11 Copyright © Cengage Learning. All rights reserved. 11 Techniques of Differentiation with Applications

22 Copyright © Cengage Learning. All rights reserved Implicit Differentiation

33 Consider the equation y 5 + 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

44 Implicit Differentiation Because we cannot solve for y explicitly in terms of x, we say that the equation y 5 + 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.

55 Example 1 – Implicit Differentiation Find given that y 5 + 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

66 Example 1 – Solution cont’d Now we must be careful. The derivative with respect to x of y 5 is not 5y 4. Rather, because y is a function of x, we must use the chain rule, which tells us that Derivative rules

77 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 5y

88 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 2x 2 + y 2 = 2.

99 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

10 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

11 Application

12 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.

13 Application The so-called Cobb-Douglas model uses a function of the form: P = Kx a y 1 – 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

14 Example 5 – Cobb-Douglas Production Function The surfboard company you own has the Cobb-Douglas production function P = x 0.3 y 0.7 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.

15 Example 5(a) – Solution We are given the equation P = x 0.3 y 0.7, in which P is constant. We find by implicit differentiation 0 = [x 0.3 y 0.7 ] 0 = 0.3x –0.7 y x 0.3 (0.7)y –0.3 d / dx of both sides Product and chain rules

16 Example 5(a) – Solution –0.7x 0.3 y –0.3 = 0.3x –0.7 y 0.7 Bring term with dy / dx to left. Solve for dy/dx. Simplify. cont’d

17 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

18 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

19 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