3 DERIVATIVES DERIVATIVES
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
3DERIVATIVESDERIVATIVES
In this section, we will learn about:
Differentiating composite functions
using the Chain Rule.
DERIVATIVES
3.5 The Chain Rule
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
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.
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.
( )y f u u
CHAIN RULE
It seems plausible if we interpret derivatives as rates
of change. 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
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
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
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
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
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
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
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
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
Since
we have
1/ 212
1'( ) and '( ) 2
2f u u g x x
u
Solution:
2
2
'( ) '( ( )) '( )
12
2 1
1
F x f g x g x
xxx
x
E. g. 1—Solution 1
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
Solution: E. g. 1—Solution 2
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
Differentiate:
a. y = sin(x2)
b. y = sin2 x
Example 2CHAIN RULE
If y = sin(x2), the outer function is the sine
function and the inner function is the squaring
function.
So, the Chain Rule gives:
2 2
outer derivative of derivative ofevaluated at evaluated atfunction outer function inner functioninner function inner function
2
sin ( ) cos ( ) 2
2 cos( )
dy dx x x
dx dx
x x
Example 2 aSolution:
Note that sin2x = (sin x)2. Here, the outer
function is the squaring function and the inner
function is the sine function.
Therefore,
2
derivativeofderivativeofinner evaluatedat inner functionouter functionfunction inner function
(sin ) 2 . sin cos
sin(2 )
dy dx x x
dx dx
x
Example 2 bSolution:
In general, if y = sin u, where u is
a differentiable function of x, then,
by the Chain Rule,
Thus,
cosdy dy du du
udx du dx dx
(sin ) cosd du
u udx dx
COMBINING THE CHAIN RULE
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
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
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
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
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
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
Solution: Example 6
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)
dyx x x x x x
dx
Example 6Solution:
Figure 3.5.1, p. 159
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
If
Notice that we used the Chain Rule twice.
2
( ) sin(cos(tan )), then
'( ) cos(cos(tan )) cos(tan )
cos(cos(tan ))[ sin(tan )] (tan )
cos(cos(tan ))sin(tan )sec
f x x
df x x x
dxd
x x xdx
x x x
Example 7CHAIN RULE
Differentiate .
The outer function is the square root function, the middle function is the secant function, and the inner function is the cubing function.
Thus,
Example 8CHAIN RULE
3
3
2 3 33 3 3
3 3
1sec
2 sec
1 3 sec tansec tan
2 sec 2 sec
dy dx
dx dxx
d x x xx x x
dxx x
3secy x
Recall that if y = f(x) and x changes from
a to a + ∆x, we defined the increment
of y as:
∆y = f(a + ∆x) – f(a)
According to the definition of a derivative,
we have:
HOW TO PROVE THE CHAIN RULE
0lim '( )x
yf a
x
So, if we denote by ε the difference
between the difference quotient and the
derivative, we obtain:
0 0lim lim '( )
'( ) '( ) 0
x x
yf a
x
f a f a
HOW TO PROVE THE CHAIN RULE
However,
If we define ε to be 0 when ∆x = 0, then ε becomes a continuous function of ∆x .
'( ) '( )y
f a y f a x xx
HOW TO PROVE THE CHAIN RULE
Thus, for a differentiable function f, we can
write:
ε is a continuous function of ∆x.
This property of differentiable functions is what enables us to prove the Chain Rule.
'( ) where 0 as 0y f a x x x
HOW TO PROVE THE CHAIN RULE Equation 5
Suppose u = g(x) is differentiable at a and
y = f(u) at b = g(a).
If ∆x is an increment in x and ∆u and ∆y
are the corresponding increments in u and y,
then we can use Equation 5 to write
∆u = g’(a) ∆x + ε1 ∆x = [g’(a) + ε1] ∆x
where ε1 → 0 as ∆x → 0
PROOF OF THE CHAIN RULE Equation 6
Similarly,
∆y = f’(b) ∆u + ε2 ∆u = [f’(b) + ε2] ∆u
where ε2 → 0 as ∆u → 0.
Equation 7PROOF OF THE CHAIN RULE
If we now substitute the expression for ∆u
from Equation 6 into Equation 7, we get:
So,
2 1[ '( ) ][ '( ) ]y f b g a x
PROOF OF THE CHAIN RULE
2 1[ '( ) ][ '( ) ]y
f b g ax
As ∆x→ 0, Equation 6 shows that
∆u→ 0.
So, both ε1 → 0 and ε2 → 0 as ∆x→ 0.
PROOF OF THE CHAIN RULE
Therefore,
This proves the Chain Rule.
0
2 10
lim
lim[ '( ) ][ '( ) ]
'( ) '( )
'( ( )) '( )
x
x
dy y
dx xf b g a
f b g a
f g a g a
PROOF OF THE CHAIN RULE
Related Documents
