Implicit Differentiation

3.6

Implicit DifferentiationSo far, all the equations and functions we looked at were all stated explicitly in terms of one variable:

53 xy tts 3016 2

xy

1

In this function, y is defined explicitly in terms of x. If we re-wrote it as xy = 1, y is now defined

implicitly in terms of x.

It is easy to find the derivative of an explicit function, but what about:

534 23 yyyx

2 2 1x y

This is not a function, but it would still be nice to be able to find the slope.

2 2 1d d dx y

dx dx dx Do the same thing to both sides.

2 2 0dy

x ydx

Note use of chain rule.

2 2dyy xdx

2

2

dy x

dx y

dy x

dx y

22 siny x y

22 sind d d

y x ydx dx dx

This can’t be solved for y.

2 2 cosdy dy

x ydx dx

2 cos 2dy dy

y xdx dx

22 cosdy

xydx

2

2 cos

dy x

dx y

This technique is called implicit differentiation.

1 Differentiate both sides w.r.t. x.

2 Solve for .dy

dx

4523 34 xxyy Find dy/dx if:

342 12583 xxdx

dyy

dx

dyy

342 12583 xxyydx

dy

yy

xx

dx

dy

83

1252

34

Find dy/dx if: 2222 sincoscossin xyxy

)2(cos2sin)2(sin2cos 2222 xxdx

dyyyxx

dx

dyyy

2222 sin2cos2sin2cos2 xxxxdx

dyyy

dx

dyyy

Chain Rule

2222 sin2cos2sin2cos2 xxxxyyyydx

dy

22

22

sin2cos2

sin2cos2

yyyy

xxxx

dx

dy

22

22

sincos

sincos

yyy

xxx

dx

dy

Find dy/dx if: 8453 322 yxyx

0121056 22 dx

dyy

dx

dyxyyx

Product Rule!

22 561210 yxyxydx

dy

2

2

1210

56

yxy

yx

dx

dy

We need the slope. Since we can’t solve for y, we use

implicit differentiation to solve for .dy

dx

2 2 7x xy y ( 1, 2)

2 2 7x xy y

2 2 0dydy

x yx ydxdx

Note product rule.

2 2 0dy dy

x x y ydx dx

22dy

y xy xdx

2

2

dy y x

dx y x

2 2 1

2 2 1m

2 2

4 1

4

5

Find the equations of the lines tangent and normal to the

curve at .

2 2 7x xy y ( 1, 2)

4

5m tangent:

42 1

5y x

4 42

5 5y x

4 14

5 5y x

Find the equations of the lines tangent and normal to the

curve at .

normal:

52 1

4y x

5 52

4 4y x

5 3

4 4y x

Normal line is perpendicular

to tangent

Find derivative at (1, 1)

xxy

xy

23

32

Product Rule is easier than quotient rule, so let’s cross

multiply!

3332 xxyxy

2322 3332 xydx

dyxyx

dx

dyy

232 632 xyxyydx

dy

2

23

32

6

xyy

xy

dx

dy

2

23

)1)(1(3)1(2

)1(6)1(

dx

dy

51

5

dx

dy

Higher Order Derivatives

Find if .2

2

d y

dx3 22 3 7x y

3 22 3 7x y

26 6 0x y y

26 6y y x

26

6

xy

y

2xy

y

2

2

2y x x yy

y

2

2

2x xy y

y y

2 2

2

2x xy

y

x

yy

4

3

2x xy

y y

Substitute back into the equation.

y