MATH 12002 - CALCULUS I 2.6: Implicit Differentiation Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.6: Implicit Differentiation

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Implicit Differentiation

A relation between x and y represents a graph in the xy-plane; i.e., the set of points (a, b) whose coordinates x = a, y = b satisfy the

• relation. If the relation is not a function, we will not be able to write y as

a single function of x. However, if the graph is a smooth curve, it will still have tangent lines and we should be able to find the slopes of the tangent lines, even though y is not of the form y = f (x) and the slope cannot be written as f ′(a). We will see how to find this slope (the derivative of y with respect to x) implicitly, but first let’s consider an example where we can find the slope explicitly.

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Implicit Differentiation

Example

Find the slope of the line tangent to the ellipse 9x2 + 4y2 = 36 at the point (1, −3

√ 3 2 ).

Solution 1

We can solve for y in terms of x: since 9x2 + 4y2 = 36, we have 4y2 = 36 − 9x2 y2 =

1 4(36 − 9x2)

y = ±1

2

• 36 − 9x2.

Here, y = 1

2

√ 36 − 9x2 represents the top half of the ellipse (y 0) and y = −1

2

√ 36 − 9x2 represents the bottom half (y 0). [Continued →]

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Implicit Differentiation

Solution 1 [continued]

Since our point, (1, −3

√ 3 2 ), is on the bottom half of the ellipse, the slope

is f ′(1), where f (x) = −1

2

√ 36 − 9x2. We have f ′(x) = −1

2 · 1 2(36 − 9x2)−1/2 · (−18x)

= 9x 2 √ 36 − 9x2 , and so the slope is f ′(1) = 9(1) 2

• 36 − 9(12)

= 9 2√36 − 9 = 9 2 √ 27 = 3 · 3 2 · 3 √ 3 = √ 3 2 .

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Implicit Differentiation

Solution 2

We now find the slope implicitly. We know that the top and bottom halves of the ellipse are functions. Instead of solving for the function explicitly, let’s just say y = f (x), so that 9x2 + 4[f (x)]2 = 36. Now take derivatives of both sides of the equation with respect to x: 18x + 8[f (x)]1 · f ′(x) = 0. Since “f (x) = y” this means “f ′(x) = dy

dx ” and 18x + 8y · dy dx = 0,

and we get dy

dx = − 9x 4y . The slope is then dy dx evaluated at (1, −3 √ 3 2 ), i.e.,

dy dx

• (1,− 3

√ 3 2 )

= − 9(1) 4(−3

√ 3 2 )

= 3 · 3 2 · 3 √ 3 = √ 3 2 .

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Examples

1 Let x2 + x3y5 + y−3 = 4 and find y′ = dy

dx , the derivative of y with

respect to x. Instead of writing y as f (x), we will just treat y as some function of x, so that y5, for example, is a composite function with inside function y. First, take the derivative with respect to x of both sides of the equation x2 + x3y5 + y−3 = 4, to obtain 2x + [3x2y5 + x3 · 5y4y′] − 3y−4y′ = 0. Now solve for y′ in terms of x and y: 2x + 3x2y5 = −5x3y4y′ + 3y−4y′ 2x + 3x2y5 = (−5x3y4 + 3y−4)y′ y′ = 2x + 3x2y5 −5x3y4 + 3y−4 .

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Examples

2 Let 2y3 = tan(x2 + y2) and find y′ = dy

dx , the derivative of y with

respect to x. First, take the derivative with respect to x of both sides of the equation 2y3 = tan(x2 + y2) to obtain 6y2y′ = [sec2(x2 + y2)](2x + 2yy′). Now solve for y′ in terms of x and y: 6y2y′ = [sec2(x2 + y2)] · 2x + [sec2(x2 + y2)] · 2yy′ 6y2y′ − [sec2(x2 + y2)] · 2yy′ = [sec2(x2 + y2)] · 2x

• 6y2 − [sec2(x2 + y2)] · 2y
• y′ = [sec2(x2 + y2)] · 2x

y′ = [sec2(x2 + y2)] · 2x 6y2 − [sec2(x2 + y2)] · 2y .

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