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11 Techniques of Differentiationwith Applications

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Copyright © Cengage Learning. All rights reserved.

11.6 Implicit Differentiation

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Implicit Differentiation

Consider the equation y5 + y + x = 0, whose graph is shown in Figure 10.

How did we obtain this graph? We did not solve for y as a function of x; that is impossible. In fact, we solved for x in terms of y to find points to plot.

Nonetheless, the graph in Figure 10 is the graph of a function because it passes the vertical line test: Every vertical line crosses the graph no more than once, so for each value of x there is no more than one corresponding value of y.

Figure 10

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Implicit Differentiation

Because we cannot solve for y explicitly in terms of x, we say that the equation y5 + y + x = 0 determines y as an implicit function of x.

Now, suppose we want to find the slope of the tangent line to this curve at, say, the point (2, –1) (which, you should check, is a point on the curve).

In the following example we find, surprisingly, that it is possible to obtain a formula for dy/dx without having to first solve the equation for y.

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Example 1 – Implicit Differentiation

Find given that y5 + y + x = 0.

Solution:

We use the chain rule and a little cleverness.

Think of y as a function of x and take the derivative with respect to x of both sides of the equation:

Original equation

Derivative with respect to x of both sides

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Example 1 – Solution cont’d

Now we must be careful. The derivative with respect to x of y5 is not 5y4.

Rather, because y is a function of x, we must use the chain rule, which tells us that

Derivative rules

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Example 1 – Solution cont’d

Thus, we get

We want to find dy/dx, so we solve for it:

Isolate dy/dx on one side.

Divide both sides by 5y4 + 1.

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Implicit Differentiation

This procedure we just used—differentiating an equation to find dy/dx without first solving the equation for y—is called implicit differentiation.

In Example 1 we were given an equation in x and y that determined y as an (implicit) function of x, even though we could not solve for y.

But an equation in x and y need not always determine y as a function of x. Consider, for example, the equation

2x2 + y2 = 2.

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Implicit Differentiation

Solving for y yields

The ± sign reminds us that for some values of x there are two corresponding values for y.

We can graph this equation by superimposing the graphs

of and

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Implicit Differentiation

The graph, an ellipse, is shown in Figure 12.

The graph of

constitutes the top half of the ellipse, and the graph of

constitutes the bottom half.

Figure 12

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Application

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Application

Productivity usually depends on both labor and capital. Suppose, for example, you are managing a surfboard manufacturing company.

You can measure its productivity by counting the number of surfboards the company makes each year.

As a measure of labor, you can use the number of employees, and as a measure of capital you can use its operating budget.

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Application

The so-called Cobb-Douglas model uses a function of the form:

P = Kxa y1 – a

where P stands for the number of surfboards made each year, x is the number of employees, and y is the operating budget.

The numbers K and a are constants that depend on the particular situation studied, with a between 0 and 1.

Cobb-Douglas model for productivity

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Example 5 – Cobb-Douglas Production Function

The surfboard company you own has the Cobb-Douglas production function

P = x0.3y0.7

where P is the number of surfboards it produces per year, x is the number of employees, and y is the daily operating budget (in dollars). Assume that the production level P is constant.

a. Find

b. Evaluate this derivative at x = 30 and y = 10,000, and

interpret the answer.

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Example 5(a) – Solution

We are given the equation P = x0.3y0.7, in which P is constant.

We find by implicit differentiation

0 = [x0.3y0.7]

0 = 0.3x –0.7y0.7 + x0.3(0.7)y

–0.3

d / dx of both sides

Product and chain rules

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Example 5(a) – Solution

–0.7x0.3y –0.3 = 0.3x–0.7y0.7 Bring term with dy / dx to left.

Solve for dy/dx.

Simplify.

cont’d

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Example 5(b) – Solution

Evaluating this derivative at x = 30 and y = 10,000 gives

To interpret this result, first look at the units of the derivative: We recall that the units of dy/dx are units of y per unit of x.

Because y is the daily budget, its units are dollars; because x is the number of employees, its units are employees.

cont’d

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Example 5(b) – Solution

Thus,

Next, recall that dy/dx measures the rate of change of y as x changes.

Because the answer is negative, the daily budget to maintain production at the fixed level is decreasing by approximately $143 per additional employee at an employment level of 30 employees and a daily operating budget of $10,000.

cont’d

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Example 5(b) – Solution

In other words, increasing the workforce by one worker will result in a savings of approximately $143 per day.

Roughly speaking, a new employee is worth $143 per day at the current levels of employment and production.

cont’d