§11.3 Derivatives of Products and Quotients
$11.4 The Chain Rule
■ the derivative of a product of two functions, and
■ the derivative of a quotient of two functions.
■ The student will be able to
■
form the composition of two functions
■
■
apply the general power rule.
the chain rule.
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Derivatives of Products
Theorem 1 (Product Rule)
If f (x) = F(x)  S(x), and if F ’(x) and S ’(x) exist, then
or
f ’ (x) = F(x)  S ’(x) + F ’(x)  S(x),
dS dF
f ' ( x)  F

S
dx dx
In words: The derivative of the 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.
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Example
Find the derivative of y = 5x2(x3 + 2).
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Example
Find the derivative of y = 5x2(x3 + 2).
Solution:
Let F(x) = 5x2, so F ’(x) = 10x
Let S(x) = x3 + 2, so S ’(x) = 3x2.
Then
f ’ (x) = F(x)  S ’(x) + F ’(x)  S(x)
= 5x2  3x2 + 10x  (x3 + 2)
= 15x4 + 10x4 + 20x = 25x4 + 20x.
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Derivatives of Quotients
Theorem 2 (Quotient Rule)
If f (x) = T (x) / B(x), and if T ’(x) and B ’(x) exist, then
B ( x)  T ' ( x)  T ( x)  B ' ( x)
f ' ( x) 
or
[ B ( x)] 2
dT
dB
B
T
dy
 dx 2 dx
dx
B
In words: The derivative of the quotient of two functions is
the bottom function times the derivative of the top function
minus the top function times the derivative of the bottom
function, all over the bottom function squared.
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Example
Find the derivative of y = 3x / (2x + 5).
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Example
Find the derivative of y = 3x / (2x + 5).
Solution:
Let T(x) = 3x, so T ’(x) = 3
Let B(x) = 2x + 5, so B ’(x) = 2.
Then
B ( x)  T ' ( x)  T ( x)  B ' ( x)
f ' ( x) 
[ B ( x)] 2
(2 x  5)  3  3 x  2
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

2
( 2 x  5)
(2 x  5) 2
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Tangent Lines
Let f (x) = (2x - 9)(x2 + 6). Find the equation of the line
tangent to the graph of f (x) at x = 3.
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Tangent Lines
Let f (x) = (2x - 9)(x2 + 6). Find the equation of the line
tangent to the graph of f (x) at x = 3.
Solution: First, find f ’(x):
f ’(x) = (2x - 9) (2x) + (2) (x2 + 6)
Then find f (3) and f ’(3):
f (3) = -45
f ’(3) = 12
The tangent has slope 12 and goes through the point (3, -45).
Using the point-slope form y - y1 = m(x - x1), we get
y – (-45) = 12(x - 3)
or
y = 12x - 81
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Summary
Product Rule:
d
F ( x )  S ( x )   F ' ( x )  S ( x )  F ( x )  S ' ( x )
dx
Quotient Rule:
d  T ( x)  B ( x)  T ' ( x)  T ( x)  B ' ( x)

 
dx  B( x) 
[ B( x)] 2
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Composite Functions
Definition: A function m is a composite of functions f and g if
m(x) = f [g(x)]
The domain of m is the set of all numbers x such that x is in the
domain of g and g(x) is in the domain of f.
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General Power Rule
We have already made extensive use of the power rule:
d n
x  nx n 1
dx
Now we want to generalize this rule so that we can
differentiate composite functions of the form [u(x)]n,
where u(x) is a differentiable function. Is the power rule
still valid if we replace x with a function u(x)?
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Example
Let u(x) = 2x2 and f (x) = [u(x)]3 = 8x6. Which of the
following is f ’(x)?
(a) 3[u(x)]2
(b) 3[u’(x)]2
(c) 3[u(x)]2 u’(x)
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Example
Let u(x) = 2x2 and f (x) = [u(x)]3 = 8x6. Which of the
following is f ’(x)?
(a) 3[u(x)]2
(b) 3[u’(x)]2
(c) 3[u(x)]2 u’(x)
We know that f ’(x) = 48x5.
(a) 3[u(x)]2 = 3(2x2)2 = 3(4x4) = 12 x4. This is not correct.
(b) 3[u’(x)]2 = 3(4x)2 = 3(16x2) = 48x2. This is not correct.
(c) 3[u(x)]2 u’(x) = 3[2x2]2(4x) = 3(4x4)(4x) = 48x5. This is the
correct choice.
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Generalized Power Rule
What we have seen is an example of the generalized power
rule: If u is a function of x, then
d n
n 1 du
u  nu
dx
dx
For example,
d 2
( x  3 x  5)3  3( x 2  3 x  5) 2 (2 x  3)
dx
du
2
Here u is x  3 x  5 and
 2x  3
dx
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Chain Rule
We have used the generalized power rule to find derivatives
of composite functions of the form f (g(x)) where f (u) = un
is a power function. But what if f is not a power function?
It is a more general rule, the chain rule, that enables us to
compute the derivatives of many composite functions of the
form f(g(x)).
Chain Rule: If y = f (u) and u = g(x) define the
composite function y = f (u) = f [g(x)], then
dy dy du
dy
du


, provided
and
exist .
dx du dx
du
dx
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Generalized Derivative Rules
1.
d
n
n 1
 f x   n  f x   f ' ( x)
dx
If y = u n , then
y’ = nu n - 1  du/dx
2.
d
1
ln [ f ( x)] 
 f ' ( x)
dx
f ( x)
If y = ln u, then
y’ = 1/u  du/dx
3.
d f ( x)
e
 e f ( x ) f ' ( x)
dx
If y = e u, then
y ’ = e u  du/dx
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Examples for the Power Rule
Chain rule terms are marked:
y  x , y'  5x
5
4
y  (2 x) , y '  5(2 x) (2)  160 x
5
4
4
y  (2 x 3 ) 5 , y '  5(2 x 3 ) 4 (6 x 2 )  480 x14
y  (2 x  1) 5 , y '  5(2 x  1) 4 (2)  10(2 x  1) 4
y  (e x ) 5 , y '  5(e x ) 4 (e x )  5e 5 x
y  (ln x) 5 , y '  5(ln x) 4 (1 / x)
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Examples for
Exponential Derivatives
d u
u du
e e 
dx
dx
y  e 3 x , y '  e 3 x (3)  3e 3 x
ye
3 x 1
ye
4 x 2 3 x  5
ye
ln x
, y'  e
3 x 1
, y'  e
 x, y '  e
(3)  3e
4 x 2 3 x  5
ln x
3 x 1
(8 x  3)
1
x
( )  1
x
x
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Examples for
Logarithmic Derivatives
d
1 du
ln u  
dx
u dx
1
1
y  ln( 4 x), y ' 
4 
4x
x
1
4
y  ln( 4 x  1), y ' 
4 
4x 1
4x 1
1
2
2
y  ln( x ), y '  2  (2 x) 
x
x
1
2x  2
2
y  ln( x  2 x  4), y '  2
 ( 2 x  2)  2
x  2x  4
x  2x  4
Barnett/Ziegler/Bylee Business Calculus 11e
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