Section 2.3 – Product and Quotient Rules and Higher- Order Derivatives

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”


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


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


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


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


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


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


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


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.



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.



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.



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.



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


(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


(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)

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Section 2.3 – Product and Quotient Rules and Higher- Order Derivatives