Transcript § 1-1 Functions

Learning Objectives for Section 11.5
Implicit Differentiation
The student will be able to
■ Use special functional
notation, and
■ Carry out implicit
differentiation.
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Function Review and
New Notation
So far, the equation of a curve has been specified in the form
y = x2 – 5x or f (x) = x2 – 5x (for example).
This is called the explicit form. y is given as a function of x.
However, graphs can also be specified by equations of the
form F(x, y) = 0, such as
F(x, y) = x2 + 4xy - 3y2 +7.
This is called the implicit form. You may or may not be able
to solve for y.
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Explicit and Implicit
Differentiation
Consider the equation y = x2 – 5x.
To compute the equation of a tangent line, we can use the
derivative y’ = 2x – 5. This is called explicit differentiation.
We can also rewrite the original equation as
F(x, y) = x2 – 5x – y = 0
and calculate the derivative of y from that. This is called
implicit differentiation.
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Example 1
Consider the equation x2 – y – 5x = 0.
We will now differentiate both sides of the equation
with respect to x, and keep in mind that y is supposed to
be a function of x.


d 2
d
x  y  5x  0
dx
dx
dy
2x   5  0
dx
dy
 y'  2 x  5
dx
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This is the same answer
we got by explicit
differentiation on the
previous slide.
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Example 2
Consider x2 – 3xy + 4y = 0 and differentiate implicitly.
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Example 2
Consider x2 – 3xy + 4y = 0 and differentiate implicitly.
d 2
d
d
d
x 
3 xy 
4y 
0
dx
dx
dx
dx
2 x  3x y '  3 y  4 y '  0
Solve for y’:
Notice we used the product
rule for the xy term.
3x  4 y'  2x  3 y
2x  3y
y' 
3x  4
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Example 3
Consider x2 – 3xy + 4y = 0.
Find the equation of the tangent at (1, -1).
Solution:
1. Confirm that (1, -1) is a point on the graph.
2. Use the derivative from example 2 to find the slope of the
tangent.
3. Use the point slope formula for the tangent.
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Example 3
Consider x2 – 3xy + 4y = 0.
Find the equation of the tangent at (1, -1).
Solution:
1. Confirm that (1, -1) is a point on the graph.
12 – 31(- 1) + 4(-1) = 1 + 3 – 4 = 0
2. Use the derivative from example 2 to find the slope of the
tangent.
2 1  3   1
5
m

 5
3 1  4
1
3. Use the point slope formula for the tangent.
y  (1)   5 ( x  1)
y  5 x  4
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Example 3
(continued)
This problem can also be
done with the graphing
calculator by solving the
equation for y and using the
draw tangent subroutine.
The equation solved for y is
x2
y
3x  4
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Example 4
Consider xex + ln y – 3y = 0 and differentiate implicitly.
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Example 4
Consider xex + ln y + 3y = 0 and differentiate implicitly.
d
d
d
d
x
xe 
ln y 
3y 
0
dx
dx
dx
dx
1
xe  e  y '3 y '  0 Notice we used both the product
rule (for the xex term) and the
y
x
x
Solve for y’:
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y'  3 y '   x e x  e x
y
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chain rule (for the ln y term)
or
 x ex  ex
y' 
1
3
y
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Notes
Why are we interested in implicit
differentiation? Why don’t we just solve for y
in terms of x and differentiate directly? The
answer is that there are many equations of the
form F(x, y) = 0 that are either difficult or
impossible to solve for y explicitly in terms of
x, so to find y’ under these conditions, we
differentiate implicitly. Also, observe that:
d
y  y'
dx
and
Barnett/Ziegler/Byleen Business Calculus 11e
d
x 1
dx
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