Lesson: Derivative Techniques - 4

Objective – Implicit Differentiation

These functions are relatively easy to differentiate because they are defined “EXPLICITLY”. (meaning they are solved for y, in other words y is by itself on one side of the equation.)

Differentiate:

23 5 11y x x

(4 )y Cos x

75(3 4)y x

1. Implicit Differentiation – The process of finding the derivative of a function that is not solved for y.

Implicit functions do not have y isolated on one side of the equation.

To do this, you need to use the chain rule, with a little creativity mixed in.

EX 1: Differentiate

1xy y x

This function is not defined explicitly, but can be with the aid of a little algebra.

Group all y terms on1 side of the = sign.

1xy y x

Factor out a y( 1) 1y x x Divide by (x + 1) to isolate y

1

1

xy

x

1

1

xy

x

Now use the quotient ruleTo differentiate

'

2

' 'f g f f g

g g

2

( 1)(1) ( 1)(1)'

( 1)

x xy

x

2

( 1) ( 1)'

( 1)

x xy

x

2

1 1'

( 1)

x xy

x

2

2'

( 1)y

x

Question:

O.K. but what about those that cannot be expressed explicitly or that are a nightmare to do so? I know you will be giving us some of those, won’t you, Mr. Winter!

Well, let’s work our way into those. Let meshow you first how to do the problem fromexample 1 “IMPLICITLY”.

Process of Implicit Differentiation

• Differentiate both sides of the equation with respect to x.• Apply Chain Rule whenever you have an expression involving y.• Move all terms involving dy/dx (or y’) to the left side of the equation, and everything else to the right side.• Factor out dy/dx (or y’) on the left.• Solve for dy/dx (or y’)

1xy y x Now let’s try it.Differentiate:

' 1 ' 1x y y y

' ' 1x y y y

' ' 1x y y y

'( 1) 1y x y

1'

1

yy

x

But that answer doesn’t look the same as theOne we got by defining it explicitly!

Remember from the first method that:

1

1

xy

x

Substitute that in for y andSee what happens.

EX. 2: Differentiate

5 2 33 4y x Sinx y 4 25 ' 6 12 'y y x Cosx y y 4 25 ' 12 ' 6y y y y Cosx x

4 2'(5 12 ) 6y y y Cosx x

4 2

6'

5 12

Cosx xy

y y

EX. 3: Use implicit differentiation to find dy/dx if

2 25 ( )y Sin y x

2 2[5 ( )] [ ]d d

y Sin y xdx dx

10 ' ( ) ' 2y y Cos y y x

10 ' ( ) ' 2y y Cos y y x

'(10 ( )) 2y y Cos y x

2'

10 ( )

xy

y Cos y

EX. 4: Use implicit differentiation to find the second derivative d2y/dx2 of

2 24 2 9x y

Solution: 3

9''y

y

EX. 5: Find the slopes of the tangent lines to the curve y2 – x + 1 = 0 at the points (2, -1) and (2, 1).

Solution:2

1

1

2xy

dy

dx

21

1

2xy

dy

dx

EX. 6: (a)Use implicit differentiation to find dy/dx for the “Folium of Descartes: x3 + y3 =3xy

3 3 3x y xy 2 23 3 ' 3( ' 1)x y y x y y

2 23 3 ' 3 ' 3x y y x y y Divide all terms by 3 to simplify

2 2 ' 'x y y x y y

2 2 ' 'x y y x y y

2 2' 'y y xy y x

2 2'( )y y x y x 2

2'

y xy

y x

EX. 7:

HW 4.1

Pg. 241(1 – 3, 5, 11 – 15, 21 – 24, 30*)