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In this section, we will learn about: Differentiating composite functions using the Chain Rule.

1.
3

DIFFERENTIATION RULES

DIFFERENTIATION RULES

2.
DIFFERENTIATION RULES

3.4

The Chain Rule

In this section, we will learn about:

Differentiating composite functions

using the Chain Rule.

3.4

The Chain Rule

In this section, we will learn about:

Differentiating composite functions

using the Chain Rule.

3.
CHAIN RULE

Suppose you are asked to differentiate

the function ()1Fxx=+

2

The differentiation formulas you learned

in the previous sections of this chapter

do not enable you to calculate F’(x).

Suppose you are asked to differentiate

the function ()1Fxx=+

2

The differentiation formulas you learned

in the previous sections of this chapter

do not enable you to calculate F’(x).

4.
CHAIN RULE

Observe that F is a composite function.

()yfuu==

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.

Observe that F is a composite function.

()yfuu==

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.

5.
CHAIN RULE

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.

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.

6.
CHAIN RULE

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.

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.

7.
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

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

8.
CHAIN RULE

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.

dydydudxdudx=

So, we expect that:

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.

dydydudxdudx=

So, we expect that:

9.
THE 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)

dydydudxdudx=

are both differentiable functions, then:

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)

dydydudxdudx=

are both differentiable functions, then:

10.
COMMENTS ON THE PROOF

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)

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)

11.
COMMENTS ON THE PROOF Equation 1

It is tempting to write:

000000limlimlimlimlimlimxxxxuxdyydxx

It is tempting to write:

000000limlimlimlimlimlimxxxxuxdyydxx

12.
COMMENTS ON THE PROOF

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.

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.

13.
COMMENTS ON THE PROOF

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.

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.

14.
CHAIN RULE Equations 2 and 3

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:

dydydudxdudx=

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:

dydydudxdudx=

15.
CHAIN RULE

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

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

16.
CHAIN RULE 2 ()1Fxx=+ E. g. 1—Solution 1

Find F’(x) if

One way of solving this is by using Equation 2.

At the beginning of this section, we()fuu=

expressed F

as F(x) = (f ◦ g))(x) = f(g(x)) where

and g(x) = x2 + 1.

Find F’(x) if

One way of solving this is by using Equation 2.

At the beginning of this section, we()fuu=

expressed F

as F(x) = (f ◦ g))(x) = f(g(x)) where

and g(x) = x2 + 1.

17.
1/2121'()and'()22fuugxxu−===

CHAIN RULE E. g. 1—Solution 1

Since

22 '()'(())'()12211Fxfgxgxxxxx=⋅=⋅+=+

we have

CHAIN RULE E. g. 1—Solution 1

Since

22 '()'(())'()12211Fxfgxgxxxxx=⋅=⋅+=+

we have

18.
CHAIN RULE E. g. 1—Solution 2

We can also solve by using Equation 3.

,yu=

If we let u = x2 + 1 and

then: 221'()(2)21(2)211dyduFxxdxdxuxxxx====+

We can also solve by using Equation 3.

,yu=

If we let u = x2 + 1 and

then: 221'()(2)21(2)211dyduFxxdxdxuxxxx====+

19.
CHAIN RULE

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)

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)

20.
CHAIN RULE

For instance, in Example 1, y2(1)yx=+

can be

considered as a function of()yu=

x

and also as a function of u .

Note that:

2 1'()whereas'()21dyxdyFxfudxduux====+

For instance, in Example 1, y2(1)yx=+

can be

considered as a function of()yu=

x

and also as a function of u .

Note that:

2 1'()whereas'()21dyxdyFxfudxduux====+

21.
In using the Chain Rule, we work from

the outside to the inside.

Equation 2 states that we differentiate the outer

function f [at the inner function g(x)] and then

we multiply by the derivative of the inner function.

{ { { outerfunctionderivativederivativeevaluatedevaluatedofouterofinneratinneratinn

the outside to the inside.

Equation 2 states that we differentiate the outer

function f [at the inner function g(x)] and then

we multiply by the derivative of the inner function.

{ { { outerfunctionderivativederivativeevaluatedevaluatedofouterofinneratinneratinn

22.
CHAIN RULE Example 2

a. y = sin(x2)

b. y = sin2 x

a. y = sin(x2)

b. y = sin2 x

23.
CHAIN RULE Example 2 a

If y = sin(x2), the outer function is

the sine function and the inner function is

the squaring function.

{ {the

{ {Chain Rule gives:

22outerderivativeofderivativeofevaluatedatevaluatedatfunctionouterfunctioninn

If y = sin(x2), the outer function is

the sine function and the inner function is

the squaring function.

{ {the

{ {Chain Rule gives:

22outerderivativeofderivativeofevaluatedatevaluatedatfunctionouterfunctioninn

24.
CHAIN RULE Example 2 b

Note that sin2x = (sin x)2. Here, the outer

function is the squaring function and the inner

function is the sine function.

(){ 2derivativeofderivativeofinnerevaluatedatinnerfunctionouterfunctionfunctioninne

Note that sin2x = (sin x)2. Here, the outer

function is the squaring function and the inner

function is the sine function.

(){ 2derivativeofderivativeofinnerevaluatedatinnerfunctionouterfunctionfunctioninne

25.
CHAIN RULE Example 2 b

The answer can be left as 2 sin x cos x

or written as sin 2x (by a trigonometric

identity known as the double-angle

The answer can be left as 2 sin x cos x

or written as sin 2x (by a trigonometric

identity known as the double-angle

26.
COMBINING THE CHAIN RULE

In Example 2 a, we combined the Chain

Rule with the rule for differentiating the

sine function.

In Example 2 a, we combined the Chain

Rule with the rule for differentiating the

sine function.

27.
COMBINING THE CHAIN RULE

In general, if y = sin u, where u is

a differentiable function of x, then,

cosdydyduduudxdudxdx==

by the Chain Rule,

In general, if y = sin u, where u is

a differentiable function of x, then,

cosdydyduduudxdudxdx==

by the Chain Rule,

28.
COMBINING THE CHAIN RULE

In a similar fashion, all the formulas for

differentiating trigonometric functions can

be combined with the Chain Rule.

In a similar fashion, all the formulas for

differentiating trigonometric functions can

be combined with the Chain Rule.

29.
COMBINING CHAIN RULE WITH POWER RULE

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: 11 [()]'()nndydydudynungxgxdxdudxdx−−===

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: 11 [()]'()nndydydudynungxgxdxdudxdx−−===

30.
POWER RULE WITH CHAIN RULE Rule 4

If n is any real number1()and u = g(x) −=

nndduunudxdx

is differentiable, then

[()][()].'()nndgxngxgxdx−=

1

If n is any real number1()and u = g(x) −=

nndduunudxdx

is differentiable, then

[()][()].'()nndgxngxgxdx−=

1

31.
POWER RULE WITH CHAIN RULE

Notice that the derivative in Example 1

could be calculated by taking n = ½

in Rule 4.

Notice that the derivative in Example 1

could be calculated by taking n = ½

in Rule 4.

32.
POWER RULE WITH CHAIN RULE Example 3

Differentiate y = (x3 – 1)100

Taking u = g(x) = x3 – 1 and n = 100 in the rule,

3100399339922399(1)100(1)(1)100(1)3300(1)dydx

we have:

Differentiate y = (x3 – 1)100

Taking u = g(x) = x3 – 1 and n = 100 in the rule,

3100399339922399(1)100(1)(1)100(1)3300(1)dydx

we have:

33.
321.()1fxxx=++

POWER RULE WITH CHAIN RULE Example 4

Find f’ (x) if

First, rewrite f as f(x) = (x2 + x + 1)-1/3

1'()(1)(1)31(1)(21)3dfxxxxxdxxxx−−=−++

24/3224/3

Thus,

POWER RULE WITH CHAIN RULE Example 4

Find f’ (x) if

First, rewrite f as f(x) = (x2 + x + 1)-1/3

1'()(1)(1)31(1)(21)3dfxxxxxdxxxx−−=−++

24/3224/3

Thus,

34.
92()21tgtt−

POWER RULE WITH CHAIN RULE ⎛⎞5=⎜⎟+⎝⎠

Example

Find the derivative of

Combining the Power Rule, Chain Rule,

and Quotient Rule, we get:

22'()921212(21)12(2)45(2)921(21)(21)tdtgttdtttttt

888210

POWER RULE WITH CHAIN RULE ⎛⎞5=⎜⎟+⎝⎠

Example

Find the derivative of

Combining the Power Rule, Chain Rule,

and Quotient Rule, we get:

22'()921212(21)12(2)45(2)921(21)(21)tdtgttdtttttt

888210

35.
CHAIN RULE Example 6

y = (2x + 1)5 (x3 – x + 1)4

In this example, we must use the Product Rule

before using the Chain Rule.

y = (2x + 1)5 (x3 – x + 1)4

In this example, we must use the Product Rule

before using the Chain Rule.

36.
CHAIN RULE Example 6

(21)(1)(1)(21)(21)4(1)(1)(1)5(21)(21)4(21)(

(21)(1)(1)(21)(21)4(1)(1)(1)5(21)(21)4(21)(

37.
CHAIN RULE 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:

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:

38.
CHAIN RULE Example 7

Differentiate y = esin x

Here, the inner function is g(x) = sin x and the outer

function is the exponential function f(x) = ex.

So, by the Chain Rule:

sinsinsin ()(sin)cosxxxdydedxdxdexexdx===

Differentiate y = esin x

Here, the inner function is g(x) = sin x and the outer

function is the exponential function f(x) = ex.

So, by the Chain Rule:

sinsinsin ()(sin)cosxxxdydedxdxdexexdx===

39.
CHAIN RULE

We can use the Chain Rule to

differentiate an exponential function

with any base a > 0.

Recall from Section 1.6 that a = eln a.

So, ax = (eln a)x = e(ln a)x.

We can use the Chain Rule to

differentiate an exponential function

with any base a > 0.

Recall from Section 1.6 that a = eln a.

So, ax = (eln a)x = e(ln a)x.

40.
CHAIN RULE

Thus, the Chain Rule gives

(ln)(ln)(ln) ()()(ln)lnlnxaxaxaxxdddaeeaxdxdxdxea

because ln a is a constant.

Thus, the Chain Rule gives

(ln)(ln)(ln) ()()(ln)lnlnxaxaxaxxdddaeeaxdxdxdxea

because ln a is a constant.

41.
CHAIN RULE Formula 5

Therefore, we have the formula:

()lnxxdaaadx=

Therefore, we have the formula:

()lnxxdaaadx=

42.
CHAIN RULE Formula 6

In particular, if a = 2, we get:

(2)2ln2xxddx=

In particular, if a = 2, we get:

(2)2ln2xxddx=

43.
CHAIN RULE

In Section 3.1, we gave the estimate

(2)(0.69)2xxddx≈

This is consistent with the exact Formula 6

because ln 2 ≈ 0.693147

In Section 3.1, we gave the estimate

(2)(0.69)2xxddx≈

This is consistent with the exact Formula 6

because ln 2 ≈ 0.693147

44.
CHAIN RULE

The reason for the name ‘Chain Rule’

becomes clear when we make a longer

chain by adding another link.

The reason for the name ‘Chain Rule’

becomes clear when we make a longer

chain by adding another link.

45.
CHAIN RULE

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:

dydydxdydudxdtdxdtdudxdt==

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:

dydydxdydudxdtdxdtdudxdt==

46.
CHAIN RULE Example 8

()sin(cos(tan)),then'()cos(cos(tan))cos(tan)cos(c

2

Notice that we used the Chain Rule twice.

()sin(cos(tan)),then'()cos(cos(tan))cos(tan)cos(c

2

Notice that we used the Chain Rule twice.

47.
CHAIN RULE Example 9

Differentiate y = esec 3θ

The outer function is the exponential function,

the middle function is the secant function and

the inner function is the tripling function.

sec3sec3sec3 (sec3)sec3tan3(3)3sec3tan3dy

Thus, we have:

Differentiate y = esec 3θ

The outer function is the exponential function,

the middle function is the secant function and

the inner function is the tripling function.

sec3sec3sec3 (sec3)sec3tan3(3)3sec3tan3dy

Thus, we have:

48.
HOW TO PROVE THE CHAIN RULE

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)

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)

49.
HOW TO PROVE THE CHAIN RULE

According to the definition of a derivative,

we have: 0lim'()xyfaxΔ→ Δ=Δ

According to the definition of a derivative,

we have: 0lim'()xyfaxΔ→ Δ=Δ

50.
HOW TO PROVE THE CHAIN RULE

So, if we denote by ε the difference

between the difference quotient and the

derivative, we obtain:

00 limlim'()'()'()0xxyfaxfafaεΔ→Δ

So, if we denote by ε the difference

between the difference quotient and the

derivative, we obtain:

00 limlim'()'()'()0xxyfaxfafaεΔ→Δ

51.
HOW TO PROVE THE CHAIN RULE

If we define ε to be 0 when ∆x = 0, then ε

becomes a continuous function of ∆x .

If we define ε to be 0 when ∆x = 0, then ε

becomes a continuous function of ∆x .

52.
HOW TO PROVE THE CHAIN RULE Equation 7

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.

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.

53.
PROOF OF THE CHAIN RULE Equation 8

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 7 to write

∆u = g’(a) ∆x + ε1 ∆x = [g’(a) + ε1] ∆x

where ε1 → 0 as ∆x → 0

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 7 to write

∆u = g’(a) ∆x + ε1 ∆x = [g’(a) + ε1] ∆x

where ε1 → 0 as ∆x → 0

54.
PROOF OF THE CHAIN RULE Equation 9

∆y = f’(b) ∆u + ε2 ∆u = [f’(b) + ε2] ∆u

where ε2 → 0 as ∆u → 0.

∆y = f’(b) ∆u + ε2 ∆u = [f’(b) + ε2] ∆u

where ε2 → 0 as ∆u → 0.

55.
PROOF OF THE CHAIN RULE

If we now substitute the expression for ∆u

from Equation 8 into Equation 9, we get:

21 ['()]['()]yfbgaxεεΔ=++Δ

['()]['()]yfbgaxεεΔ=++Δ

21

If we now substitute the expression for ∆u

from Equation 8 into Equation 9, we get:

21 ['()]['()]yfbgaxεεΔ=++Δ

['()]['()]yfbgaxεεΔ=++Δ

21

56.
PROOF OF THE CHAIN RULE

As ∆x→ 0, Equation 8 shows that

∆u→ 0.

So, both ε1 → 0 and ε2 → 0 as ∆x→ 0.

As ∆x→ 0, Equation 8 shows that

∆u→ 0.

So, both ε1 → 0 and ε2 → 0 as ∆x→ 0.

57.
PROOF OF THE limlim['()]['()]'()'()'(())'()

0210CHAIN RULE xxdyydxxfbg

This proves the Chain Rule.

0210CHAIN RULE xxdyydxxfbg

This proves the Chain Rule.