in this section, we will learn about: differentiating composite functions using the chain rule
DESCRIPTION
DERIVATIVES. In this section, we will learn about: Differentiating composite functions using the Chain Rule. 3.5 The Chain Rule. CHAIN RULE. Suppose you are asked to differentiate the function - PowerPoint PPT PresentationTRANSCRIPT
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In this section, we will learn about:
Differentiating composite functions
using the Chain Rule.
DERIVATIVES
3.5 The Chain Rule
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Suppose you are asked to differentiate
the function
The differentiation formulas you learned in the previous sections of this chapter do not enable you to calculate F’(x).
2( ) 1F x x
CHAIN RULE
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Observe that F is a composite function.
In fact, if we let and
let u = g(x) = x2 + 1, then we can write
y = F(x) = f (g(x)).
That is, F = f ◦ g.
( )y f u u
CHAIN RULE
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We know how to differentiate both f and g.
So, it would be useful to have a rule that
shows us how to find the derivative of F = f ◦ g
in terms of the derivatives of f and g.
CHAIN RULE
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It turns out that the derivative of the composite
function f ◦ g is the product of the derivatives
of f and g.
This fact is one of the most important of the
differentiation rules. It is called the Chain Rule.
CHAIN RULE
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It seems plausible if we interpret
derivatives as rates of change.
Regard: du/dx as the rate of change of u with respect to x dy/du as the rate of change of y with respect to u dy/dx as the rate of change of y with respect to x
CHAIN RULE
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If u changes twice as fast as x and y changes
three times as fast as u, it seems reasonable
that y changes six times as fast as x.
So, we expect that:dy dy du
dx du dx
CHAIN RULE
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If g is differentiable at x and f is differentiable
at g(x), the composite function F = f ◦ g
defined by F(x) = f(g(x)) is differentiable at x
and F’ is given by the product:
F’(x) = f’(g(x)) • g’(x)
In Leibniz notation, if y = f(u) and u = g(x) are both differentiable functions, then:
THE CHAIN RULE
dy dy du
dx du dx
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Let ∆u be the change in corresponding to
a change of ∆x in x, that is,
∆u = g(x + ∆x) - g(x)
Then, the corresponding change in y is:
∆y = f(u + ∆u) - f(u)
COMMENTS ON THE PROOF
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It is tempting to write:
0
0
0 0
0 0
lim
lim
lim lim
lim lim
x
x
x x
u x
dy y
dx xy u
u xy u
u xy u dy du
u x du dx
COMMENTS ON THE PROOF Equation 1
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The only flaw in this reasoning is that,
in Equation 1, it might happen that ∆u = 0
(even when ∆x ≠ 0) and, of course, we
can’t divide by 0.
COMMENTS ON THE PROOF
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Nonetheless, this reasoning does at least
suggest that the Chain Rule is true.
A full proof of the Chain Rule is given
at the end of the section.
COMMENTS ON THE PROOF
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The Chain Rule can be written either in
the prime notation
(f ◦ g)’(x) = f’(g(x)) • g’(x)
or, if y = f(u) and u = g(x), in Leibniz notation:
dy dy du
dx du dx
CHAIN RULE Equations 2 and 3
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Equation 3 is easy to remember because,
if dy/du and du/dx were quotients, then we
could cancel du.
However, remember: du has not been defined du/dx should not be thought of as an actual quotient
CHAIN RULE
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Find F’(x) if
One way of solving this is by using Equation 2.
At the beginning of this section, we expressed F as F(x) = (f ◦ g))(x) = f(g(x)) where and g(x) = x2 + 1.
2( ) 1F x x
( )f u u
E. g. 1—Solution 1CHAIN RULE
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Since
we have
1/ 212
1'( ) and '( ) 2
2f u u g x x
u
CHAIN RULE
2
2
'( ) '( ( )) '( )
12
2 1
1
F x f g x g x
xxx
x
E. g. 1—Solution 1
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We can also solve by using Equation 3.
If we let u = x2 + 1 and then:
,y u
2 2
1'( ) (2 )
21
(2 )2 1 1
dy duF x x
dx dx ux
xx x
CHAIN RULE E. g. 1—Solution 2
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When using Equation 3, we should bear
in mind that:
dy/dx refers to the derivative of y when y is considered as a function of x (called the derivative of y with respect to x)
dy/du refers to the derivative of y when considered as a function of u (the derivative of y with respect to u)
CHAIN RULE
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For instance, in Example 1, y can be
considered as a function of x
and also as a function of u .
Note that:
( y x2 1)
( y u )
dy
dxF '(x)
x
x2 1whereas
dy
du f '(u)
1
2 u
CHAIN RULE
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In using the Chain Rule, we work from
the outside to the inside.
Equation 2 states that we differentiate the outerfunction f [at the inner function g(x)] and then we multiply by the derivative of the inner function.
outer function derivative derivativeevaluated evaluated
of outer of innerat inner at innerfunction functionfunction function
( ( )) ' ( ( )) . '( ) d
f g x f g x g xdx
NOTE
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Let’s make explicit the special case of
the Chain Rule where the outer function is
a power function.
If y = [g(x)]n, then we can write y = f(u) = un where u = g(x).
By using the Chain Rule and then the Power Rule, we get:
1 1[ ( )] '( )n ndy dy du dynu n g x g x
dx du dx dx
COMBINING CHAIN RULE WITH POWER RULE
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If n is any real number and u = g(x)
is differentiable, then
Alternatively,
1( )n nd duu nu
dx dx
POWER RULE WITH CHAIN RULE
1[ ( )] [ ( )] . '( )n ndg x n g x g x
dx
Rule 4
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Notice that the derivative in Example 1
could be calculated by taking n = ½
in Rule 4.
POWER RULE WITH CHAIN RULE
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Differentiate y = (x3 – 1)100
Taking u = g(x) = x3 – 1 and n = 100 in the rule, we have:
3 100
3 99 3
3 99 2
2 3 99
( 1)
100( 1) ( 1)
100( 1) 3
300 ( 1)
dy dx
dx dxd
x xdx
x x
x x
Example 3POWER RULE WITH CHAIN RULE
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Find f’ (x) if
First, rewrite f as f(x) = (x2 + x + 1)-1/3
Thus,
3 2
1 .( )1
f xx x
Example 4POWER RULE WITH CHAIN RULE
2 4 / 3 2
2 4 / 3
1'( ) ( 1) ( 1)
31
( 1) (2 1)3
df x x x x x
dx
x x x
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Find the derivative of
Combining the Power Rule, Chain Rule, and Quotient Rule, we get:
92
( )2 1
tg t
t
Example 5POWER RULE WITH CHAIN RULE
8
8 8
2 10
2 2'( ) 9
2 1 2 1
2 (2 1) 1 2( 2) 45( 2)9
2 1 (2 1) (2 1)
t d tg t
t dt t
t t t t
t t t
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Differentiate:
y = (2x + 1)5 (x3 – x + 1)4
In this example, we must use the Product Rule before using the Chain Rule.
Example 6CHAIN RULE
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Thus,
5 3 4 3 4 5
5 3 3 3
3 4 4
5 3 3 2
3 4 4
(2 1) ( 1) ( 1) (2 1)
(2 1) 4( 1) ( 1)
( 1) 5(2 1) (2 1)
4(2 1) ( 1) (3 1)
5( 1) (2 1) 2
dy d dx x x x x x
dx dx dxd
x x x x xdx
dx x x x
dx
x x x x
x x x
CHAIN RULE Example 6
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Noticing that each term has the common
factor 2(2x + 1)4(x3 – x + 1)3, we could factor
it out and write the answer as:
4 3 3 3 22(2 1) ( 1) (17 6 9 3)dy
x x x x x xdx
Example 6CHAIN RULE
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The reason for the name ‘Chain Rule’
becomes clear when we make a longer
chain by adding another link.
CHAIN RULE
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Suppose that y = f(u), u = g(x), and x = h(t),
where f, g, and h are differentiable functions,
then, to compute the derivative of y with
respect to t, we use the Chain Rule twice:
dy dy dx dy du dx
dt dx dt du dx dt
CHAIN RULE