x and y are related by above equation. As we
know, the change in y caused by a 1-unit increase in x can be
approximated by the derivative dy/dx. Using implicit differentiation,
dy
dy
we have
0  6x2  x2
 2 xy  3 y 2
dx
 (x2  3y2 )
dx
dy
 6 x 2  2 xy
dx
dy
6 x 2  2 xy
 2
dx
x  3y2
Now evaluate this derivative when x=30 and y=20 to conclude that
dy
Change in y 
dx
x 30
y  20
6(30) 2  2(30)(20)

 3.14 hours
(30) 2  3(20) 2
78
In certain practical problems, x and y are related
by an equation and can be regarded as a function
of a third variable t, which often represents time.
Then implicit differentiation can be used to relate
dx/dt to dy/dt. This kind of problem is said to
involve related rates.
A procedure for solving related rates problems
1. Find a formula relating the variables.
2. Use implicit differentiation to find how the rates are related.
3. Substitute any given numerical information into the equation in
step 2 to find the desired rate of change.
79
Example 29
The manager of a company determines that when
q hundred units of a particular commodity are
produced, the cost of production is C thousand
dollars, where C 2  3q3  4275. When 1500 units
are being produced, the level of production is
increasing at the rate of 20 units per week.
What is the total cost at this time and at what rate
is it changing?
80
Solution:
We want to find dC/dt when q=15 and dq/dt=0.2. Differentiating
the equation C 2  3q 3  4275implicitly with respect to time, we
get
dC
 2 dq 
2C
 33q
0

dt
dt 

so that
dC 9q 2 dq

dt
2C dt
When q=15, the cost C satisfies
C 2  3(15)3  4275 C 2  4275 3(15)3  14400 C  120
and by substituting q=15, C=120 and dq/dt=0.2 into the formula
for dC/dt, we obtain
dC  9(15) 2 

(0.2)  1.6875 thousand dollars per week.
dt  2(120) 
81
Example 30
A storm at sea has damaged an oil rig. Oil spills
from the rupture at the constant rate of 60 ft 3 / min ,
forming a slick that is roughly circular in shape
and 3 inches thick.
a. How fast is the radius of the slick increasing
when the radius is 70 feet?
b.Suppose the rupture is repaired in such a way that
flow is shut off instantaneously. If the radius of
the slick is increasing at the rate of 0.2 ft/min
when the flow stops. What is the total volume of
oil that spilled onto the sea?
82
Solution:
We can think of the slick as a cylinder of oil of radius r feet and
thickness h=3/12=0.25 feet. Such a cylinder will have volume
V  r 2 h  0.25r 2 ft3
Differentiating implicitly in this equation with respect to time t,
and since dV/dt=60 at all times, we get
dV
dr
dr
 dr 
 0.25  2r
 60  0.5r
  0.5r
dt
dt 
dt
dt

a. We want to find dr/dt when r=70, so that
dr
60

 0.55 ft /min
dt
(0.5) (70)
60
 191 feet
b. Since dr/dt=0.2 at that instant, we have r 
0.5 (0.2)
Therefore, the total amount of oil spilled is
V  0.25 (191) 2  28652ft3
83
Exercise
A lake is polluted by waste from a plant located on
its shore. Ecologists determine that when the level
of pollute is x parts per million (ppm), there will be
F fish of a certain species in the lake, where
32000
F 
3 x
When there are 4000 fish left in the lake, the
pollution is increasing at the rate of 1.4 ppm/year.
At what rate is the fish population changing at this
time?
84
Summary

Definition of the Derivative
f ( x  h)  f ( x )
f ( x)  lim
h 0
h
 Interpretation of the Derivative
Slope as a Derivative: The slope of the tangent line to
the curve y=f(x) at the point (c,f(c)) is mtan  f (c)
Instantaneous Rate of Change as a Derivative: The
rate of change of f(x) with respect to x when x=c is
given by f’(c)
85
 Sign
of The Derivative
If the function f is differentiable at x=c, then
f is increasing at x=c if f (c ) >0
f is decreasing at x=c if f (c ) <0
Techniques of Differentiation
d
d
d
d
[ x n ]  nx n 1
c   0
cf ( x)  c  f ( x)
dx
dx
dx
dx
d
d
d
[ f ( x)  g ( x)] 
[ f ( x)] 
[ g ( x)]
dx
dx
dx
d
 f ( x) g ( x)  f ( x) d [ g ( x)]  g ( x) d [ f ( x)] The Product Rule
dx
dx
dx
d
d
g ( x) [ f ( x)]  f ( x) [ g ( x)]
d f ( x)
dx
dx
[
]
if g ( x)  0 The Quotient Rule
2
dx g ( x)
g ( x)
86
The Chain Rule
dy
dy du

dx du dx
dy
 f ( g ( x )) g ( x )
dx
d
n
n 1 d
[h( x)]  n[h( x)]
[h( x)] The General Power Rule
dx
dx
The Higher -order Derivative
d2y
dx2

or
f ( x)
The Second Derivative
dny
(n)
or
f
( x) The nth Derivative
n
dx
Application of Derivative
Tangent line, Rectilinear Motion, Projectile Motion
87
Marginal Analysis and Approximation by increments
The marginal cost is C ( x0 ) , it approximates C( x0  1)  C( x0 ) ,
the additional cost generated by producing one more unit.
C ( x0  1)  C ( x0 )
C ( x0  h)  C ( x0 )

 C ( x0  1)  C ( x0 )  C ( x0 )  lim
h 0
1
h
f ( x0  x)  f ( x0 )  f ( x0 )x Approximation by Increment
Marginal Revenue
Marginal Profit
Implicit Differentiation and Related Rate
88