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Implicit dierentiation

Statement

Strategy for dierentiating implicitly

Examples

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Home Page23.Implicit dieren tiation

23.1.

Statemen t

The equationy=x2+ 3x+ 1 expresses a relationship between the quantitiesxandy. If a value ofxis given, then a corresponding value ofyis determined. For instance, ifx= 1, theny= 5. We say that the equation expressesyexplicitlyas a function ofx, and we write y=y(x) (read \yofx") to indicate thatydepends onx. The derivative of this function is denotedy0, so thaty0= 2x+ 3. The equationx2+y2= 2 (circle of radiusp2) also expresses a relationship between the quantitiesxandy. Solving fory, we get y=p2x2: There are two functions here; the one with the positive sign gives the top half of the circle, while the one with the negative sign gives the bottom half. We say that the equation x

2+y2= 2 expresses each of these functionsimplicitlyas a function ofx. We can nd the

derivatives of both functions simultaneously, and without having to solve the equation for y, by using the method of \implicit dierentiation."

Implicit dierentiation

Statement

Strategy for dierentiating implicitly

Examples

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Home PageMethod of implicit differentiation.Given an equation involving the variablesxandy, the derivative ofyis found usingimplicit dier- entiationas follows: ?Applyddx to both sides of the equation. (In the process of applying the derivative rules,y0will appear, possibly more than once.) ?Solve fory0.23.1.1 ExampleGivenx2+y2= 2, ndy0and use it to nd the slopes of the lines tangent to the graph of the equation at the points (1;1) and (1;1) as follows: (a) use implicit dieren tiation, (b) solv ef oryrst. Also, sketch the graph of the equation and the tangent lines.

Solution

(a) Using the metho dof implicit dieren tiation,w eapply ddx to both sides of the equation

Implicit dierentiation

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Strategy for dierentiating implicitly

Examples

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Home Pageand then solve fory0:

ddx x2+y2=ddx [2] ddx x2+ddx y2= 0

2x+ 2yy0= 0

y 0=xy (The chain rule was used in the next to the last step.) The slopes of the tangent lines at the points (1;1) and (1;1) are, respectively, y

0j(1;1)=11

=1 andy0j(1;1)=11= 1: (b) The p oint(1 ;1) is on the graph ofy=p2x2(top half of circle). The derivative of this function is y 0=ddx h p2x2i 12 (2x2)1=2 2x; so the slope at (1;1) isy0j1=12 (2(1)2)1=2(2)(1) =1. Similarly, the point (1;1) is on the graph ofy=p2x2(bottom half of circle). The derivative of this function is y 0=ddx h p2x2i =12 (2x2)1=2 2x; so the slope at (1;1) isy0j1=12 (2(1)2)1=2(2)(1) = 1.

Here is the sketch:

Implicit dierentiation

Statement

Strategy for dierentiating implicitly

Examples

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Home PageThe example illustrates the fact that it is usually much easier to use implicit dierentiation

than it is to rst solve the equation fory. When an equation givesyexplicitly as a function ofx, meaning that the equation hasy on one side and an expression involving onlyx's on the other, then the derivativey0equals an expression involving onlyx's, so to nd the slope of the line tangent to the graph of the equation at a point, one needs only thex-coordinate of the point (see solution to (b) in last example). By contrast, when an equation givesyimplicitly as a function ofx, the formula for the derivativey0typically involves bothx's andy's, sobothcoordinates of a point are required in order to nd the slope of the tangent at that point (see solution to (a)). This is un-

Implicit dierentiation

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Strategy for dierentiating implicitly

Examples

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Home Pagederstandable since an equation giving a function implicitly usually gives more than one function (for instancex2+y2= 2 gives the top half of the circle and also the bottom half); anx-coordinate alone does not determine which of the functions is intended, so the y-coordinate must also be supplied. 23.2.

Strategy for dieren tiatingimplicitly

In carrying out implicit dierentiation, one needs to keep in mind thatyrepresents a func- tion ofx(although an explicit formula might not be known). In deciding which derivative rules to apply, it is useful to think what you would do for a particulary, say,y= sinx. For instance, in the next example, in order to nd the derivative ofxyone should use the product rule sincexsinxrequires the product rule; in order to nd the derivative ofy3one should use the chain rule since (sinx)3requires the chain rule.

23.2.1 ExampleGivenx+xyy3= 7, ndy0.

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Home PageSolutionUsing the method of implicit dierentiation, we have ddx x+xyy3=ddx [7] ddx [x] +ddx [xy]ddx y3= 0 1 + ddx [x]y+xddx [y]

3y2y0= 0

1 + (y+xy0)3y2y0= 0

y

0x3y2=1y

y

0=1yx3y2=1 +y3y2x:

(The third line was obtained using the product rule and the chain rule.)23.3.Examples The next example shows the usefulness of implicit dierentiation for situations where there is no obvious way to solve the equation fory.

23.3.1 ExampleGivenex2y=x+y, ndy0.

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Home PageSolutionUsing the method of implicit dierentiation, we have ddx h ex2yi =ddx [x+y] e x2yddx x2y=ddx [x] +ddx [y] e x2yddx x2y+x2ddx [y] = 1 +y0 e x2y2xy+x2y0= 1 +y0 y 0 x2ex2y1 = 12xyex2y y

0=12xyex2yx

2ex2y1:23.3.2 ExampleGiven cos(xy) =2yx

3, ndy0.

Implicit dierentiation

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Home PageSolutionUsing the method of implicit dierentiation, we have ddx [cos(xy)] =ddx 2yx 3 sin(xy)ddx [xy] =x 3ddx [2y]2yddx x3(x3)2 sin(xy)(1y+xy0) =x3(2y(ln2)y0)2y(3x2)x 6 y 0 xsin(xy)2yln2x 3 =ysin(xy)32yx 4 y

0=ysin(xy)32yx

4xsin(xy)2yln2x

3 y

0=32yx4ysin(xy)x

5sin(xy) +x2yln2:

(In the last step, the complex fraction was simplied by multiplying numerator and denom- inator byx4. Also, numerator and denominator were multiplied by1 in order to reduce

the number of negative signs.)23.3.3 ExampleFind all points on the graph ofx4+y4+ 2 = 4xy3at which the

tangent line is horizontal. SolutionA horizontal line has slope zero, so the horizontal tangent lines occur at points on the graph where the derivative is zero. We compute the derivative using the method of

Implicit dierentiation

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Home Pageimplicit dierentiation:

ddx x4+y4+ 2=ddx 4xy3

4x3+ 4y3y0= 4y3+ 4x(3y2y0)

y

0(4y312xy2) = 4y34x3

y

0=4y34x34y312xy2=y3x3y

33xy2:

Settingy0= 0, we get

0 = y3x3y 33xy2
y

3x3= 0

y 3=x3 y=x; So a horizontal tangent line occurs at the point (x;y) on the graph if and only ify=x. In order for the point to be on the graph, its coordinates must satisfy the equation: x

4+y4+ 2 = 4xy3

x

4+x4+ 2 = 4x4

x 4= 1 x=1: The only candidates for such points are (1;1) and (1;1). Both of these points lie on the graph, so the answer is (1;1) and (1;1).

Implicit dierentiation

Statement

Strategy for dierentiating implicitly

Examples

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Home Page23{Exercises

23
{1Giv eny2=x, ndy0and use it to nd the slopes of the lines tangent to the graph of the equation at the points (4;2) and (4;2) as follows: (a) use implicit dieren tiation, (b) solv ef oryrst. Also, sketch the graph of the equation and the tangent lines. 23
{2Giv en2 xy+y2=x+y, use implicit dierentiation to ndy0. 23
{3Let px+y= 1 +x2y2. (a)

Find y0.

(b) Find an equation of the line tan gentto the graph of the giv enequation at the p oint (0;1). 23
{4Giv enxsiney= lny, ndy0.quotesdbs_dbs33.pdfusesText_39

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