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11 Techniques of Differentiationwith Applications

Copyright © Cengage Learning. All rights reserved.

11.3 The Product and Quotient Rules

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The Product and Quotient Rules

Product Rule

If f (x) and g (x) are differentiable functions of x, then so is their product f (x)g (x), and

Product Rule in Words

The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.

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The Product and Quotient Rules

Quick Example

f (x) = x2 and g (x) = 3x – 1 are both differentiable functions of x, and so their product x2(3x – 1) is differentiable, and

Quotient Rule

If f (x) and g(x) are differentiable functions of x, then so is their quotient f (x)/g(x) (provided g(x) 0), and

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The Product and Quotient Rules

Quotient Rule in Words

The derivative of a quotient is the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

Quick Example

f (x) = x3 and g (x) = x2 + 1 are both differentiable functions of x, and so their quotient x3/(x2 + 1) is differentiable, and

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Derivation of the Product Rule

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Derivation of the Product Rule

To calculate the derivative of the product f (x)g (x) of twodifferentiable functions, we go back to the definition of the derivative:

We now rewrite this expression so that we can evaluate the limit: Notice that the numerator reflects a simultaneous change in f [from f (x) to f (x + h)] and g [from g (x) to g (x + h)].

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Derivation of the Product Rule

To separate the two effects, we add and subtract a quantity in the numerator that reflects a change in only one of the functions:

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Derivation of the Product Rule

1010

Derivation of the Product Rule

Now we already know the following four limits:

Putting these limits into the one we’re calculating, we get

which is the product rule.

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Example 5 – Using the Product Rule

Compute the following derivatives.

Solution:

a. We can do the calculation in two ways.

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Example 5 – Solution

Not Using the Product Rule: First, expand the given expression.

(x3.2 + 1)(1 – x) = –x4.2 + x3.2 – x + 1

Thus,

[(x3.2 + 1)(1 – x)] = (–x4.2 + x3.2 – x + 1)

= –4.2x3.2 + 3.2x2.2 – 1

cont’d

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Example 5 – Solution

b. Here we have a product of three functions, not just two. We can find the derivative by using the product rule twice:

[(x + 1)(x2 + 1)(x3 + 1)]

= (x + 1) [(x2 + 1)(x3 + 1)] + (x + 1) [(x2 + 1)(x3 + 1)]

= (1)(x2 + 1)(x3 + 1) + (x + 1)[(2x)(x3 + 1) + (x2 + 1)(3x2)]

= (1)(x2 + 1)(x3 + 1) + (x + 1)(2x)(x3 + 1) + (x + 1)(x2 + 1)(3x2)

cont’d

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Example 5 – Solution

We can see here a more general product rule:

(fgh) = f gh + fgh + fgh

Notice that every factor has a chance to contribute to the rate of change of the product. There are similar formulas for products of four or more functions.

cont’d

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Derivation of the Product Rule

Calculation Thought Experiment

The calculation thought experiment is a technique to determine whether to treat an algebraic expression as a product, quotient, sum, or difference.

Given an expression, consider the steps you would use in computing its value. If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient; and so on.

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Derivation of the Product Rule

Quick Example

(3x2 – 4)(2x + 1) can be computed by first calculating the expressions in parentheses and then multiplying.

Because the last step is multiplication, we can treat the expression as a product.

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Applications

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Example 8 – Applying the Product and Quotient Rules: Revenue and Average Cost

Sales of your newly launched miniature wall posters for college dorms, iMiniPosters, are really taking off. (Those old-fashioned large wall posters no longer fit in today’s “downsized” college dorm rooms.) Monthly sales to students at the start of this year were 1,500 iMiniPosters, and since that time, sales have been increasing by 300 posters each month, even though the price you charge has also been going up.

a. The price you charge for iMiniPosters is given by:

p(t) = 10 + 0.05t2 dollars per poster,

where t is time in months since the start of January of this year.

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Example 8 – Applying the Product and Quotient Rules: Revenue and Average Cost

Find a formula for the monthly revenue, and then compute its rate of change at the beginning of March.

b. The number of students who purchase iMiniPosters in a month is given by

n(t) = 800 + 0.2t,

where t is as in part (a). Find a formula for the average number of posters each student buys, and hence estimate the rate at which this number was growing at the beginning of March.

cont’d

2020

Example 8(a) – Solution

To compute monthly revenue as a function of time t, we use

R(t) = p(t)q(t).

We already have a formula for p(t). The function q(t) measures sales, which were 1,500 posters/month at time t = 0, and rising by 300 per month:

q(t) = 1,500 + 300t.

Therefore, the formula for revenue is

R(t) = p(t)q(t)

R(t) = (10 + 0.05t2)(1,500 + 300t).

Revenue = Price Quantity

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Example 8(a) – Solution

Rather than expand this expression, we shall leave it as a product so that we can use the product rule in computing its rate of change:

R(t) = p(t)q(t) + p(t)q(t) = [0.10t ][1,500 + 300t ] + [10 + 0.05t

2][300].

Because the beginning of March corresponds to t = 2, we have

R(2) = [0.10(2)][1,500 + 300(2)] + [10 + 0.05(2)2][300]

= (0.2)(2,100) + (10.2)(300)

= $3,480 per month.

Therefore, your monthly revenue was increasing at a rate of $3,480 per month at the beginning of March.

cont’d

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Example 8(b) – Solution

The average number of posters sold to each student is

The rate of change of k(t) is computed with the quotient rule:

cont’d

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Example 8(b) – Solution

so that

Therefore, the average number of posters sold to each student was increasing at a rate of about 0.37 posters/student per month.

cont’d