LESSON 11 IMPLICIT DIFFERENTIATION
Definition An implicit function of x is a relationship involving x and y for which y
is a function of x but which you might not be able to algebraically solve the
equation for y in terms of x.
Examples Here are two examples of implicit functions.
1.
x2  y2  1
This equation is from trigonometry. It is called the Unit Circle. We can
almost solve this equation for y:
x2  y2  1  y 2  1  x2 
y 
1  x2  y  
We have that either y 
y
2.
y2 
1  x2 
1  x2
1  x 2 , which is the top-half of the circle, or
1  x 2 , which is the bottom-half of the circle.
x 2 y  6 xy3  2 y 5  x  3 y
If you can solve this for y, please give me the answer.
Even if we can not solve an implicit function for y in terms of x, we can
differentiate the implicit function using a method called implicit differentiation
where we assume that y is some function of x, say y  f ( x ) . In the Calculus III
course, you will learn about the Implicit Function Theorem. This will provide
information about when y is a function of x for an implicit function. In the same
way, x could be a function of y for an implicit function. We will only consider the
case where y is a function of x for an implicit function in this lesson.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Implicit differentiation is an application of the Chain Rule.
2
2
Example Differentiate x  y  1 using implicit differentiation assuming that y
is a function of x.
In the first example above, we showed that either y 
Thus, y  f ( x ) , where either f ( x ) 
1  x 2 or y  
1  x 2 or f ( x )  
1  x2 .
1  x2 .
In order to see how the Chain Rule is used, let’s replace y in the equation
x 2  y 2  1 by f ( x ) . Thus, we have that x 2  [ f ( x ) ]2  1 . Note, that by the
2
Chain Rule, we have that D x [ f ( x ) ]  2 [ f ( x ) ]  Dx f ( x ) = 2[ f ( x ) ]  f ( x ) .
X
X
2
2
Since y  f ( x ) , then Dx [ f ( x ) ]  2 [ f ( x ) ]  f ( x )  D x ( y )  2 y y  .
2
2
Since D x ( x )  2 x , D x ( y )  2 y y  , Dx (1)  0 , then we have that
x2  y
X
2
 1  2x  2 y y  0  2 ( x  y y )  0  x  y y  0 
y y   x  y  
Answer: y   
x
y
x
y
Examples Differentiate the following functions (assuming that y is a function of
x.)
1.
3x 2  4 y 5  12
3x 2  4 y
X
5
 12  6 x  20 y 4 y   0  6 x  20 y 4 y  
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
y 
6x
3x


y

20 y 4
10 y 4
Answer: y  
2.
3x
10 y 4
5x 2  xy  3 y 4  x 3
NOTE: When it comes time to differentiate the term xy , you will need to
use the Product Rule to differentiate this product. We will not need to use the
Chain Rule to find Dx ( y ) , the derivative of y with respect to x, since by
dy
 y .
definition, D x ( y ) 
dx
5x 2  x y  3 y
X
4
 x 3  10 x  (1 y  xy  )  12 y 3 y   3x 2 
10 x  y  xy  12 y3 y  3x 2   xy   12 y 3 y   3x 2  10 x  y 
3x 2  10 x  y
 y  ( x  12 y )  3x  10 x  y  y   
x  12 y 3
3
2
3x 2  10 x  y
Answer: y   
x  12 y 3
3.
x3  2x 2 y 3  y 5  x 4  y
x3  2x 2 y
3
 y
X X
5
 x4  y 
3x 2  2 ( 2 xy3  x 2 3 y 2 y  )  5 y 4 y   4 x 3  y  
3x 2  4 xy3  6 x 2 y 2 y  5 y 4 y  4 x 3  y 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
6 x 2 y 2 y   5 y 4 y   y   4 x 3  3x 2  4 xy3 
y  ( 6 x 2 y 2  5 y 4  1)  4 x 3  3x 2  4 xy3 
4 x 3  3x 2  4 xy3
y 
6x 2 y 2  5 y 4  1
Answer:
4.
4 x 3  3x 2  4 xy3
y 
6x 2 y 2  5 y 4  1
x 4  6 xy7  3 y 2  2 x 5  y 5  8
x 4  6 xy
X
7
 3y
X
2
 2x5  y
X
5
8 
4 x 3  6 ( y 7  x 7 y 6 y  )  6 yy   10 x 4  5 y 4 y  
4 x 3  6 y 7  42 xy6 y   6 yy   10 x 4  5 y 4 y  
4 x 3  6 y 7  10 x 4  5 y 4 y   42 xy6 y   6 yy  
4 x 3  6 y 7  10 x 4  y ( 5 y 4  42 xy6  6 y ) 
4 x 3  6 y 7  10 x 4
y  4
5 y  42 xy6  6 y
4 x 3  6 y 7  10 x 4
Answer: y  
5 y 4  42 xy6  6 y
5.
( y 2  9 ) 4  ( 4 x 2  3 x  1) 3
2
4
2
3
NOTE: Dx ( y  9 )  4 ( y  9 )  Dx ( y
X
X
2
 9 ) = 4 ( y 2  9 ) 3 2 yy  =
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
8 yy ( y 2  9 ) 3 and Dx ( 4 x 2  3x  1) 3  3 ( 4 x 2  3x  1) 2 ( 8 x  3 ) . Thus,
X
( y 2  9 ) 4  ( 4 x 2  3 x  1) 3 
8 yy  ( y 2  9 ) 3  3 ( 4 x 2  3x  1) 2 ( 8x  3 ) 
3 ( 4 x 2  3 x  1) 2 ( 8 x  3 )
y 
8 y ( y 2  9 )3
3 ( 4 x 2  3 x  1) 2 ( 8 x  3 )
Answer: y  
8 y ( y 2  9 )3
6.
6x 2
 y  3x  2
y4  5
6x 2
 y  3x  2 
y 4 5
X
12 x ( y 4  5 )  6 x 2 4 y 3 y 
 y  3 
( y 4  5)2
 12 x ( y 4  5 )  24 x 2 y 3 y 

4
2

( y  5) 

y
  3( y  5) 
4
2
( y  5)


4
2
12 x ( y 4  5 )  24 x 2 y 3 y   y ( y 4  5 ) 2  3 ( y 4  5 ) 2 
 24 x 2 y 3 y   y ( y 4  5 ) 2  3 ( y 4  5 ) 2  12 x ( y 4  5 ) 
y [ ( y 4  5 ) 2  24 x 2 y 3 ]  3 ( y 4  5 ) 2  12 x ( y 4  5 ) 
3 ( y 4  5 ) 2  12 x ( y 4  5 )
y 
( y 4  5 ) 2  24 x 2 y 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
3 ( y 4  5 ) 2  12 x ( y 4  5 )
Answer: y  
( y 4  5 ) 2  24 x 2 y 3
7.
4
x4  y 4  7x3 y 6
( x 4  y 4 )1 / 4  7 x 3 y 6
4
4 1/ 4

NOTE: Dx ( x  y )
X
1 4
( x  y 4 )  3 / 4  Dx ( x 4  y 4 ) =
4
X
1 4
( x  y 4 )  3 / 4 ( 4 x 3  4 y 3 y  ) = ( x 4  y 4 )  3 / 4 ( x 3  y 3 y )
4
3 6
2 6
3
5
2 6
3 5
and Dx ( 7 x y )  21x y  7 x 6 y y  = 21x y  42 x y y Thus,
( x 4  y 4 )1 / 4  7 x 3 y 6 
( x 4  y 4 )  3 / 4 ( x 3  y 3 y )  21x 2 y 6  42 x 3 y 5 y 
( x 4  y 4 ) 3 / 4 [ ( x 4  y 4 )  3 / 4 ( x 3  y 3 y  ) ]  ( x 4  y 4 ) 3 / 4 ( 21x 2 y 6  42 x 3 y 5 y  ) 
x 3  y 3 y   21x 2 y 6 ( x 4  y 4 ) 3 / 4  42 x 3 y 5 y( x 4  y 4 ) 3 / 4 
x 3  21x 2 y 6 ( x 4  y 4 ) 3 / 4  42 x 3 y 5 y( x 4  y 4 ) 3 / 4  y 3 y 
x 3  21x 2 y 6 ( x 4  y 4 ) 3 / 4  y[ 42 x 3 y 5 ( x 4  y 4 ) 3 / 4  y 3 ] 
x 3  21x 2 y 6 ( x 4  y 4 ) 3 / 4
y 
42 x 3 y 5 ( x 4  y 4 ) 3 / 4  y 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
x 3  21x 2 y 6 ( x 4  y 4 ) 3 / 4
Answer: y  
42 x 3 y 5 ( x 4  y 4 ) 3 / 4  y 3
8.
3
xy 2  27  x 4 y
Since
x1 / 3 y
X
3
xy 2  ( xy 2 )1 / 3  x1 / 3 y 2 / 3 , then we have that
2/3
 27  x 4 y 
1  2/3 2/3
2
x
y  x1 / 3 y  1 / 3 y    ( 4 x 3 y  x 4 y  ) 
3
3
2
1

3 x 2 / 3 y 1 / 3  x  2 / 3 y 2 / 3  x1 / 3 y  1 / 3 y     3 x 2 / 3 y 1 / 3 ( 4 x 3 y  x 4 y  ) 
3
3

y  2 xy   12 x11/ 3 y 4 / 3  3x14 / 3 y1 / 3 y 
2 xy   3x14 / 3 y1 / 3 y    12 x11/ 3 y 4 / 3  y 
y ( 2 x  3x14 / 3 y1 / 3 )   12 x11/ 3 y 4 / 3  y 
 12 x11/ 3 y 4 / 3  y
y 
2 x  3x14 / 3 y 1 / 3
 12 x11/ 3 y 4 / 3  y
12 x11/ 3 y 4 / 3  y
Answer: y  
or y   
14 / 3 1 / 3
2 x  3x14 / 3 y1 / 3
2 x  3x y
Here is another way that this problem can be worked:
3
xy 2  27  x 4 y  ( xy2 )1 / 3  27  x 4 y
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
2 1/ 3
NOTE: Dx ( xy ) 
X
1
1
( xy 2 )  2 / 3  Dx ( xy 2 ) = ( xy 2 )  2 / 3 ( y 2  x 2 yy  ) =
3
3
X
1  2/3  4/3
1
(x
y
) ( y 2  2 xyy  ) = ( x  2 / 3 y  4 / 3 ) y ( y  2 xy  ) =
3
3
1  2 / 3  1/ 3
(x
y
) ( y  2 xy  ) Thus,
3
( xy2 )1 / 3  27  x 4 y 
1  2 / 3  1/ 3
(x
y
) ( y  2 xy  )   ( 4 x 3 y  x 4 y  ) 
3
1

3x 2 / 3 y 1 / 3  ( x  2 / 3 y  1 / 3 ) ( y  2 xy  )    3x 2 / 3 y 1 / 3 ( 4 x 3 y  x 4 y  ) 
3

y  2 xy   12 x11/ 3 y 4 / 3  3x14 / 3 y1 / 3 y 
2 xy   3x14 / 3 y1 / 3 y    12 x11/ 3 y 4 / 3  y 
y ( 2 x  3x14 / 3 y1 / 3 )   12 x11/ 3 y 4 / 3  y 
 12 x11/ 3 y 4 / 3  y
y 
2 x  3x14 / 3 y 1 / 3
Example Find the equation (in point-slope form) of the tangent line to the graph of
2 x 3  x 2 y  y 3  1  0 at the point ( 2 ,  3 ) .
2 x 3  x 2 y  y 3  1  0  6 x 2  ( 2 xy  x 2 y  )  3 y 2 y  0
x  2 , y   3 : 24  (  12  4 y  )  27 y   0 NOTE: y  in this equation is
actually y  ( 2 ,  3 ) . However, to write y  ( 2 ,  3 ) in the equation would produce
an equation that would be more difficult to work with. Thus, we write y  instead.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
24  (  12  4 y  )  27 y   0  24  12  4 y   27 y   0 
23y    36  y   
Answer: y  3  
36
23
Thus, mtan  y  ( 2 ,  3 )  
36
23
36
( x  2)
23
Example Find the slope of the normal line to the graph of
y 4  3 y  4 x 3  5x  31 at the point (  1,  2 ) .
y 4  3 y  4 x 3  5x  31  4 y 3 y   3 y   12 x 2  5
x   1, y   2 :  32 y   3 y   12  5 NOTE: y  in this equation is actually
y  (  1,  2 ) . However, to write y  (  1,  2 ) in the equation would produce an
equation that would be more difficult to work with. Thus, we write y  instead.
 32 y   3 y   12  5   29y   17  y   
Thus, mtan  y  (  1,  2 )  
17
29
17
29
 mnormal 
29
17
29
Answer:
17
The proof of the Power Rule for rational exponents: We will now give the
proof of the Power Rule for rational exponents using implicit differentiation and
the Power Rule for positive and negative integer exponents. The Power Rule for
positive integer exponents was proved in Lesson 8, and the Power Rule for
negative integer exponents was proved in Lesson 9.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Let y 
x m  x m / n , where n is a positive integer and m is an integer (positive or
m m/n  1
m/n
n
m
negative). We want to show that y   x
. Since y  x , then y  x .
n
m xm  1
n 1
m 1
 y 
Using implicit differentiation, we have that n y y   m x
.
n yn  1
n
Since y  x
m xm  1
nx
m
( n  1)
n
m/n
m xm  1
m xm  1
 y 
, then we have that y  
=
n yn  1
n ( xm / n )n  1
m m  1  (m  m/ n)
m m  1  m  m/n
m m/n  1
m xm  1
x
x
=
= n
= n
= n x
.
n xm  m/ n
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850