Implicit Differentiation
copyright asu 7/15/2012
Implicit Differentiation
Suppose you wanted to
determine the slope of the circle
with center at (0,0), radius = 1
at any point along the circle?
Caution: There are two functions
that are needed to graph a circle.
x  y 1
2
2
Which equation would you choose?
copyright asu 7/15/2012
Implicit Differentiation
Up until now, the functions we have encountered can be described by expressing
one variable explicitly in terms of another variable.
But many curves have equations that can not be expressed explicitly.
It is difficult to impossible to solve for one variable in terms of the other.
To determine dy/dx we use implicit differentiation.
x  y  2y
2
3
e
xy
 5y
x ln y  y  3
copyright asu 7/15/2012
Implicit Differentiation
Implicit Differentiation:
 Differentiate both sides of the equation wrt x.
Remember to use the Chain Rule on each variable y term.
 Solve for dy/dx .
x  y 1
2
2
2x  2 y
dy
dy
dx
0
dx
2y
dy
(1,0 )
dy
dx
 2 x
( 0 ,1)
 
1
0
 
0

dx
dy
dx
2x
2y

 1
dy
dx
0
1
dx
dy
u n d efin ed
1
1 

,


2
2


 1
2
 1
2
x
y
copyright asu 7/15/2012
Implicit Differentiation
Find dy/dx. x  y  2 y
2
3
x  y  2y
2
3
2x  3y
2
dy
2
dx
3y
2
dy
2
dx
dy
dx

dx
 2 x
Use the Chain Rule
Move dy/dx to one side of the equation
dx
3y

dx
dy
dy
dy
2
 2   2 x
Factor out dy/dx
2 x
3y  2
2
Divide both sides of the equation by the same term
copyright asu 7/15/2012
Implicit Differentiation
Find dy/dx.
xy
e
e
xy
 5y
 5y
e  '  e
u
u
Use the Chain Rule which
includes the Product Rule
u'
dy
u  xy ; u '  x
e
xy
xe
xy
dx
dy
xy
dy
 dy

x

1

y

5


dx
 dx

dy
 ye
dy
 ye
xy
dx
xe
xy
 1 y
dy
dx
5
dy
dx
5
dy
dx
  ye
xy
xe

dx
dx
Distribute
exy
Bring dy/dx terms
to one side
copyright asu 7/15/2012

xe
 5    ye
xy
xy
5
xy
Factor out the
common factor
Isolate dy/dx
Implicit Differentiation
x ln y  y  3
Find dy/dx.
x ln y  y  3
 fg  ' 
f ' g  fg '
Use the Product Rule
f  x ; g  ln y
1  ln y  x
1 dy
y dx
x
1 dy
y dx

dy

dy
dy  x  y 

   ln y Simplify
dx  y 
0
dx
  ln y
dx

dy  x

1

   ln y
dx  y

dy
Bring dy/dx terms
to one side
Factor out the
common factor
copyright asu 7/15/2012
dx

y ln y
x y
Isolate dy/dx