Chapter 2
The Concepts
The Derivative
Product and Quotient Rules Higher-Order Derivatives
The Chain Rule
Marginal Analysis
Why We Learn Differentiation?
Calculus is the mathematics of change, and the primary tool for studying change is a procedure called differentiation.
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
Illustration
If air resistance is neglected, an object dropped from a great height will fall feet in t seconds.
What is the object’s instantaneous velocity after t=2 seconds?
Average rate of change of s(t) over the time period [2,2+h] is
Compute the instantaneous velocity by the limit
That is, after 2 seconds, the object is traveling at the rate of 64 feet per second.
2
Rates of Change
How to determine
instantaneous rate of change
or rate
of change of
f(x)
at
x=c
?
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
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.
1 2
1 2
x in change
y in change
change of
rate Slope
x x
y y
x y
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
The Slope of Secant Line
The secant line
is a line that intersects the curve at the
point
x
and point
x+h
The Derivative
A difference quotient
for the function
f
(
x
) is the
expression
The derivative of a function
f
(
x
) with respect to
x
is the
function
f’
(
x
) given by
The process of computing the derivative is called
differentiation
.
f
(
x
) is
differentiable
at
x=c
if
f
’
(
c
) exists
h
x
f
h
x
f
x
f
h
)
(
)
(
lim
)
(
0
h
x
f
h
x
Example
Find the derivative of the function
f
(
x
)
2
x
2
16
x
35
The difference quotient for
f(x)
is
16
Thus, the derivative of
f(x)
is the function
Slope as a Derivative
: The slope of the tangent
line to the curve
y=f(x)
at the point
(c,f(c))
is
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)
)
(
tan
f
c
m
Remarks: Since the slope of the tangent line at
(a,f(a))
is
f’(a)
, the equation of the tangent line is
)
)(
(
)
(
a
f
a
x
a
f
Example
What is the equation of the tangent line to the curve
at the point where
x=4
?
x
y
According to the definition of derivative, we have
x
Example
A manufacturer determines that when x thousand units of a
particular commodity are produced, the profit generated will be
12000
dollars. At what rate is profit changing with respect to the level of production x when 9 thousand units are produced?
Thus, when the level of production is x=9, the profit is changing at the rate of dollars per thousand units.
400
6800
)
9
(
800
)
9
(
p
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
Significance of the Sign of the Derivative
If the function
f
is differentiable at
x=c
, then
f
is increasing at
x=c
if
f’(c)>0
and
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
.
Differentiability and Continuity
If the function
f(x)
is differentiable at
x=c
, then it is also
continuous at
x=c
.
For any constant c, we have
That is, the derivative of a constant is zero
c
0
dx
d
Since
f
(
x+h
)
=c
for all
x
0 lim
) ( )
( lim )
(
0
0
h
c c h
x f h
x f x
f
h h
The Power Rule
Examples.
For any real number
n
1
]
[xn nxn dx
d
2 1 2
1 2
3
2
1
)
(
)
(
3
)
(
x
x
dx
d
x
dx
d
x
x
The Constant Multiple Rule
If
c
is a constant and
f(x)
is differentiable then so is
cf
(
x
) and
That is, the derivative of a multiple is the multiple of
the derivative
The Sum Rule
If f(x) and g(x) are differentiable then so is the sum of
s(x)=f(x)+g(x) and
Example
Find the equation of the tangent line to the function
at
x=16
.
x
x
x
f
(
)
4
8
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:
16 3
) 16 (
3
32
x y x
Example
It is estimated that
x
months form now, the
population of a certain community will be
8000
20
)
(
x
x
2
x
p
a. At what rate will the population be changing with
respect to time
15
months from now?
a. The rate of change of the population with respect to time
is the derivative of the population function. That is
20
2
)
(
change
of
Rate
p
x
x
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 16
thmonth
is
p
(
16
)
p
(
15
)
8576
8525
51
people
The
relative rate of change
of a quantity
Q(x)
with
respect to
x
is given by the ratio
The corresponding
percentage rate of change
of
Q(x)
with respect to
x
is
Example
The gross domestic product (GDP) of a certain country was
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?
106
5
)
(
t
t
2
t
N
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
year per
% 77 . 9 256
25 100
) 10 (
) 10 (
100
Example
Based on the techniques of differentiation, we have
So the solution is
p’(x)=0 at x=-1 and 2
p’(x)>0 at x<-1 and x>2
Rectilinear Motion
Motion of an object along a straight-line 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
and
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
dt ds t
s t
v( ) ( )
velocity
dt dv t
v t
a( ) ( ) on
Example
The position at time t of an object moving along a line is given by
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.
5
9
6
)
The motion of an object:
5 9 6
)
(t t3 t2 t s
Interval Sign of v(t)
Description of Motion
0<t<1 +
Advancing
from s(0)=5 to s(1)=9
1<t<3
-Retreating
from s(1)=9 to s(3)=5
3<t<4 +
Advancing
from s(3)=5 to s(4)=9
a. The velocity is . The object will be stationary when
9 12
3 )
( t2 t dt
ds t
v
0 )
3 )(
1 (
3 9
12 3
)
(t t2 t t t v
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
12
c. The acceleration of the object is
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.
Product and Quotient Rules
The Product Rule
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
or equivalently,
( ) ( )
( ) [ ( )] ( ) [ f (x)]Use the product rule to find the derivative of the function
Example
A manufacturer determines that t months after a new
product is introduced to the market, hundred units can be produced and then sold at a price of
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?
t t
t
x( ) 2 3
30 2
)
( 2
3
t
a. The revenue is given by
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:
Thus, after 4 months, the revenue is changing at the rate of 14
hundred dollars per month. It is decreasing at that time since
The Quotient Rule
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
or equivalently,
0
Use the quotient rule to find the derivative of the function
A Word of Advice
The quotient rule is somewhat cumbersome, so don’t use it
unnecessarily.
Differentiate the function . y x x xx1
Don’t use the quotient rule! Instead, rewrite the function as
1
and then apply the power rule term by term to get
The Second Derivative
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
The second derivative gives the rate of change of the rate of change of the original function.
) ( or
2 2
x f
dx y
d
Alternate Notation
There is some alternate notation for higher order derivatives as well.
2 2
) ( )
(
dx f d x
f dx
df x
f
Note: Before computing the second derivative of a function, always take time to simplify the first derivative as much as possible.
Example.
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 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.?
t t
t t
a. The worker’s rate of production is the first derivative
b. The rate of change of the rate of production is the second
derivative of the output function. At 11:00 A.M., this rate is
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.
Acceleration as Second Order Derivative
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 thought of as the second derivative of position; that is ( ) 22
dt s d t
a
If the position of an object moving along a straight line is given by at time s(t) t3 3t2 4 t, find its velocity and acceleration.
Example.
The velocity of the object is
and its acceleration is
4 6
3 )
( t2 t
dt ds t
v
6 6
)
( 2
2
t
dt s d dt
dv t
The
n
thDerivative
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
Find the fifth derivative of the function y=1/x
Illustration
respect to with respect to
with
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
The product of these two rates is the rate of change of cost with respect to time that is
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
or, equivalently, by
dx
du
du
dy
dx
dy
) ( )) (
(g x g x
f dx
dy
Example
and according to the chain rule,
Example
The cost of producing
x
units of a particular commodity
is
dollars, and the production level
t
hours into a particular
production run is
53
4
3
1
)
(
x
x
2
x
C
units
03
.
0
2
.
0
)
(
t
t
2t
x
At what rate is cost changing with respect to time after
4
We find that
and by substituting
t=4
and
x=3.32
into the formula for
, we get
Let’s look at the functions
then we can write the function as a composition.
differentiate a composition function using the Chain Rule.
In general, we differentiate the outside function
leaving the inside function alone and multiple all
of this by the derivative of the inside function
function inside
of derivative alone
function inside leave
function outside
of derivative
)
(
))
(
(
)
(
x
f
g
x
g
x
F
Example.
a. Based on the chain rule, find the derivative of
with respect to
x
b. Find the derivative of
32
)
(
)
(
x
x
x
f
1 1
1 )
(
2
x x
The General Power Rule
For any real number
n
and differentiable function
h
, we
have
Think of [
h(x)
]
nas the composite function
2.5 Marginal Analysis and
Approximations Using
For instance, suppose C(x) is the total cost of producing x units of a particular commodity. If units are currently being produced, then the derivative
Marginal Analysis
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
Marginal Cost
If C(x) is the total cost of producing x units of a commodity. Then the marginal cost of producing units is the derivative , which approximates the additional cost incurred
when the level of production is increased by one unit, from to
+1, assuming >>1.
0
x C(x0)
) ( )
1
(x0 C x0
C
0
x
x00
The
marginal revenue
is , it approximates
, the additional revenue generated by
producing one more unit.
)
(
x
0R
) (
) 1
(x0 R x0
R
) (x0 P
)
(
)
1
(
x
0P
x
0P
The
marginal profit
is , it approximates
, the additional profit obtained by
producing one more unit, assuming >>1.
Marginal Revenue and Marginal Profit
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=x
0units are
being produced, then
Example
A manufacturer estimates that when x units of a particular commodity are produced, the total cost will be dollars, and furthermore, that all x units will be sold when the price is dollars per unit.
98 3
8 1 )
(x x2 x C
) 75 ( 3 1 )
(x x
p
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.
a. The marginal cost is C’(x)=(1/4)x+3. The total revenue is
, 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.
3 / 25
) 3 / ) 75
(( )
( )
(x xp x x x x x2
Marginal analysis is an important example
of a
general incremental approximation
Approximation by Increment
Based on the fact that since
Percentage of Change
The
percentage of change
of a quantity expresses the
change in that quantity as a percentage of its size prior to
the change. In particular,
quantity
of
size
quantity
in
change
100
change
of
Example
The GDP of a certain country was billion dollars t
years after 1997. Use calculus to estimate the percentage change in the GDP during the first quarter of 2005.
200 5
)
(t t2 t N
Use the formula
Differentials
Example
In each case, find the differential of
y=f(x)
.
Explicit and Implicit Forms of Functions
2 3
2
1
and 3 2
1
1
3 y x
x x y
x x
y
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.
are all functions in explicit form.
Sometimes practical problems will lead to equations in which the function y is not written explicitly in terms of the independent variable x.
y x
y y
x xy
y y
x2 3 6 5 3 and 2 2 3 3 2
Example.
Find
dy/dx
if
x
2y
y
2
x
3We 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 equation as
.
Secondly, we differentiate bothImplicit 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.
Example.
Example
25
2 2
y x
y x dx
dy dx
dy y
x 2 0
2
4 3 4
3
) 4 , 3 (
dx dy
Thus, the slope at (3,4) is
The slope at (3,-4) is
Differentiating both sides of the equation with respect to x, we have Find the slope of the tangent line to the circle at the point (3,4). What is the slope at the point (3,-4)?
4 3
) 4 , 3 (
An Application
3 2
3
2x x y y
Q
Suppose the output at a certain factory is
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
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
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, we have
2
Now evaluate this derivative when x=30 and y=20 to conclude that
Related Rates
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.
Example
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 . 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?
4275
3
32
We want to find dC/dt when q=15 and dq/dt=0.2. Differentiating
and by substituting q=15, C=120 and dq/dt=0.2 into the formula for dC/dt, we obtain
Example
A storm at sea has damaged an oil rig. Oil spills from
the rupture at the constant rate of
60
, 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?
min /
3
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
3
Differentiating implicitly in this equation with respect to time t,
and since dV/dt=60 at all times, we get
Therefore, the total amount of oil spilled is
Summary
Definition of the Derivative
h
x
f
h
x
f
x
f
h
)
(
)
(
lim
)
(
0
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
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)
) (
tan f c
Sign of The Derivative
If the function f is differentiable at x=c, then
f is increasing at x=c if >0f (c)
f is decreasing at x=c if <0f (c)
Techniques of Differentiation
c 0 The Product Rule
0
The Chain Rule
The General Power RuleThe Higher -order Derivative
Applications of Derivative
Tangent line, Rectilinear Motion )
The Second Derivative
The marginal cost is , it approximates , the additional cost generated by producing one more unit.
) (x0
C C(x0 1) C(x0)
Marginal Analysis and Approximation by Increments
h
