# The Chain Rule

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In this section, we will learn about: Differentiating composite functions using the Chain Rule.
1. 3
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. 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).
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.
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.
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.
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
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:
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:
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
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.
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=
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
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.
17. 1/2121'()and'()22fuugxxu−===
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====+
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)
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====+
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
22. CHAIN RULE Example 2
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
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
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
26. COMBINING THE CHAIN RULE
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,
28. COMBINING 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−−===
30. POWER RULE WITH CHAIN RULE Rule 4
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.
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:
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,
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
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.
36. CHAIN RULE Example 6
(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:
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===
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.
40. CHAIN RULE
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:
42. CHAIN RULE Formula 6
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
44. CHAIN RULE
The reason for the name ‘Chain Rule’
becomes clear when we make a longer
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==
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.
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:
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)
49. HOW TO PROVE THE CHAIN RULE
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
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 .
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.
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
54. PROOF OF THE CHAIN RULE Equation 9
∆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
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.
57. PROOF OF THE limlim['()]['()]'()'()'(())'()
0210CHAIN RULE xxdyydxxfbg
This proves the Chain Rule.