When a function operates on the output of another function, the combination is a third func-tion known as a composite funcfunc-tion. Sometimes it’s called a funcfunc-tion of a funcfunc-tion. We can use a procedure called the chain rule for differentiation to find the derivative of a composite function.
Definition and notation
To determine a function of a function, we apply the second function to the output of the first.
In other words, we apply the first (or “inner”) function to its independent variable, and then we apply the second (or “outer”) function to the value of the first function. For example, if
f(x ) = 6x 3 and
g ( y ) = −3y then
g[ f (x )] = −3× (6x 3) = −18x 3
Chain Rule 99
Here’s another example. If
f(x ) = (x 3+ 1) and
g ( y )= (4y 2− 3y ) then
g[ f (x )] = 4× (x 3+ 1)2− 3 × (x 3+ 1)
= 4 × (x6+ 2x 3+ 1) − 3x 3− 3 = 4x6+ 5x 3+ 1
To make things look simpler, we can write g ( f ) without including the first function’s independent variable. But we don’t want to give our readers the idea that we mean the product g f or g ×f. Some texts put a small letter o, which stands for “of,” between the two functions.
In this situation they’d write g ( f ) as g o f.
Note that g ( f ) is rarely the same as f ( g). The function-of-a-function operation is not, in general, commutative.
The rule in brief
The chain rule can be stated informally like this:
• To differentiate a function of a function, we multiply the derivative of the “outer”
function by the derivative of the “inner” function.
For the chain rule to apply, both functions must be differentiable at every point that interests us. Stated symbolically, if f and g are differentiable functions of the same single variable, then
[g ( f )]′ =g ′ ( f ) ×f ′
If we want to include an independent variable (say x ) in the notation, then we can write the above statement as
{g [ f (x )]}′ =g ′ [ f (x )] ×f ′(x )
taking care not to confuse the multiplication symbol with the variable x.
Are you confused?
The above notation can baffle some people, because the difference between the two expressions {g [ f (x )]}′
and
g ′ [ f (x )]
is subtle. Let’s clarify this before we go any further! The first expression, stated as an ordered list of instruc-tions, reads like this:
• Apply the function f to the variable x.
• Apply the function g to that output.
• Differentiate the function g with respect to x to get the final result.
The second expression tells us to do these things in order:
• Apply the function f to the variable x.
• Differentiate the function g with respect to some arbitrary variable other than x. (We can call it anything we want, such as y .)
• Plug in f (x ) as that other variable.
• Apply the derivative function g ′ to that output to get the final result.
Example
Let’s see how this rule works. Consider the first of the two examples we saw a short while ago:
f(x ) = 6x 3 and
g ( y ) = −3y
When we differentiate the composite function directly, we get {g [ f (x )]}′ = d /dx (−18x 3) = −54x 2 The derivatives of the individual functions are
f ′(x ) = 18x 2 and
g ′( y ) = −3 When we apply the chain rule, we get
{g [ f (x )]}′ = g′ [ f (x )] ×f ′(x ) = −3×d /dx (6x 3) = −3× 18x 2 = −54x 2
Chain Rule 101
Another example
Now let’s see what happens with the composite of the two functions f(x ) = (x 3+ 1)
and
g ( y ) = (4y 2− 3y ) The derivatives of the individual functions are
f ′(x ) = 3x 2 and
g ′( y ) = 8y− 3 Differentiating the composite function directly gives us
{g [ f (x )]}′ =d /dx (4x6+ 5x 3+ 1) = 24x 5+ 15x 2 When we apply the chain rule, we get
{g [ f (x )]}′ = g ′ [ f (x )] ×f ′ (x ) = [8 × (x 3+ 1) − 3] ×d /dx (x 3+ 1) = (8x 3+ 5) (3x 2) = 24x 5+ 15x 2
Are you confused?
If you aren’t certain that you understand how the chain rule works, make up some more examples in the same format as the previous two. First, differentiate the composite function as a whole. Then, use the chain rule. You should always get the same final answer either way. Problems like this are “self-checking.”
If you make a mistake anywhere, you’ll probably get answers that don’t agree with each other.
Here’s a challenge!
Show that the following statement is not true in general:
g (f ) =f ( g)
Solution
It’s easy to prove that a statement isn’t universally true. We simply find a situation called a counterexample where it’s false. Consider
f(x ) = (x 3+ 1)
and
g ( y ) = (4y 2− 3y ) We’ve already determined that
g[ f (x )] = 4x6+ 5x 3+ 1 Working the other way, we get
f [ g ( y )] = (4y 2− 3y )3+ 1 = 64y6− 144y5+ 108y4− 27y3+ 1
The function g ( f ) operates on a variable we call x, and the function f ( g) operates on a variable we call y. But the names of the variables aren’t important. The composite functions g ( f ) and f ( g) do different things. That’s what matters!
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
1. Find the derivative of the polynomial function
f (x ) = −4x 4+ 2x 3−x 2−x+ 1 Then find the derivative of
g (x ) = 2× (−4x 4+ 2x 3−x 2−x+ 1) = −8x 4+ 4x 3− 2x 2− 2x+ 2 Verify that g ′ = 2f ′.
2. Find the derivative of the polynomial function
f (x ) = −40x 4+ 20x 3− 10x 2− 10x+ 10 Then find the derivative of
g (x ) = (1/5)× (−40x 4+ 20x 3− 10x 2− 10x+ 10) = −8x 4+ 4x 3− 2x 2− 2x+ 2 Verify that g ′ = (1/5) f ′.
Practice Exercises 103
3. Show that the two-function product rule for differentiation is commutative. That is, show that if f and g are differentiable functions, then
(f g )′ = ( g f )′
4. Based on the two-function product rule for differentiation, derive a three-function product rule for differentiation. Assume that all three functions are differentiable.
5. Find the derivative of the function
p ( y ) = 1 / ( y 2+ 1)
in the set of real numbers. Indicate the real values of y, if any, for which this function is nondifferentiable.
6. Find the derivative of the function
r ( y ) = 1 / ( y 2− 1)
in the set of real numbers. Indicate the real values of y, if any, for which this function is nondifferentiable.
7. Find the derivative of the function
s ( z)= (z4− 1) / (z2+ 1)
in the set of real numbers. First differentiate s (z) directly after dividing out the
quotient. Then differentiate s (z) using the quotient rule. Indicate the real values of z, if any, for which this function is nondifferentiable.
8. Find the derivative of the function
t (z)= (z4− 1) / (z2− 1)
in the set of real numbers. First differentiate t (z) directly after dividing out the
quotient. Then differentiate t (z) using the quotient rule. Indicate the real values of z, if any, for which this function is nondifferentiable.
9. Consider the functions
f (x ) =x 3− 4x 2 and
g ( y ) =y 2+ 5y
Differentiate the composite function g ( f ) directly. Then do it using the chain rule.
10. Consider the functions
f(x ) =x 2− 4x and
g ( y ) = 2y 2+ 7y
Differentiate the composite function g ( f ) directly. Then do it using the chain rule.
Practice Exercises 105
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