7'4 Implicit Differentiation

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7.4 Implicit Differentiation
The student will learn about
Special functional notation and
implicit differentiation.
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Function Review and a New Notation
We have defined a function as y x 2 5x .
We have also used the notation f(x) x 2 5x .
In both situations y was the dependent variable
and x was the independent variable.
However, a function may have two (or more)
independent variable and is usually specified
as F(x,y) x 2 4 xy - 3 y 2 7 .
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Explicit Differentiation
Consider the equation y x 2 5x.
Then y
2x - 5
This is what we have been doing and is called
explicit differentiation.
If we rewrite the original equation, y x 2
5x, as x 2 y 5x 0 it is the same
equation. We can differentiate this equation
implicitly.
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Implicit Differentiation
Again consider the equation x 2 y 5x 0
We will now implicitly differentiate both sides
of the equation with respect to x
And solving for dy/dx
The same answer we got by explicit
differentiation on the previous slide.
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Example 2
Consider x 2 3 xy 4y 0 and
differentiate implicitly.
Solve for y
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Example 3
Consider x 2 3 xy 4y 0. Find the
equation of the tangent at (1, -1).
1. Confirm that (1, -1) is a point on the
function.
1 2 3 ? 1 ? - 1 4 ? - 1 0 or 1 3 4
0 or 0 0
2. Use the derivative from example 2 to find the
slope of the tangent.
3. Use the point slope formula for the tangent.
y - 5 x 4
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Example 4
Consider x e x ln y 3y 0 and
differentiate implicitly.
Solve for y
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Summary.
We learned how to differentiate implicitly.
NOTE