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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.

1 1

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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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