Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 3: Rules Of Differentiation – Section 3.3: Differentiation Of Compositions Of Functions – The Chain Rule

3.3
Differentiation Of Compositions Of Functions –
The Chain Rule

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1. Compositions Of Functions

Consider an example. Let f(x) = x² and g(x) = 3x + 1. Then f( g(x)) = ( g(x))² = (3x + 1)². We've obtained a new
function, whose value at x is (3x + 1)². This new function is obtained by combining or composing functions f and g, and
thus is called the composition, or the composite function, of f and g. It's denoted by f o g, pronounced “ f circle g ” or
“ f round g ”. Hence, ( f o g)(x) = f( g(x)) = (3x + 1)².

In the notation f( g(x)), f is the outside function and g the inside one. To compute ( f o g)(x) = (3x + 1)², first we
compute 3x + 1 = g(x), then we compute (3x + 1)² = f( g(x)) = ( f o g)(x). Therefore, the inside function g precedes

Fig. 1.1 Composition Of Functions f And g.

the outside function f in the order of computation of ( f o g)(x) or f( g(x)). See Fig. 1.1. Keep in mind that here we have
three different functions: f, g, and f o g. The domain of f o g is the set of all x such that f( g(x)) exists or makes sense.
So it's the set of all x in the domain of g such that g(x) is in the domain of f.

Definition 1.1

Let f and g be functions and D the set of all x in dom( g) such that g(x) is in dom( f ). Then the composition, or
composite function, of f and g is the function from D to range( f ) that is denoted by f o g, pronounced “ f circle
g ” or “ f round g ”, and that is defined by:

( f o g)(x) = f( g(x))

for all x in D. Note that D = dom( f o g).

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2. The Chain Rule

We're going to establish in the following theorem the formula for the derivative of the composition f o g of differentiable
functions f and g in terms of the derivatives of f and g. Let u = g(x) and y = f(u) = f( g(x)) = ( f o g)(x). See Fig. 2.1.
Recall the interpretation of the derivative as the rate of change as discussed in Section 2.3. Suppose du/dx = 3 and

Fig. 2.1 y = f(u) = f( g(x)) = ( f o g)(x).

dy/du = 2. As x changes, u changes 3 times as fast as x does and y changes 2 times as fast as u does, thus y changes
2 x 3 = 6 times as fast as x does. We've just observed intuitively that the rate of change of y with respect to x is equal to
the rate of change of y with respect to u multiplied by the rate of change of u with respect to x: dy/dx = (dy/du)(du/dx).
That's the same as to say that the derivative of f o g with respect to x is equal to the derivative of f with respect to g(x)
multiplied by the derivative of g with respect to x: ( f o g)'(x) = f '( g(x))g'(x).

Theorem 2.1 – The Chain Rule

If u = g(x) is differentiable at the point x and y = f(u) is differentiable at the point g(x), then y = f( g(x)) =
( f o g)(x) is differentiable at x and:

( f o g)'(x) = f '( g(x))g'(x).

In Leibniz notation:

Proof

Hence:

EOP

Note On The Proof

Remark 2.1

In the Leibniz notation:

the du's appear to cancel from the numerator and denominator of the two fractions. The notations dy/dx, dy/du, and
du/dx appear as normal fractions. This is useful in remembering the formula.

A Simplified Notation

The formula ( f o g)'(x) = f '(g(x))g'(x) can be written as ( f(g(x)))' = f '(g(x))g'(x). Now let u = u(x) = g(x). Then
this last formula can be written in the following somehow simplified form, which is perhaps easier to remember:

( f(u))' = f '(u)u'(x),

where ( f(u))' = (d/dx) f(u) (derivative of f(u) with repsect to x), f '(u) = (d/du) f(u) (derivative of f(u) with respect to
u), and u'(x) = g'(x).

Example 2.1

Differentiate g(x) = (3x + 4)².

Solution
g'(x) = 2(3x + 4)(3) = 6(3x + 4).
EOS

We think of 3x + 4 as u (u = 3x + 4) and f(u) as u² ( f(u) = u²). Then g(x) = f(u) and so g'(x) = (d/dx) f(u) = ( f(u))' =
f '(u)u'(x) = 2u(3) = 6(3x + 4). For this particular simple function g we can check: g(x) = (3x + 4)² = 9x² + 24x + 16;
thus g'(x) = 18x + 24 = 6(3x + 4), the same as found by the chain rule.

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3. The Power Rule – Differentiation Of Integer Powers Of Functions

In Section 3.2 Corollary 4.1 we have that for any integer n, the derivative of xⁿ is nx^n–1: (d/dx)xⁿ = nx^n–1. That's the
derivative of integer powers of variables. We now extend that to integer powers of functions.

Corollary 3.1 – The Power Rule For Integer Exponents

Proof
Let y = (u(x))ⁿ. Using the chain rule we obtain:

EOP

Example 3.1

Differentiate g(x) = (3x + 4)².

Solution
g'(x) = 2(3x + 4)(3) = 6(3x + 4).
EOS

This is the same function g as in Example 2.1, where we used the chain rule directly to differentiate g.

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4. The Power Rule – Differentiation Of Rational Powers Of Functions

We now extend the power rule to rational powers of functions. Recall that a rational number is a number that can be
written as a ratio or fraction m/n, where m is an integer and n a positive integer.

Corollary 4.1 – The Power Rule For Rational Exponents

Proof
There exist an integer m and a positive integer n such that r = m/n. Utilizing the power rule for integer exponents we
have:

EOP

We'll see in Section 6.3 Eq. [4.1] that (d/dx) x^a = ax^a–1 for any real number a, from which we obtain, by the chain rule,
(d/dx) (u(x))^a = a(u(x))^a–1(du/dx) for any real number a, rational or irrational.

Example 4.1

Solution

EOS

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5. Differentiation Of Square Roots Of Functions

In summary, for all x > 0 or for all x where u(x) > 0:

Remark that the first formula was also obtained in Section 3.2 Corollary 2.1.

Example 5.1

Solution

EOS

This function h(t) was also differentiated in Example 4.1 using the power rule. As a matter of fact for the square root
function the square root rule as seen here is simpler than the power rule.

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6. Applying The Chain Rule More Than Once In One Step

Example 6.1

Solution 1

EOS

In the above solution, we apply the chain rule twice in two different steps: first to differentiate the 10th power, and then
to differentiate the 15th power. We can and it's better to apply all the instances of the chain rule in just one step, as
shown in Solution 2 below.

Solution 2

EOS

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Problems & Solutions

1. Differentiate the following functions. You needn't simplify your answers.

Solution

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2. Let y = f(u) = (u² + 3u – 4)^3/2 and u = g(x) = x³ – 3. Find ( f o g)'(2) by:
a. Expressing y directly as a function of x and differentiating.
b. Using the chain rule.

Solution

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3. a. Show that (d/dx) |x| = sgn x, where sgn is the signum function defined by:

b. Find f '(x) if f(x) = |2 + x³|.

Solution

b. f '(x) = (sgn (2 + x³))(3x²) = 3x² sgn (2 + x³).

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4. Use the formulas (d/dx)sin x = cos x, (d/dx) cos x = – sin x, and (d/dx) ln x = 1/x.
a. Find f '(x) if f(x) = sin cos sin³ x.
b. Find v' if v = cos² (5 – 4y³).
c. Calculate (d/dt) ln (a ln(bt + c)).

Solution

a. f '(x) = (cos cos sin³ x)(– sin sin³ x)(3 (sin² x)(cos x)) = –3 sin² x cos x cos cos sin³ x sin sin³ x.

b. v' = 2 (cos (5 – 4y³))(– sin (5 – 4y³))(–12y²) = 24y² cos (5 – 4y³) sin (5 – 4y³).

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5. Let y = (u + 1)/u. Suppose u = g(x) and g(3) = 2, where g is a differentiable function, and suppose (dy/dx)|_x=3 = – 5.
Find g'(3).

Solution

We have:

or – 5 = (–1/2²)g'(3) = (–1/4)g'(3). Thus, g'(3) = 20.

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6. Let f(x) = (x – a)^m(x – b)ⁿ, where a < b and m and n are positive integers. Prove that there exists c where a < c < b
such that the derivative of f vanishes at c.

Solution

We have:

Using the facts that a < b, m > 0, and n > 0 we get:

a – b < 0 < b – a,
m(a – b) < 0 < n(b – a),
ma – mb < 0 < nb – na,
ma – mb + na – na < 0 < nb – na + mb – mb,
a(m + n) – (mb + na) < 0 < b(m + n) – (mb + na),
a(m + n) < mb + na < b(m + n),

completes the proof.

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