Calculus 12 – Ch. 2 Derivatives
Lesson 7: Implicit Differentiation
 If a relation is written such that y is explicitly in terms of x [ie. y = f (x) ] then we apply our
basic derivative rules.
 In many cases, it is either difficult or impossible to solve an equation for y in terms of x .
Ex.
x 2 + y 2 = 16
or
x 3 + 2x 2 y + y !
1
=1
y2
 Nonetheless, we still need to be able to find the derivative to interpret the slope of the
tangent line at particular points.
 For relations not written (or impossible to write) in y = f (x) we use a process called:
IMPLICIT DIFFERENTIATION
STEPS:
1) Differentiation both sides of the equation with respect to x
dy
dy
2) Solve the resulting equation for y! or
(ie. isolate y! or
)
dx
dx
Examples:
1) Use implicit differentiation to find y! or
a) x 2 + y 2 = 16
2x + 2yi y! = 0
2yi y! = "2x
"2x
y! =
2y
x
y! = "
y
dy
for each of the following:
dx
b) 3x 2 + y = y 3 + 5
6x + (1) y! = 3y 2 i y!
6x = 3y 2 i y! " y!
6x = y! ( 3y 2 "1)
6x
= y!
3y 2 "1
c) 2x 3 + xy = y 2
d) x 2 + y = x 2 y 3 !10
1
2x +
1
1 !2
y i y" ! 3x 2 y 2 i y" = 2xy 3 ! 2x
2
# 1
&
y" %%
! 3x 2 y 2 (( = 2xy 3 ! 2x
$2 y
'
6x 2 + (1) y + xy! = 2yi y!
xy! " 2yi y! = "6x 2 " y
y! ( x " 2y) = "6x 2 " y
y! =
1 !2
y i y" = 2xy 3 + x 2 i3y 2 i y"
2
"6x 2 " y
x " 2y
y" =
y" =
(2xy
! 2x ) 2 y
# 1
&
! 3x 2 y 2 (2 y
%
$2 y
'
(y
3
3
!1) 4x y
2
1 ! 6x y
5
2
2) Find the equation of the tangent line to the curve 2x 5 + x 4 y + y 5 = 36 at the point (1, 2) .
Slope m = y!
10x 4 + 4x 3 y + x 4 y! + 5y 4 y! = 0
x 4 y! + 5y 4 y! = "10x 4 " 4x 3 y
y! ( x 4 + 5y 4 ) = "10x 4 " 4x 3 y
y! =
"10x 4 " 4x 3 y
x 4 + 5y 4
Slope at the point (1, 2) :
!10(1)4 ! 4(1)3 (2)
(1)4 + 5(2)4
!10 ! 8
=
1+ 80
!18
=
81
2
=!
9
m=
Tangent line at the point (1, 2) :
y ! 2 !2
=
x !1 9
9y !18 = !2x + 2
2x + 9y ! 20 = 0
3) Find the slope of the tangent line to the curve
x + y + xy = 4 at the point (2, 2) .
1
1
1
1
!
!
( x + y) 2 (1+ y") + ( xy) 2 ( y + xy") = 0
2
2
1
1
1
1
1
1
1
1
!
!
!
!
( x + y) 2 + ( x + y) 2 y" + ( xy) 2 y + ( xy) 2 xy" = 0
2
2
2
2
1
1
1
1
1
1
1
1
!
!
!
!
( x + y) 2 y" + ( xy) 2 xy" = ! ( x + y) 2 ! ( xy) 2 y
2
2
2
2
1
1
1
1
#1
&
1
1
1
!
!
!
!
y" % ( x + y) 2 + ( xy) 2 x ( = ! ( x + y) 2 ! ( xy) 2 y
$2
'
2
2
2
1
1
1
1
!
!
! ( x + y) 2 ! ( xy) 2 y
2
y" = 2
1
1
1
1
!
!
( x + y) 2 + ( xy) 2 x
2
2
Slope at (2, 2) :
1
1
1 "1
"
(2 + 2) 2 " ( 4 ) 2 (2)
2
y! = 2
1
1
1
1
"
"
(2 + 2) 2 + ( 4 ) 2 (2)
2
2
1#1& 1#1&
" % ( " % ( (2)
2$2' 2$2'
=
1#1& 1#1&
% ( + % ( (2)
2$2' 2$2'
1 1
" "
= 4 2
1 1
+
4 2
= "1
"
4) Find the slope of the tangent line to the curve
!3 $
1 1
+ = 1 at the point # , 3& .
"2 %
x y
x !1 + y!1 = 1
!1x !2 !1y!2 y" = 0
y" 1
! 2= 2
y
x
y" = !
y2
x2
!3 $
Slope at # , 3& :
"2 %
( 3)
y! = "
2
2
# 3&
% (
$2'
= "4