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