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Section 2.3 – Product and Quotient Rules and Higher-
Order Derivatives
The Product RuleThe derivative of a product of functions is NOT the
product of the derivatives.
If f and g are both differentiable, then:
In other words, the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.
' 'ddx f x g x g x f x f x g x
Another way to write the Rule: ' 'd
u v v u u vdx
Example 1Differentiate the function:
f ' x ddx 3x 2 1 7 2x 3
3 2 2 37 2 3 1 3 1 7 2d d d ddx dx dx dxx x x x
3 2 27 2 6 0 3 1 0 6x x x x 4 4 242 12 18 6x x x x
Product Rule
Sum/Difference Rule
Power Rule
Simplify
f x 3x 2 1 7 2x 3
3 2 2 37 2 3 1 3 1 7 2d ddx dxx x x x
u v
30x 4 6x 2 42x
v u' u v'
Example 2If h(x) = xg(x) and it is known that g(3) = 5 and g'(3) =2, find h'(3).
h' x ddx xg x
xg' x g x 1
xg' x g x
h' 3 3g' 3 g 3
Product Rule
Find the derivative:
Evaluate the derivative:
x ddx g x g x d
dx x
u v
32 5
u v' u v'
11
The Quotient RuleThe derivative of a quotient of functions is NOT the
quotient of the derivatives.
If f and g are both differentiable, then:
In other words, the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
ddx
f x g x g x f ' x f x g ' x
g x 2
Another way to write the Rule:2
' 'd u vu uv
dx v v
“Lo-d-Hi minus
Hi-d-Lo”
Example 1Differentiate the function:
24 73
' d xdx x
f x
2
22
3 4 4 7 2
3
x x x
x
2 2
22
12 4 8 14
3
x x x
x
2
22
4 14 12
3
x x
x
Quotient Rule
f x 4 x 73 x 2
2 2
22
3 4 7 4 7 3
3
d ddx dxx x x x
x
uv
u v'v u'
v2
Example 2Find an equation of the tangent line to the curve
at the point (1,½).
y x
1x 2
y'1x 2 d
dx x1 2 x ddx 1x 2
1x 2 2
1x 2 1
2 x 1 2 x 2x
1x 2 2
1x 2 1
2 x 2x x
1x 2 2
2 x
2 x
1x 2 4 x 2
2 x 1x 2 2
1 3x 2
2 x 1x 2 2
Fin
d th
e D
eriv
ativ
e
Find the derivative (slope of the tangent line) when x=1
1 3 1 2
2 1 1 1 2 2
28
14
Use the point-slope formula to find an equation
y 12 1
4 x 1
Example 3Differentiate the function:
1 21 12 3' d
dxf x x x
1 21 12 3
d ddx dxx x
2 31 12 31 2x x
2 31 2
2 3x x
Sum and Difference Rules
Constant Multiple Rule
Power Rule
Simplify
21 1
2 3x xf x
1 21 12 3
d ddx dxx x
The Quotient Rule is long, don’t forget to rewrite if possible.
Rewrite to use the power rule
33
xx 2
2
34 36
xx
3 3
3 46 6
xx x
Try to find the Least Common
Denominator
Derivative of SecantDifferentiate f(x) = sec(x).
secddx x 1
cosddx x
2
cos 1 1 cos
cos
d ddx dxx x
x
2
cos 0 1 sin
cos
x x
x
2
sin
cos
x
x
sin1cos cos
x
x x
sec tanx x
The derivation for tangent is in
the book.Rewrite as a
Quotient
Quotient Rule
Rewrite to use Trig Identities
More Derivatives of Trigonometric Functions
We will assume the following to be true:
2tan secd
x xdx
sec sec tand
x x xdx
2cot cscd
x xdx
csc csc cotd
x x xdx
Example 1Differentiate the function:
f ' dd
sec1tan
1tan sec tan sec sec2
1tan 2
sec tan 1 1tan 2
Quotient Rule
f sec1tan
1tan dd sec sec d
d 1tan 1tan 2
u
v
u v'v u'
Use the Quotient
Rule
sec tan sec tan 2 sec3 1tan 2
Always look to simplify
sec tan tan 2 sec2
1tan 2Trig Law: 1 + tan2 = sec2
Example 2Differentiate the function:
f ' dd 3sec
3sectan sec3
3 sectan 3sec
Product Rule
f 3 sec
3 dd sec sec d
d 3
u v
u v' v u'
Use the Product
Rule
Example 1Let
a. Find the derivative of the function.
f x 2x 4 9x 3 5x 2 7
ddx 2x 4 9x 3 5x 2 7
ddx 2x 4 d
dx 9x 3 ddx 5x 2 d
dx 7
2 ddx x 4 9 d
dx x 3 5 ddx x 2 d
dx 7
24x 4 1 93x 3 1 52x 2 1 0
8x 3 27x 2 10x
Example 1 (Continued)Let
b. Find the derivative of the function found in (a).
f x 2x 4 9x 3 5x 2 7
ddx 8x 3 27x 2 10x
ddx 8x 3 d
dx 27x 2 ddx 10x1
8 ddx x 3 27 d
dx x 2 10 ddx x1
83x 3 1 272x 2 1 101x1 1
24x 2 54x 10
Example 1 (Continued)Let
c. Find the derivative of the function found in (b).
f x 2x 4 9x 3 5x 2 7
ddx 24x 2 54 x 10
ddx 24x 2 d
dx 54 x1 ddx 10
24 ddx x 2 54 d
dx x1 ddx 10
242x 2 1 541x1 1 0
48x 54
Example 1 (Continued)Let
d. Find the derivative of the function found in (c).
f x 2x 4 9x 3 5x 2 7
ddx 48x 54
ddx 48x1 d
dx 54
48 ddx x1 d
dx 54
481x1 1 0
48
Example 1 (Continued)Let
e. Find the derivative of the function found in (d).
f x 2x 4 9x 3 5x 2 7
ddx 48
0
We have just differentiated the derivative of a function. Because the derivative of a function is a function,
differentiation can be applied over and over as long as the derivative is a differentiable function.
First Derivative
Second Derivative
Third Derivative
Fourth Derivative
nth Derivative
Higher-Order Derivatives: Notation
ddx f x
2
2ddx
f x
3
3ddx
f x
4
4ddx
f x
n
nddx
f x
Notice that for derivatives of higher order than the third, the parentheses distinguish a derivative from a power. For
example: .
f 4 x f 4 x
'y''y'''y 4y ny
'f x
''f x
'''f x 4f x nf x
ddx y
dyddx dx
2
2
d yddx dx
3
3
d yddx dx
1
1
n
n
d yddx dx
dydx
2
2
d y
dx3
3
d y
dx4
4
d y
dxn
n
d y
dx
Example 1 (Continued)Let
f. Define the derivatives from (a-e) with the correct notation.
f x 2x 4 9x 3 5x 2 7
5 0f x
You should note that all higher-order derivatives of a
polynomial p(x) will also be polynomials, and if p has degree n, then p(n)(x) = 0 for
k ≥ n+1.
f 4 x 48
f ' ' ' x 48x 54
f ' ' x 24x 2 54x 10
f ' x 8x 3 27x 2 10x
Example 2If , find . Will ever equal 0?
f x 1x
ddx x 1
1x 1 1
x 2
f 3 x
f n x Find the first derivative:
2ddx x
1 2x 2 1
2x 3
Find the second derivative:
32ddx x
2 3x 3 1
6x 4
Find the third derivative:
f ' ' ' x 6x 4
No higher-order derivative will equal 0 since the power of the function will never be 0. It decreases by one each time.
Example 3Find the second derivative of .
s t t 2 sin t 3t
s' t ddt t 2 sin t 3t
ddt t 2 sin t d
dt 3t
t 2 ddt sin t sin t d
dt t 2 ddt 3t
t 2cos t sin t2t 3
t 2 cos t 2t sin t 3
Find the first derivative:
s' ' t ddt t 2 cos t 2t sin t 3
ddt t 2 cos t d
dt 2t sin t ddt 3
t 2 ddt cos t cos t d
dt t 2 2t ddt sin t sin t d
dt 2t ddt 3
t 2 cos t cos t2t 2tcos t sin t2 0
t 2 cos t 2t cos2t cos t 2sin t
Find the second derivative:
Graphs of a Function and its Derivatives
What can we say about g, g', g'' for the segment of the graph of y = g(x)?
g :
g' :
g'' :
Increasing
Positive, Increasing
Positive
As the graph increases, the tangent lines are getting
steeper.
Since the first derivative is increasing, the second derivative must be positive.
Graphs of a Function and its Derivatives
What can we say about g, g', g'' for the segment of the graph of y = g(x)?
g :
g' :
g'' :
Decreasing
Negative, Decreasing
Negative
As the graph decreases, the tangent lines
are getting less steep.
Since the first derivative is decreasing, the second derivative
must be negative.
Graphs of a Function and its Derivatives
What can we say about g, g', g'' for the segment of the graph of y = g(x)?
g :
g' :
g'' :
Decreasing
Negative, Increasing
Positive
As the graph decreases, the tangent lines are steeper.
Since the first derivative is increasing, the second derivative
must be positive.
Graphs of a Function and its Derivatives
What can we say about g, g', g'' for the segment of the graph of y = g(x)?
g :
g' :
g'' :
Increasing
Positive, Decreasing
Negative
As the graph increases, the tangent lines
are getting less steep.
Since the first derivative is decreasing, the second derivative must be negative.
Graphs of a Function and its Derivatives
What can we say about g, g', g'' for the segment of the graph of y = g(x)?
g :g' :g'' :
Decreasing
Negative, Decreasing
Negative
Find the pieces of this graph that compare to the previous
graphs.
On the left side :g :g' :g'' :
Decreasing
Negative, Increasing
Positive
On the right side :
Average Acceleration
Example: Estimate the velocity at time 5 for graph of velocity at time t below.
1 2 3 4 5 6
2
-2
-4
v(t)
t
Acceleration is the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.
Find the average rate of change of velocity for times that are close and enclose time 5.
6 4
6 4
v v 4 6
6 4
5
Instantaneous AccelerationIf s = s(t) is the position function of an object that moves in
a straight line, we know that its first derivative represents the velocity v(t) of the object as a function of time.
The instantaneous rate of change of velocity with respect to time is called the acceleration a(t) of an object. Thus, the acceleration function is the derivative of the velocity function and is therefore the second derivative of the position function.
a t v ' t s' ' t d 2sdt 2
v t s' t dsdt
Position, Velocity, and Acceleration
Position, Velocity, and Acceleration are related in the following manner:
Position:
Velocity:
Acceleration:
( )s t
'( ) ( )s t v t
Units = Measure of length (ft, m, km, etc)
The object is…Moving right/up when v(t) > 0Moving left/down when v(t) < 0Still or changing directions when v(t) = 0
Units = Distance/Time (mph, m/s, ft/hr, etc)
Speed = absolute value of v(t)
s' ' t v ' t a t Units = (Distance/Time)/Time (m/s2)
Example
1 2 3 4 5 6
2
-2
-4
v(t)
t
Example: The graph below at left is a graph of a particle’s velocity at time t. Graph the object’s acceleration where it exists and answer the questions below
1 2 3 4 5 6
2
-2
-4
a(t)
t
m = 2
Cornerm = 0
m = -4
m = 4
When is the particle speeding up?
When is the particle traveling at a constant speed?
When is the function slowing down?
Positive acceleration and positive velocity
(0,2)
Negative acceleration and Negative velocity
U (5,6)
(2,4)
0 accelerationAnd constant velocity
Negative acceleration and Positive velocity
(4,5) U (6,7)
Positive acceleration and Negative velocity
Moving away from x-axis.
HorizontalMoving towards the x-axis.
Speeding Up and Slowing DownAn object is SPEEDING UP when the following occur:• Algebraic: If the velocity and the acceleration agree in sign• Graphical: If the velocitycurve is moving AWAY from the x-axis
An object is traveling at a CONSTANT SPEED when the following occur:
• Algebraic: Velocity is constant and acceleration is 0.• Graphically: The velocity curve is horizontal
An object is SLOWING DOWN when the following occur:• Algebraic: Velocity and acceleration disagree in sign• Graphically: The velocity curve is moving towards the x-axis
Example 3The position of a particle is given by the equation
where t is measured in seconds and s in meters.
(a) Find the acceleration at time t.
s t t 3 6t 2 9t
s' t v t ddt t 3 6t 2 9t
v t ddt t 3 d
dt 6t 2 ddt 9t
v t ddt t 3 6 d
dt t 2 9 ddt t1
v t 3t 3 1 62t 2 1 9t1 1
v t 3t 2 12t 9
The derivative
of the position
function is the
velocity function.
s' ' t v ' t a t ddt 3t 2 12t 9
a t ddt 3t 2 d
dt 12t ddt 9
a t 3 ddt t 2 12 d
dt t1 ddt 9
a t 32t 2 1 121t1 1 0
a t 6t 12
The derivative of the velocity function is
the acceleration
function.
Example 3 (continued)The position of a particle is given by the equation
where t is measured in seconds and s in meters.
(b) What is the acceleration after 4 seconds?
(c) Is the particle speeding up, slowing down, or traveling at a constant speed at 4 seconds?
s t t 3 6t 2 9t
a 4
6 4 12 12 m/s2
v 4
3 4 2 12 4 9
9 m/s
a 4 12 m/s2
Since the velocity and acceleration agree in signs, the particle is
speeding up.
Example 3 (continued)The position of a particle is given by the equation
where t is measured in seconds and s in meters.
(d) When is the particle speeding up? When is it slowing down?
s t t 3 6t 2 9t
Velocity:
Acceleration:
+ +–
+–
Speeding Up:
(1,2) U (3,∞)
Slowing Down:
(0,1) U (2,3)