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§3.1 Implicit Differentiation
The student will learn about
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 sometimes specified as
F (x, y) = x 2 + 4 xy - 3 y 2 +7 .
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Explicit DifferentiationConsider 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.
Note: we normally do implicit differentiation when explicit differentiation is difficult.
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Implicit Differentiation
2d d d dx – y – 5x 0, and
dx dx dx dx
2d dx – y – 5x 0 , and
dx dx
2 x dx
dx
Again consider the equation x 2 – y – 5x = 0
We will now implicitly differentiate both sides of the equation with respect to x
The same answer we got by explicit differentiation on the previous slide.
And solving for dy/dxdy
y ' 2 x 5dx
– 1dy
dx– 5
dx
dx Discuss
dy2 x - 1 5 0
dx
0
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Implicit Differentiation
2d d d dx – y – 5x 0, and
dx dx dx dx
2d dx – y – 5x 0 , and
dx dx
2 x
Let’s examine a short cut where we ignore the dx/dx.
And solving for dy/dx
dyy ' 2 x 5
dx
dy– 1
dx– 5 0.
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Explicit DifferentiationBe careful! Remember that x and y play different roles; x is the independent variable while y is the dependent variable.
Therefore we must include a (from the
generalized power rule) when we differentiate y n .
dy
dx
We don’t need to include a when
differentiating x n since = 1.
dx
dx
dx
dx
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Example 2
2 2d d d d dx y 3x 4y 0
dx dx dx dx dx
dy dy2y 4 2x 3
dx dx
Consider x 2 + y 2 + 3x + 4y = 0 and differentiate implicitly.
Solve for dy/dx
dy 2x 3
dx 2y 4
dx2x
dx
dx3
dx
dy4
dx 0dy
2ydx
dy2y 4 2x 3
dx
This equation would be difficult to differentiate explicitly.
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Example 2 continued
We just differentiated x 2 + y 2 + 3x + 4y = 0 and got:
We could have differentiated with respect to y
Indeed, we could have differentiated with respect to t.
The last two derivatives are presented to help you understand implicit differentiation.
dx dy dx dy dx2x 2y 3 4 0 Note 1
dx dx dx dx dx
dy dy dy
dx dy dx d
dy
y dy2x 2y 3 4 0 Note 1
dy
dt dt
dx dy dx dy2x 2y 3
dt4 0
dt
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Example 3
2d d d dx 3xy 4y 0
dx dx dx dx
Consider x 2 – 3 xy + 4y = 0 and differentiate implicitly.
Solve for dy/dx
dy4 3x 3y 2x
dx
dy 3y 2x
dx 4 3x
Notice we used the product rule for the - 3xy term.
dx2x
dx
dx3(y) (1)
dx
dy4
dx 0
dy3(x) (1)
dx1 1
dy dy4 3x 3y 2x
dx dx
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Example 3 (Again)
2d d d dx 3xy 4y 0
dx dx dx dx
Consider x 2 – 3 xy + 4y = 0 and differentiate implicitly.
Solve for y’ dy4 3x 3y 2x
dx
dy 3y 2x
dx 4 3x
Notice we used the product rule for the - 3xy term.
2x 3ydy
4dx
0dy
3xdx
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Example 3 Continued
We just differentiate implicitly x 2 – 3 xy + 4y = 0 to get
We could evaluate this derivative at a point on the original function, say (1, - 1).
3y 2xy '
4 3x
3( 1) 2(1) 5y ' 5
4 3(1) 1
That means that the slope of the tangent line (or any of the other meanings of the derivative such as marginal profit) at (1, - 1) is – 5.
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Example
2 2 3 5d d d d d3x y 7x x y 0
dx dx dx dx dx
differentiate implicitly 3x 2 + y 2 – 7x + x 3 y 5 = 0
Solve for dy/dx
3 4 2 5dy2y 5x y 7 6x 3x y
dx
2 5
3 4
dy 7 6x 3x y
dx 2y 5x y
Notice we used the product rule.
dx6x
dx3 4 dy
x 5ydx
5 2 dxy 3x
dx 0
dx7
dx
dy2y
dx
3 4 2 5dy dy2y 5x y 7 6x 3x y
dx dx
1 1 1
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Finding by implicit differentiation.
Implicit Differentiation Review
dx
dy
dx
dy1. Differentiate both sides of the equation with respect to x. when differentiating a y, include
dx
dy2. Collect all terms involving on one side, and all other terms on the other side.
dx
dy3. Factor out the and solve for it by dividing.
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Summary.
• We learned how to implicitly differentiate in order to find derivatives of difficult functions.
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ASSIGNMENT
§3.1 on my website
Test Review
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§ 2.1
Know the basic derivative formula.
If f (x) = C then f ’ (x) = 0.
If f (x) = xn then f ’ (x) = n xn – 1.
If f (x) = k • u (x) then f ’ (x) = k • u’ (x) = k • u’.If f (x) = u (x) ± v (x), then
f ’ (x) = u’ (x) ± v’ (x).
Test Review
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§ 2.2
Know the Product Rule. If f (x) and s (x), then
f • s ' + s • f ' df s
dx
Know the Quotient Rule. If t (x) and b (x), then
2
d t b t ' t b'
dx b b
continued
Test Review
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§ 2.2
Know
Marginal average costd
C'(x) C(x)dx
Marginal average revenue dR'(x) R (x)
dx
Marginal average profit dP'(x) P(x)
dx
Test Review
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§ 2.2
Know how to find second derivative and the applications associated with them.
Test Review
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§ 2.3Know the chain rule.
n n 1d duu nu
dx dx
Know that some functions are not differentiable.
Test Review
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§ 2.4
Know how to optimize a function including tax revenue.
There are a lot of applied problems on the test. It would be worth your time to go over
the ones assigned for homework!
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