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UNIT 2 LESSON 9 IMPLICIT DIFFERENTIATION 1

UNIT 2 LESSON 9

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UNIT 2 LESSON 9. IMPLICIT D IFFERENTIATION. IMPLICIT DIFFERENTIATION. So far, we have been differentiating expressions of the form y = f ( x ), where y is written explicitly in terms of x. y = ( x – 4) 3 (2 x + 5) 5. y = 2 x 2 + 5 x – 7. - PowerPoint PPT Presentation

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Page 1: UNIT 2 LESSON 9

1

UNIT 2 LESSON 9

IMPLICIT DIFFERENTIATION

Page 2: UNIT 2 LESSON 9

2

IMPLICIT DIFFERENTIATION

So far, we have been differentiating expressions of the form y = f(x), where y is written explicitly in terms of x.

It is not always convenient or possible to isolate the y, and in these cases, we must differentiate with respect to x without first isolating y.

This is called implicit differentiation.

y = 2x2 + 5x – 7 y = (x – 4)3(2x + 5)5562

xxy

2y = y2 + 3x3

Page 3: UNIT 2 LESSON 9

3

IMPLICIT DIFFERENTIATION

We represent the derivative of y as dxdy

The derivative of x is equal to 1; therefore, it is not

necessary to write it as dxdx

That is, if y = x then

xdxdy

dxd

dxdx

dxdy

1dxdy

Page 4: UNIT 2 LESSON 9

4

If y is a function of x then its derivative is

y2 is a function of y, which in turn is a function of x.

22 2d d dyy y ydx dx dx

12d y

dx

121

2y dy

dx

Using the chain rule:

Find the following derivative with respect to x

IMPLICIT DIFFERENTIATIONdydx

Page 5: UNIT 2 LESSON 9

5

xy xy dx

xddxyd

2

21

21

xdxdy 2

1

21

xdxdy

12 dxdyy

xdxdy

21

xdx

dy2

1

ydxdy

21

METHOD I: Rearrange for y

EXAMPLE 1: a) Differentiate y 2 = x

METHOD II: Implicitly

OR

y = - x ½y = x ½

Page 6: UNIT 2 LESSON 9

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EXAMPLE 1: b) Find the slopes of the tangent lines on y 2 = x at (4, 2) and (4, -2)

41

421

dxdy

41

421

dxdy

xdxdy

21

xdxdy

21

(4, 2)

(4, -2)

Using the x value of 4 we need the graph to know which slope goes with which point.

14

m

14

m

Page 7: UNIT 2 LESSON 9

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EXAMPLE 1 b) Find slopes of the tangents on y 2 = x at (4, 2) and (4, -2)

41

221

dxdy

ydxdy

21

41

221

dxdy

(4, 2)

(4, -2)

Using the y values of -2 and 2 we know which slope goes with which point.

Page 8: UNIT 2 LESSON 9

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undefineddxdy

02

1

undefineddxdy

02

1

undefineddxdy

02

1

EXAMPLE 1 c) Find slope of the tangent on y 2 = x at (0, 0)

ydxdy

21

(0, 0)

xdxdy

21

xdxdy

21

Using the x value of 0

Using the y value of 0

Page 9: UNIT 2 LESSON 9

9

EXAMPLE 2a: x 2 + y 2 = 25 This is not a function, but it would still be nice to be able to find the slope for any tangent line.

In order to graph this on our calculators we have to rearrange and isolate y

y2 = 25 – x2

225y x 225y x OR

Page 10: UNIT 2 LESSON 9

10

EXAMPLE 2b: x 2 + y 2 = 25

Take derivative of both sides.

2 2 0dyx ydx

2 2dyy xdx

22

dy xdx y

dy xdx y

2 2 1d d dx ydx dx dx

25

We could differentiate each of the above explicitly but it would be more difficult than using implicit differentiation.

225y x 225y x OR

Page 11: UNIT 2 LESSON 9

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dy xdx y

EXAMPLE 2 c) Find the Equation of the tangent line at (3, 4)

So at (3, 4)

x2 + y 2 = 25 (3, 4)

Slope = 43

m

34 (3)4

254

b

b

3 254 4

y x

Page 12: UNIT 2 LESSON 9

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EXAMPLE 3: Differentiate 2y = y 2 + 3x 2

22 32 xydxdy

dxd

22 32 xdxdy

dxdy

dxd

xydxdy 622

xdxdyy

dxdy 622

xdxdyy

dxdy 622

yx

yx

dxdy

13

226

Page 13: UNIT 2 LESSON 9

13

Assignment Questions

Do Questions 1-9 on pages 3 & 4