Implicit Differentiation

Report
11
Techniques of Differentiation
with Applications
Copyright © Cengage Learning. All rights reserved.
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11.6 Implicit Differentiation
Copyright © Cengage Learning. All rights reserved.
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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.
Figure 10
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.
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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
Derivative rules
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
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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:
Cobb-Douglas model for productivity
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.
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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
0=
by implicit differentiation
[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
cont’d
Bring term with dy/dx to left.
Solve for dy/dx.
Simplify.
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Example 5(b) – Solution
cont’d
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.
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Example 5(b) – Solution
cont’d
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.
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Example 5(b) – Solution
cont’d
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.
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