Section 2.5 – Implicit Differentiation

Explicit Equations

The functions that we have differentiated and handled so far can be described by expressing one variable explicitly in terms of another variable. For example:

Or, in general, y = f(x).

y x 3 1

y x sin x

Implicit EquationsSome functions, however, are defined

implicitly ( not in the form y = f(x) ) by a relation between x and y such as:

x 2 y 2 25

x 3 y 3 6xyIt is possible to solve some Implicit Equations for y, then differentiate:

x 2 y 2 25

y 2 25 x 2

y 25 x 2

Yet, it is difficult to rewrite most Implicit Equations

explicitly. Thus, we must be introduced to a new

technique to differentiate these implicit functions.

Can we take the derivative of these functions?

White Board ChallengeSolve for y:

2 3 6 8xy y x y x

2 6 84

x xxy

*Reminder*Technically the Chain Rule can be applied to every

derivative: 3y x

3dy ddx dx x

u

f u

u'

f ' u x

u3

1

3u2

23 1dydx u

23x

Derivatives Involving the Dependent Variable (y)

Find the derivative of each expression

ddx y

y'

dydx

y

y 3

ddx y 3

u

f u

u'

f ' u

y

u3

dydx

3u2

ddx y 3 3u2dy

dx

3y 2dydx

a.

The derivative of y with respect to

x is…

the derivative of y.

This is another way to write y

prime.

b.The Chain Rule is

Required.

Instructions for Implicit DifferentiationIf y is an equation defined implicitly as a differentiable

function of x, to find the derivative:1. Differentiate both sides of the equation with respect to x. (Remember that y is really a function of x for part of the curve and use the Chain Rule when differentiating terms containing y)

2. Collect all terms involving dy/dx on the left side of the equation, and move the other terms to the right side.

3. Factor dy/dx out of the left side

4. Solve for dy/dx

Example 1If is a differentiable function of x such that

find .

Chain Rule

u

f u

u' 'f u

y

u3

dydx

3u2

ddx x 2y 2y 3 d

dx 3x 2y

ddx x 2y d

dx 2y 3 ddx 3x d

dx 2y

x 2 ddx y y d

dx x 2 2 ddx y 3 3 d

dx x 2 ddx y

x 2 dydx y2x 23u2dy

dx 312 dydx

x 2 dydx 2xy 6y 2 dy

dx 32 dydx

Differentiate both sides.

y f x

dydx

Product AND Constant

Multiple Rules

x 2 dydx 6y 2 dy

dx 2 dydx 3 2xy

dydx x 2 6y 2 2 3 2xy

dydx 3 2xy

x 2 6y 2 2

x 2y 2y 3 3x 2y

Solve for dy/dx

Example 2Find if .

Chain Rule Twice

sin x y y 2 cos x

u1

f1 u1

u1'

f1' u1

x y

sin u1

1 dydx

cosu1

ddx sin x y d

dx y 2 cos x

ddx sin x y y 2 d

dx cos x cos x ddx y 2

u2

f2 u2

y

u22

u2 '

f2 ' u2

dydx

2u2

cosu1 1 dydx y 2 sin x cos x2u2

dydx

cos x y 1 dydx y 2 sin x cos x2ydy

dx

Differentiate both sides

y'

Product Rule

cos x y cos x y dydx y 2 sin x 2y cos xdy

dx

cos x y dydx 2y cos xdy

dx y 2 sin x cos x y

dydx cos x y 2y cos x y 2 sin x cos x y

dydx y 2 sinx cos xy

cos xy 2y cosx

Example 3Find if .

Chain Rule

x 2 y 2 10

u

f u

u'

f1' u

y

u2

dydx

2u

ddx x 2 y 2 d

dx 10

ddx x 2 d

dx y 2 ddx 10

2x 2udydx 0

2x 2ydydx 0

Find the first derivative by Differentiating both sides.

d 2ydx 2

2ydydx 2x

dydx 2x

2y

dydx x

y

Now Find the Second Derivative

d 2ydx 2 d

dx xy

d 2ydx2 y d

dx x x ddx y

y 2

d 2ydx2 y 1 x dy

dx

y 2

d 2ydx 2 yx dy

dx

y 2

d 2ydx2

yx xy

y 2

Quotient Rule

d 2ydx 2

y x2y

y 2

yy

dydx y 2 x 2

y 3

dydx

y 2 x 2 y 3

dydx 10

y 3

Remember:

dydx x

y

Remember:

x 2 y 2 10

Example 4Find the slope of a line tangent to the circle

at the point .

Chain Rule

P 5,4

u

f u

u' 'f u

y

u2

dydx

2u

ddx x 2 y 2 d

dx 5x 4y

ddx x 2 d

dx y 2 ddx 5x d

dx 4y

ddx x 2 d

dx y 2 5 ddx x 4 d

dx y

2x 2udydx 5 4 dy

dx

2x 2y dydx 5 4 dy

dx

Find the derivative by differentiating

both sides.

x 2 y 2 5x 4 y

2y dydx 4 dy

dx 5 2x

dydx 2y 4 5 2x

dydx 5 2x

2y 4

Evaluate the derivative at x=5 and y=4.

dydx 5,4

5 2524 4

54

White Board Challenge Find the derivative of:

2 23 4 4x y

34

dy xdx y

Example 5If and , find .

Chain Rule

f 3 1

u

f u

u' 'f u

f x

u3

f ' x

3u2

36d d

dx dxx f x xf x 3

6d d ddx dx dxx f x xf x

3 36d d d d d

dx dx dx dx dxx f x f x x x f x f x x

323 ' 1 ' 1 0x u f x f x x f x f x 2 3

3 ' ' 0x f x f x f x xf x f x

Find the derivative by differentiating both sides.

x f x 3 xf x 6

2 33 ' 'x f x f x xf x f x f x

2 3' 3f x x f x x f x f x

f ' x f x 3 f x 3x f x 2 x

f ' 3

Example 5 (continued)If and , find .

f 3 1

f ' 3 f 3 3 f 3 33 f 3 2 3

1 3 1

9 1 2 3

Evaluate the derivative with the given information.

x f x 3 xf x 6

f ' 3

f ' x f x 3 f x 3x f x 2 x

1 193

212

16

Example 6

Chain Rule

u

f u

u' 'f u

y

u2

dydx

2u

ddx x 2 y 2 d

dx 25

ddx x 2 d

dx y 2 ddx 25

2x 2udydx 0

2x 2ydydx 0

Find the derivative by differentiating both sides.

2ydydx 2x

dydx 2x

2y

dydx x

y

Now evaluate the derivative at x=3 and y=4.

Find an equation of the tangent to the circle at the point .

3,4

dydx 3,4

34

Use the Point-Slope Formula to find the

equation of the tangent line

y 4 34 x 3

2 2 25x y

White Board Challenge Find the second derivative of:

2 23 4 4x y 34

dy xdx y

2

2 33

4d ydx y