§2.3 Techniques of Differentiation
The derivative of a product of function is not the product
of separate derivative!!
Suppose we have two function f(x)=x3 and g(x)=x6
§2.3 Techniques of Differentiation
The Product Rule If the two functions f(x) and g(x)
are differentiable at x, then so is the product
P(x)=f(x)g(x) and
d
d
d
[ f ( x ) g( x )]  f ( x ) [ g( x )]  g( x ) [ f ( x )]
dx
dx
dx
( fg )'  fg '  gf '
§2.3 Techniques of Differentiation
The Quotient Rule If the two functions f(x) and g(x)
are differentiable at x, then so is the quotient
Q(x)=f(x)/g(x) and
d
d
g( x ) [ f ( x )]  f ( x ) [ g( x )]
d f ( x)
dx
dx
[
]
dx g( x )
g 2 ( x)
f ' gf '  fg '
( ) 
g
g2
if g( x )  0
Example: 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 p(t )  2t 3 / 2  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?
Solution: The revenue is given by
R(t )  x(t ) p(t )  (t 2  3t )(2t 3 / 2  30)
d
d 2
3/2
3/ 2
[2t  30]  ( 2t  30) [t  3t ]
dt
dt
3
 ( t 2  3t )[2( t 1 / 2 )]  ( 2t 3 / 2  30)[2t  3]
2
R ( t )  ( t 2  3 t )
R(4)  14
Decreasing
Example: A biologist models the effect of introducing a toxin
to a bacterial colony by the function
t 1
p( t )  2
t t4
Where p is the population of 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?
Solution: The rate of change of population with respect
to time is given by
p( t ) 
( t 2  t  4)
d
d
[t  1]  ( t  1) [t 2  t  4]
dt
dt
( t 2  t  4) 2
- t 2  2t  3
 2
( t  t  4) 2
The toxin is introduced when t=0, and at that time the population
is changing at the rate
p(0) 
( 0  0  3)
3

 0.1875
2
16
( 0  0  4)
The population is decreasing when P’(t)<0. Since
We can write
 t 2  2t  3  (t  1)(t  3)
- (t - 1)(t  3) For 0≤t<1 P’(t)>0 and P(t) is increasing
p( t )  2
for t>1
P’(t)<0 and P(t) is decreasing
( t  t  4) 2
§2.3 The second Derivative
Consider the following function
By differentiating this function, we get
This is a function and so it can be differentiated. Here is
the notation that we’ll use for that as well as the derivative.
This is called the second derivative and f ( x ) is now
called the first derivative
§2.3 The Higher -order Derivative
Alternate Notation
There is some alternate notation for higher order derivatives
as well.
Again,
This is a function as so we can differentiate it again. This
will be called the third derivative
Continuing,
.
§2.3 The Higher -order Derivative
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
dx n
f ( n) ( x )
1
y
x
dy d
1

( x 1 )   x  2   2
dx dx
x
3
d y d
6
3
4

(
2
x
)


6
x


dx 3 dx
x4
d2y
d
2
2
3

(

x
)

2
x

dx 2 dx
x3
d4y
d
24
4
5

( 6 x )  24x  5
4
dx
dx
x
§2.4 The Chain Rule
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 rate of change of cost 

dq  with respect to output 
(dollars per unit)
dq rate of change of output 


dt  with respect to time

(units per hour)
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)
§2.4 The Chain Rule (IMPORTANT)
The Chain Rule If y=f(u) is a differentiable function
of u and u=g(x) is in turn a differentiable 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
dy
Example : Find
if y  ( x 2  2) 3  3( x 2  2) 2  1
dx
Solution:
Note that y  u 3  3u 2  1 , where u  x 2  2
dy
du
2
and
 3u  6u
 2x
du
dx
And according to the chain rule,
dy dy du

 ( 3u 2  6u)(2 x )
dx du dx
dy
 ( 3u 2  6u)(2 x )
dx
 [3( x 2  2)2  6( x 2  2)](2 x ) Replace u with x2+2
 6 x 3 ( x 2 2 )
§2.4 The Chain Rule
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
In general, We differentiate the outside function leaving the
inside function alone and multiply all of this by the derivative of
the inside function,
§2.4 The General Power Rule
The General Power Rule
differentiable function h.
For any real number n and
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  un
g( u)  nu n1 and h( x ) 
By the chain rule
d
[h( x )]
dx
d
d
n
n 1 d


[h( x )] 
g[h( x )]  g [h( x )]h ( x )  n[h( x )]
[h( x )]
dx
dx
dx
Example:
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 2  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?
Solution:
dc
The goal is to find
when t=3. Since
dt
dc 1
1
 (0.5 2  17) 1 / 2 [0.5( 2 p)]  p(0.5 2  17) 1 / 2
dp 2
2
and
dp
 0.2t
dt
It follows from the chain rule that
dc dc dp 1
0.1 pt
2
1 / 2

 p(0.5  17) (0.2t ) 
dt dp dt 2
0.5 p 2  17
When t=3, p(3)=3.1+0.1(3)2=4, and so
dc

dt
0.1(4)(3)
0.5(4)2  17
 0.24 partspe r mil lionpe r ye ar
§2.5 Marginal Analysis
边际效益是经济学中的一个概念，它大体可以这样理解:
即一个市场中的经济实体为追求最大的利润，多次进行扩大生产，每一次投资所
产生的效益都会与上一次投资产生的效益之间要有一个差，这个差就是边际效益。
Example: 你肚子很饿了，你只有钱可以买5个馒头吃。
第一个馒头的边际效益最大,因为你那时候最饿，最需要，你多花一点钱也愿意买；
第二个的边际效益就递减了，因为有1个馒头进肚了...不是那么饿了。
第五个的边际效益最小，因为那个时候你几乎已经快饱了，馒头如果卖的贵的话，
你一定不会买了。
每支出1个馒头的价钱产生的效益，也就是你感觉花钱买来的价值。
从第一个向最后一个递减！这就是边际效益了。
假如用П表示厂商的利润，则П＝TR－TC。
那么利润最大化就可以用П＝TR－TC的最大化来表示。
什么时候总收益和总成本之差最大呢？答案是当边际收益等于边际成本时。
§2.5 Marginal Analysis
Suppose a business owner is operating a plant that manufactures a
certain product at a known level. Sometimes the business owner
will want to know how much it costs to produce one more
unit of this product.
Example 1: Suppose the total cost in dollars per week by
ABC Corporation for producing its best-selling product is
given by
Find the actual
cost of producing the 101st item
The cost of producing the (x + 1)st item can be found by computing
the average rate of change, that is by computing :
c(101)  c(100)
 ( 210939.7  209000) / 1  1939.7
101 100
Note that
where x = 100 and h = 1
This will give us the actual cost of producing the next item.
However, it is often inconvenient to use. For this reason, It
is usually approximated by the instantaneous rate of change of
the total cost function evaluated at the specific point of interest.
C ( x0 )  0.6 x  2000
C ( x 0  1)  C ( x 0 ) 
C (100)  1940  c(101)  c(100)  1939.7
101 100
C ( x 0  1)  C ( x 0 )
C ( x 0  h)  C ( x 0 )
 C ( x 0 )  lim
h 0
1
h
So, we’ll define the marginal cost function as the derivative
of the total cost function.
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
§2.5 Marginal Cost
 Marginal Cost: 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 ( x 0 ) , 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
§2.5 Marginal Analysis
Marginal Revenue and Marginal Profit:
Suppose R(x) is the revenue generated when x units
of a particular commodity are produced, and P(x) is
the corresponding profit. 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
Example: A manufacturer estimates that when x units of
a particular commodity are produced, the total cost will be
1 2
x  3 x  98 dollars, and furthermore, that all x units
8
1
will be sold when the Price is p( x )  (75  x ) dollars per unit
3
C ( x) 
a. Find the marginal cost and the marginal revenue.
The marginal cost is C ( x ) 
1
x  .3
4
R(x)=(number of units sold )(price per unit)
1
1 2
 xp( x )  x[ (75  x )]  25x  x
3
3
The marginal revenue is
2
R ( x )  25  x
3
b. Use marginal cost to estimate the cost of producing the ninth unit.
The cost of producing the ninth units is the change in cost as x
increases from 8 to 9 and can be estimated by the marginal cost
C ( 8 ) 
1
(8)  3  $5
4
c. What is the actual cost of producing the ninth unit?
C (9)  c(8)  $5.13
d. Use marginal revenue to estimate the revenue derived from
the sale of the ninth unit
2

R (8)  25  (8)  $19.67
3
e. What is the actual revenue derived from the sale of the ninth unit?
R(9)  R(8)  $19.33
§2.5 Marginal Analysis (Review)
 Marginal Cost: 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 ( x 0 ) , 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
C ( x 0  1)  C ( x 0 ) 
C ( x 0  1)  C ( x 0 )
C ( x 0  h)  C ( x 0 )
 C ( x 0 )  lim
h 0
1
h
Marginal analysis is an important example of a general
Incremental approximation procedure
§2.5 Approximation by increments
§2.5 Approximation by increments
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
Example: During a medical procedure, the size of a roughly
tumor is estimated by measuring its diameter and using the
4 3
V

R to compute its volume. If the diameter is
formula
3
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  8.181cm3
3
3
2
6
6
V  V ( 2.5  x )  V ( 2.5)  V ( 2.5)x
to be continued
Maxi mume rrori n vol ume V  [V ( 2.5)]x
 [V ( 2.5)](0.02( 2.5))
1
1 2
2
V ( x )   ( 3 x )  x
6
2
1
V ( 2.5)   ( 2.5) 2  9.817
2
Maximume rrorin volume (9.817)(0.05)  0.491
7.690  V  8.672
§2.5 Approximation of percentage
Change
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 quantity
Pe rce ntageof change 100
siz eof quantity
§2.5 Differentials
Differentials 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
§2.6 Implicit Differentiation
Explicit form: y=f(x)
f ( x )   x 2  20x  8000
x3  1
f ( x) 
2x  3
Implicit form
x2 y3  6  5 y3  x
x 2 y  2 y 3  3x  2 y
Implicit Differentiation Suppose an equation defines y implicitly
as a differentiable function of x. To find
dy
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 d y
dx
Example: Find
dy
if
dx
x2 y  y2  x3
Solution: Differentiate both sides of the equation with respect to
x. Don’t forget that y is actually a function of x.
x 2 f ( x)  ( f ( x))2  x 3
d 2
d 3
[ x f ( x )  ( f ( x ))2 ] 
[x ]
dx
dx
df
d
df
[x2
 f ( x ) ( x 2 )]  2 f ( x )
 3
x2
dx
dx

dx

 
 d ( x3 )
d
[ x 2 f ( x )]
dx
df
df
x
 f ( x )(2 x )  2 f ( x )
 3x2
dx
dx
2
df 3 x  2 xf ( x )
 2
dx
x  2 f ( x)
2
d
[( f ( x )) 2 ]
dx
dx
df
[ x  2 f ( x )]  3 x 2  2 xf ( x )
dx
2
dy 3 x 2  2 xy
 2
dx
x  2y
§2.6 Computing the slope of a
Tangent Line by Implicit differentiation
Find the slope of the tangent line to the circle x  y  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
2
dy
2x  2 y
0
dx
2
dy  x

dx
y
dy
The slope at (3,4) is the value of
dx
when x=3 and y=4
dy
x

dx ( 3,4 )
y
x3
y4

3
4
Similarly, at (3,-4)
dy
x

dx ( 3, 4 )
y
x3
y  4
3

4
§2.6 Related Rates
The manager of a company determines that when
q hundred units of a particular commodity are produced, the
total cost of production is C thousand dollars, where
C 2  3q 3  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?
Example:
 C and q are related by equation C 2  3q 3  4275
Both C and q can be regarded as function of a third variable t
 Implicit differentiation can be used to relate
dC
dt
dq
to dt
This kind of problem is said to involve related rates.
Solution:
dC
We want to find
when q=15 (1500 units) and dq  0.2( 20 / 100)
dt
dt
Differentiating implicitly in the equation C 2  3q 3  4275
with respect to time, we get
dC
2 dq
2C
 3[3q
] 0
dt
dt
dC
2 dq
2C
 9q
dt
dt
When q=15, the cost C satisfies
C 2  3(15) 3  4275
dC 9q 2 dq

dt 2C dt
C=120
dq
dC
Substituting q=15,c=120 and
 0.2 into the formula for
dt
dt
we obtain
dC
9(15) 2
[
](0.2)  1.6875
dt
2(120)
§2.6 Related Rates
Example 2. A lake is polluted by waste form a plant located
on its shore. Ecologist determine that when the level of
pollutant is x parts per million (ppn). There will be F fish
of a certain species in the lake. Where
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
dF
dx
Solution: We want to find dt when F=4000 and dt  1.4
When there are 4000 fish in the lake, the level of pollution x
satisfies
x=25
According to the chain rule
Substituting F=4000, x=25 and
dx
 1.4 , we obtain
dt
dF
16000
[
](1.4)  70
2
dt
25(3  25)
Summary
 Definition of the Derivative
f ( x )  lim
h 0
f ( x  h)  f ( x )
h
 Interpretation of the Derivative
Slope as a Derivative : The slope of the tangent
line to the curve y=f(x) at 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 byf (c)
Summary
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
f is decreasing at x=c if f (c ) <0
 Sign
Techniques of Differentiation
d
d
d
d n
n 1
[
cf
(
x
)]

c
[ f ( x )]
[ x ]  nx
[c ]  0
dx
dx
dx
dx
d
d
d
[ f ( x )  g( x )]  [ f ( x )]  [ g( x )]
dx
dx
dx
d
d
d
[ f ( x ) g( x )]  f ( x ) [ g( x )]  g( x ) [ f ( x )] The Product Rule
dx
dx
dx
d f ( x)
[
]
dx g( x )
g( x )
d
d
[ f ( x )]  f ( x ) [ g( x )]
dx
dx
g 2 ( x)
if g( x )  0
The Quotient Rule
Summary
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
 Application of Derivative
Tangent line, Rectilinear Motion, Projectile Motion
Summary
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 ( x 0  1)  C ( x 0 )
C ( x 0  h)  C ( x 0 )
C ( x 0  1)  C ( x 0 ) 
 C ( x 0 )  lim
h 0
1
h
f ( x 0  x )  f ( x 0 )  f ( x 0 )x
Approximation by Increment
Implicit Differentiation and Related Rate