3.7

Implicit Differentiation

x f(x) g(x) f ‘ (x) g‘ (x)

2 8 2 1/3 3

3 3 4 2 5

2at 2 . xxfa 3at . xxgxfb

3at . xxgxfc 2at / . xxgxfd

x f(x) g(x) f ‘ (x) g‘ (x)

2 8 2 1/3 3

3 3 4 2 5

2at . xxgfe 2at . xxff

3at /1 . 2 xxgg 2at . 22 xxgxfh

In Exercises 1 –5, sketch the curve defined by the equation and find two functions y

1 and y

2 whose graphs will combine to give the curve.

Quick Review

,1 xy

0 .1 2 yx 3694 .2 22 yx

04 .3 22 yx 9 .4 22 yx

32 .5 22 xyx

xy 2 ,93

2 21 xy 2

2 93

2xy

,21

xy

22

xy ,9 2

1 xy 22 9 xy

,32 21 xxy 2

2 32 xxy

In Exercises 6 –8, solve for y ‘ in terms of y and x.Quick Review

2

24

x

xyyxy

yxxyyx 42 .6 2

yyxxxxy cossin .7

yxyyyx 22 .8

xx

xxyy

sin

cos

xyx

xyy

2

2

In Exercises 9 and 10, find an expression for the function using rational powers rather than radicals.

Quick Review

6

5

2

3

xx 3 .9 xxx

3

3 2

.10x

xx 6

5

2

1 xx

What you’ll learn about Implicitly Defined Functions Lenses, Tangents, and Normal Lines Derivatives of Higher Order Rational Powers of Differentiable Functions

Essential Question

How does implicit differentiation allows us to find derivatives of functions that are not defined or written explicitly as a function of a single variable?

Implicitly Defined Functions

An important problem in Calculus is how to find the slope when the

function can't conveniently be solved for . Instead, is treated as

a differentiable function of and both sides of the equation

y y

x are

differentiated with respect to , using the appropriate rules for sums,

products, quotients and the Chain rule. Then solve for in terms of

and together to obtain a formula that calculates

x

dyx

dxy

3 3

1 2

the slope at any

point , on the graph.

The process by which we find is called .

The phrase derives from the fact that the equation

9 0

defines the functions , a

x y

dy

dx

x y xy

f f

implicit differentiation

3nd implicitly (i.e., hidden inside the

equation), without giving us formulas to work with.

f

explicit

Implicitly Defined Functions

Implicit Differentiation Process1. Differentiate both sides of the equation with respect to .

2. Collect the terms with on one side of the equation.

3. Factor out .

4. Solve for .

x

dy

dxdy

dxdy

dx

Example Implicitly Defined Functions6for Find .1 22 xyyx

dx

dy

2x

dx

dyy x2 x

dx

dyy2 2y 1 0

dx

dy xyx 22 22 yxy

dx

dy 22 yxy xyx 22

Example Implicitly Defined Functions 0tanfor Find .2 xyx

dx

dy

1 xy2sec xdx

dyy 1 0

xydx

dyx 2sec1 xyy 2sec 0

xyxdx

dy 2sec 1 xyy 2sec

dx

dy xyy 2sec1 xyx 2sec

Example Implicitly Defined Functionsxxy

dx

yd

dx

dy2for find then Find .3 22

2

2

y2dx

dyx2 2

dx

dy 22 x

y2

2

2

dx

yd y 1

y

21 x2

1

y

1

xy

1 x dx

dy

2y

y

x 1 12 2 xx

3

1

y

2y1

xx 22

xxy 2

12

Lenses, Tangents and Normal LinesIn the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular to the surface of the lens at the point of entry (angles A and B in Figure 3.50). This line is called the normal to the surface at the point of entry. In a profile view of a lens, the normal is a line perpendicular to the tangent to the profile curve at the point of entry.

Implicit differentiation is often used to find the tangents and normals of lenses described as quadratic curves.

Example Lenses, Tangents and Normal Lines4. Find the lines that are (a) tangent and (b) normal to the curve at the

given point. 3 ,1 ,922 yx

2x

dx

dyy2 2y x2 0

dx

dy yx22 22xy

dx

dy 22xyyx22

x

y

x

y

3 ,1at 1

3

3

3y 3 1x3y 33 x

63 xy

3y3

1 1x

93 y 1 x

83 xy

3

8

3

1 xy

Rule 9 Power Rule For Rational Powers of x

1

If is any rational number, then

.

If 1, then the derivative does not exist at 0.

n n

n

dx nx

dxn x

3

2

for Find .5 xydx

dy

3

2

dx

dy3

1

x 3/13

2

x

33

2

x

Rule 9 Power Rule For Rational Powers of x

1

If is any rational number, then

.

If 1, then the derivative does not exist at 0.

n n

n

dx nx

dxn x

2

1

52for Find .6 xydx

dy

2

1

dx

dy 2

3

52 x 2

3

521 x

352

1

x

52

1

x

2

Pg. 162, 3.7 #2-42 even, 50, 61-64 all

Find the x-value of the position(s) of the horizontal tangents of each equation.Review for Quiz

1353

1 .1 23 xxxxf

552

72 .2 23 xxxxf

Find the x-value of the position(s) of the horizontal tangents of each equation.Review for Quiz

32

49

4

1 .3 24 xxxf

Find the x-value of the position(s) of the horizontal tangents of each equation.Review for Quiz

7302

11

3

4

4

1 .4 234 xxxxxf