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4 Techniques of Differentiationwith Applications
Copyright © Cengage Learning. All rights reserved.
4.4 The Chain Rule
3
The Chain Rule
We can now find the derivatives of expressions involving powers of x combined using addition, subtraction, multiplication, and division, but we still cannot take the derivative of an expression like (3x + 1)0.5.
For this we need one more rule.
The function h(x) = (3x + 1)0.5 is not a sum, difference, product, or quotient.
4
The Chain Rule
To find out what it is, we can use the calculation thought experiment and think about the last operation we would perform in calculating h(x).
1. Calculate 3x + 1.
2. Take the 0.5 power (square root) of the answer.
The last operation is “take the 0.5 power.” We do not yet have a rule for finding the derivative of the 0.5 power of a quantity other than x.
5
The Chain Rule
There is a way to build h(x) = (3x + 1)0.5 out of two simpler functions: u(x) = 3x + 1 (the function that corresponds to the first step in the calculation above) and f (x) = x0.5 (the function that corresponds to the second step):
h(x) = (3x + 1)0.5
= [u(x)]0.5
= f (u(x))
u(x) = 3x + 1
f (x) = x0.5
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The Chain Rule
We say that h is the composite of f and u. We read f (u(x)) as “f of u of x.”
To compute h(1), say, we first compute 3 1 + 1 = 4 and then take the square root of 4, giving h(1) = 2.
To compute f (u(1)) we follow exactly the same steps: First compute u(1) = 4 and then f (u(1)) = f (4) = 2.
We always compute f (u(x)) numerically from the inside out: Given x, first compute u(x) and then f (u(x)).
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The Chain Rule
Now, f and u are functions whose derivatives we know.
The chain rule allows us to use our knowledge of the derivatives of f and u to find the derivative of f (u(x)).
For the purposes of stating the rule, let us avoid some of the nested parentheses by abbreviating u(x) as u.
Thus, we write f (u) instead of f (u(x)) and remember that u is a function of x.
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The Chain Rule
Chain Rule
If f is a differentiable function of u and u is a differentiable function of x, then the composite f (u) is a differentiable function of x, and
In words The derivative of f(quantity) is the derivative of f, evaluated at that quantity, times the derivative of the quantity.
Chain Rule
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The Chain Rule
Quick Example
Take f (u) = u2. Then
The derivative of a quantity squared is two times the quantity, times the derivative of the quantity.
Because f '(u) = 2u
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The Chain Rule
The following table gives more examples.
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The Chain Rule
To motivate the chain rule, let us see why it is true in the special case when f (u) = u3, where the chain rule tells us that
But we could have done this using the product rule instead:
which gives us the same result.
Generalized Power Rule with n = 3
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The Chain Rule
A similar argument works for f (u) = un where n = 2, 3, 4, …
We can then use the quotient rule and the chain rule for
positive powers to verify the generalized power rule for
negative powers as well.
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Example 1 – Using the Chain Rule
Compute the following derivatives.
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Example 1 – Solution
a. Using the calculation thought experiment, we see that the last operation we would perform in calculating (2x2 + x)3 is that of cubing.
Thus we think of (2x2 + x)3 as a quantity cubed.
There are two similar methods we can use to calculate its derivative.
Method 1: Using the formula We think of (2x2 + x)3 as u3, where u = 2x2 + x.
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Example 1 – Solution
By the formula,
Now substitute for u:
= 3(2x2 + x)2(4x + 1)
cont’d
Generalized Power Rule
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Example 1 – Solution
Method 2: Using the verbal form If we prefer to use the verbal form, we get:
The derivative of (2x2 + x) cubed is three times (2x2 + x) squared, times the derivative of (2x2 + x).
In symbols,
as we obtained above.
cont’d
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Example 1 – Solution
b. First, the calculation thought experiment: If we were computing (x3 + x)100, the last operation we would
perform is raising a quantity to the power 100.
Thus we are dealing with a quantity raised to the power 100, and so we must again use the generalized power rule.
According to the verbal form of the generalized power rule, the derivative of a quantity raised to the power 100 is 100 times that quantity to the power 99, times the derivative of that quantity.
cont’d
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Example 1 – Solution
In symbols,
cont’d
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Example 1 – Solution
c. We first rewrite the expression as and then use the generalized power rule as in parts (a) and (b):
The derivative of a quantity raised to the 0.5 is 0.5 times the quantity raised to the –0.5, times the derivative of the quantity.
Thus,
cont’d
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Example 1 – Solution
d. The calculation thought experiment tells us that |4x2 − x| is the absolute value of a quantity, so we use the generalized rule for absolute values (above):
The derivative of the absolute value of a quantity is the absolute value of the quantity divided by the quantity, times the derivative of the quantity.
cont’d
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Example 1 – Solution
Thus,
cont’d
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Applications
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Example 4 – Marginal Product
A consultant determines that Precision Manufacturers’
annual profit (in dollars) is given by
P = –200,000 + 4,000q – 0.46q2 – 0.00001q3
where q is the number of surgical lasers it sells each year.
The consultant also informs Precision that the number of
surgical lasers it can manufacture each year depends on
the number n of assembly line workers it employs
according to the equation
q = 100n
Use the chain rule to find the marginal product
Each worker contributes 100 lasers per year.
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Example 4 – Solution
We could calculate the marginal product by substituting the expression for q in the expression for P to obtain P as a function of n and then finding dP/dn.
Alternatively—and this will simplify the calculation—we can use the chain rule.
To see how the chain rule applies, notice that P is a function of q, where q in turn is given as a function of n.
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Example 4 – Solution
By the chain rule,
Now we compute
and
cont’d
Chain Rule
Notice how the “quantities” dq appear to cancel.
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Example 4 – Solution
Substituting into the equation for gives
Notice that the answer has q as a variable. We can express dP/dn as a function of n by substituting 100n for q:
cont’d
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Chain Rule in Differential Notation
If y is a differentiable function of u, and u is a differentiable function of x, then
Notice how the units cancel:
Applications
The terms du cancel.
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Applications
Quick Example
If y = u3, where u = 4x + 1, then
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Applications
Manipulating Derivatives in Differential Notation
1. Suppose y is a function of x. Then, thinking of x as a function of y (as, for instance, when we can solve for x) we have
Quick Example
In the demand equation q = –0.2p – 8, we have
Therefore,
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Applications
2. Suppose x and y are functions of t. Then, thinking of y as a function of x (as, for instance, when we can solve for t as a function of x, and hence obtain y as a function of x) we have
Quick Example
If x = 3 – 0.2t and y = 6 + 6t, then
The terms dt appear to cancel.
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Applications
To see why the above formulas work, notice that the second formula,
can be written as
which is just the differential form of the chain rule.
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Applications
For the first formula, use the second formula with y playing the role of t:
