Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
Implicit Differentiation
Some functions can be described by expressing one variable explicitly in terms of another
variable — for example,
r
1−x
2
y=x , y=
, y = tan 2x
1 + x3
or, in general, y = f (x). Some functions, however, are defined implicitly by a relation between
x and y such that
x2 + y 2 = 1,
x3 + y 3 = 4xy,
(x2 + y 2 − 2x)2 = 4(x2 + y 2 )
In some cases it is possible to solve such an equation for x or for y, but sometimes it is impossible.
One of the main goals of Section 2.6 is to show how to find derivatives of implicitly defined
functions.
EXAMPLE: Find an equation of the tangent line to y = x2 at the point (2, 4).
Solution: The graph of the curve y = x2 is a parabola. Clearly,
y ′ = (x2 )′ = 2x
To find an equation of the tangent line to y = x2 at the point (2, 4)
we note that
m = y ′ (2) = 2 · 2 = 4
therefore
y − y0 = m(x − x0 )
=⇒
y − 4 = 4(x − 2)
=⇒
y = 4x − 4
EXAMPLE: Find equations of the tangent lines to x = y 2 at the points (4, 2) and (4, −2).
1
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
EXAMPLE: Find equations of the tangent lines to x = y 2 at the points (4, 2) and (4, −2).
Solution 1: The graph of the curve x = y 2 is a parabola. We have
( √
x
if y ≥ 0
x = y 2 =⇒ y =
√
− x if y < 0
√
Clearly, if y = x, then
1
1
1
y ′ = (x1/2 )′ = x1/2−1 = x−1/2 = √
2
2
2 x
√
Similarly, if y = − x, then
(1)
1
1
1
y ′ = (−x1/2 )′ = − x1/2−1 = − x−1/2 = − √
2
2
2 x
(2)
To find an equation of the tangent line to y 2 = x at the point (4, 2) we note that by (1) we
have
1
1
m1 = y ′ (4) = √ =
4
2 4
therefore
1
1
y − y1 = m1 (x − x1 ) =⇒ y − 2 = (x − 4) =⇒ y = x + 1
4
4
2
Similarly, to find an equation of the tangent line to y = x at the point (4, −2) we note that
by (2) we have
1
1
m2 = y ′ (4) = − √ = −
4
2 4
therefore
1
1
y − y2 = m2 (x − x2 ) =⇒ y − (−2) = − (x − 4) =⇒ y = − x − 1
4
4
2
Solution 2: To find equations of the tangent lines to x = y at the points (4, 2) and (4, −2) we
dy
first find
by differentiating both sides of x = y 2 :
dx
1
x = y 2 =⇒ x′ = (y 2 )′ =⇒ 1 = 2y · y ′ =⇒ y ′ =
2y
It follows that
′
m = y (4) =







therefore
y − y1 = m1 (x − x1 )
=⇒
1
2·2
if y = 2
1
if y = −2
2 · (−2)
1
y − 2 = (x − 4)
4
=⇒
1
y = x + 1 at (4, 2)
4
and
y − y2 = m2 (x − x2 )
=⇒
1
y − (−2) = − (x − 4)
4
2
=⇒
1
y = − x − 1 at (4, −2)
4
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
EXAMPLE: If x2 + y 2 = 5, find
at the point (2, 1).
dy
. Then find an equation of the tangent line to x2 + y 2 = 5
dx
Solution: The graph of the curve x2 + y 2 = 5 is a circle:
2
y
1
0
-2
-1
0
1
2
x
-1
-2
To find
dy
we differentiate both sides:
dx
x2 + y 2 = 5
=⇒
(x2 + y 2 )′ = 5′
=⇒
(x2 )′ + (y 2 )′ = 0
=⇒
2x · x′ + 2y · y ′ = 0
hence
2x · 1 + 2y · y ′ = 0
=⇒
2x + 2y · y ′ = 0
=⇒
2y · y ′ = −2x
=⇒
y′ = −
x
2x
=−
2y
y
To find an equation of the tangent line to x2 + y 2 = 5 at the point (2, 1) we note that
2
m = y ′ (2) = − = −2
1
therefore
y − y0 = m(x − x0 )
EXAMPLE: If 2x2 +3y 2 = 5, find
at the point (1, 1).
=⇒
y − 1 = −2 · (x − 2)
=⇒
y = −2x + 5
dy
. Then find an equation of the tangent line to 2x2 +3y 2 = 5
dx
3
Section 2.6 Implicit Differentiation
EXAMPLE: If 2x2 +3y 2 = 5, find
at the point (1, 1).
2010 Kiryl Tsishchanka
dy
. Then find an equation of the tangent line to 2x2 +3y 2 = 5
dx
Solution: The graph of the curve 2x2 + 3y 2 = 5 is an ellipse:
1
y
0.5
0
-1.5
-1
-0.5
0
0.5
1
1.5
x
-0.5
-1
To find
dy
we differentiate both sides:
dx
2x2 + 3y 2 = 5 =⇒ (2x2 + 3y 2 )′ = 5′ =⇒ 2(x2 )′ + 3(y 2 )′ = 0 =⇒ 2(2x · x′ ) + 3(2y · y ′ ) = 0
hence
2(2x · 1) + 3(2y · y ′ ) = 0 =⇒ 4x + 6y · y ′ = 0 =⇒ 6y · y ′ = −4x =⇒ y ′ = −
2x
4x
=−
6y
3y
To find an equation of the tangent line to 2x2 + 3y 2 = 5 at the point (1, 1) we note that
m = y ′ (1) = −
2·1
2
=−
3·1
3
therefore
y − y0 = m(x − x0 )
=⇒
2
y − 1 = − (x − 1)
3
=⇒
2
5
y =− x+
3
3
dy
. Then find an equation of the tangent line to x3 +y 3 = 4xy
dx
at the point (2, 2). At what point in the first quadrant is the tangent line horizontal?
EXAMPLE: If x3 +y 3 = 4xy, find
4
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
dy
. Then find an equation of the tangent line to x3 +y 3 = 4xy
dx
at the point (2, 2). At what point in the first quadrant is the tangent line horizontal?
EXAMPLE: If x3 +y 3 = 4xy, find
Solution: The graph of the curve x3 + y 3 = 4xy is the folium of Descartes:
2
1
y
x
-3
-2
-1
0
1
2
0
-1
-2
-3
(a) To find
dy
we differentiate both sides:
dx
x3 + y 3 = 4xy
=⇒
(x3 + y 3 )′ = (4xy)′
=⇒
(x3 )′ + (y 3 )′ = (4xy)′
Since (4xy)′ = 4(xy)′ = 4(x′ y + xy ′ ) = 4(1 · y + xy ′ ) = 4(y + xy ′ ), this gives us
3x2 · x′ + 3y 2 · y ′ = 4(y + xy ′ )
=⇒
3x2 + 3y 2 · y ′ = 4y + 4xy ′
=⇒
3y 2 y ′ − 4xy ′ = 4y − 3x2
hence
4y − 3x2
3y 2 − 4x
(b) To find an equation of the tangent line to x3 + y 3 = 4xy at the point (2, 2) we note that
y ′ (3y 2 − 4x) = 4y − 3x2
m = y ′ (2) =
therefore
y − y0 = m(x − x0 )
=⇒
y′ =
=⇒
4 · 2 − 3 · 22
= −1
3 · 22 − 4 · 2
y − 2 = −1 · (x − 2)
4
y
2
0
-4
-2
0
2
4
x
-2
-4
5
=⇒
y = −x + 4
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
(c) The tangent line is horizontal if y ′ = 0. Using the expression for y ′ from part (a), we see
that y ′ = 0 when 4y − 3x2 = 0 (provided that 3y 2 − 4x 6= 0). We have
2
4y−3x = 0
=⇒
3
y = x2
4
3
=⇒
which gives
both sides by x3 which implies
= 4x
3 2
x
4
=⇒
x3 +
27 6
x = 3x3
64
27 6
x = x3 . Since x 6= 0 in the first quadrant, we can divide
128
27 3
x =1
128
x=
3
27 6
x = 2x3
64
Dividing both sides by 2, we get
hence
3
x + x2
4
x3 +y 3 =4xy
r
3
128
=
27
(√
3
128
√
=
3
27
=⇒
x3 =
128
27
√ )
√
√
√
3
3
4√
432
64 · 2
64 3 2
3
√
√
=
=
=
2 ≈ 1.6798947
3
3
3
3
27
27
3
Plugging in this into y = x2 , we get
4
2
4√
3 4√
3
3
2 =
4 ≈ 2.1165347
y=
4 3
3
√
√ Finally, one can check that 3y 2 − 4x 6= 0 at 43 3 2, 34 3 4 . Thus the tangent line is horizontal
√
√ at 34 3 2, 34 3 4 , which is approximately (1.6798947, 2.1165347). Looking at the figure, we see
that our answer is reasonable.
EXAMPLE: If (x2 +y 2 −2x)2 = 4(x2 +y 2 ), find an equation of the tangent line to (x2 +y 2 −2x)2 =
4(x2 + y 2 ) at the point (0, 2).
6
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
EXAMPLE: If (x2 +y 2 −2x)2 = 4(x2 +y 2 ), find an equation of the tangent line to (x2 +y 2 −2x)2 =
4(x2 + y 2 ) at the point (0, 2).
Solution: The graph of the curve (x2 + y 2 − 2x)2 = 4(x2 + y 2 ) is the cardioid:
We differentiate both sides of (x2 + y 2 − 2x)2 = 4(x2 + y 2 ):
[(x2 + y 2 − 2x)2 ]′ = [4(x2 + y 2 )]′
=⇒
2(x2 + y 2 − 2x)2−1 (x2 + y 2 − 2x)′ = 4(x2 + y 2 )′
hence
2(x2 + y 2 − 2x)(2x + 2y · y ′ − 2) = 4(2x + 2y · y ′ )
(3)
To find slope of the tangent line to (x2 + y 2 − 2x)2 = 4(x2 + y 2 ) at the point (0, 2) we replace
x by 0 and y by 2 in (3):
2(02 + 22 − 2 · 0)(2 · 0 + 2 · 2 · y ′ − 2) = 4(2 · 0 + 2 · 2 · y ′ )
=⇒
8(4y ′ − 2) = 4(4y ′ )
hence
32y ′ − 16 = 16y ′
=⇒
32y ′ − 16y ′ = 16
=⇒
16y ′ = 16
=⇒
y′ =
16
=1
16
so m = 1, therefore
y − y0 = m(x − x0 )
=⇒
y − 2 = 1 · (x − 0)
=⇒
y =x+2
dy
for any x and then find the slope of the tangent line to (x2 + y 2 −
dx
2x)2 = 4(x2 + y 2 ) at the point (0, 2), the computations will be more complicated (see Appendix,
page 10).
REMARK: If we first find
7
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
EXAMPLE: If (x2 + y 2 )3 = 4x2 y 2 , find
dy
.
dx
Solution: The graph of the curve (x2 + y 2 )3 = 4x2 y 2 is a rose curve:
0.8
y 0.4
0
-0.8
-0.4
0
0.4
0.8
x
-0.4
-0.8
Solution: To find
dy
we differentiate both sides:
dx
(x2 + y 2 )3 = 4x2 y 2
=⇒
[(x2 + y 2 )3 ]′ = (4x2 y 2 )′
hence
3(x2 + y 2 )3−1 · (x2 + y 2 )′ = 4(x2 y 2 )′
(4)
(x2 + y 2 )′ = (x2 )′ + (y 2 )′ = 2x + 2y · y ′ = 2(x + y · y ′ )
(5)
(x2 y 2 )′ = (x2 )′ y 2 + x2 (y 2 )′ = 2xy 2 + x2 · 2y · y ′ = 2(xy 2 + x2 y · y ′ )
(6)
Note that
and
Substituting (5) and (6) into (4), we obtain
3(x2 + y 2 )2 · 2(x + y · y ′ ) = 4 · 2(xy 2 + x2 y · y ′ )
3(x2 + y 2 )2 (x + y · y ′ ) = 4(xy 2 + x2 y · y ′ )
3(x4 + 2x2 y 2 + y 4 )(x + y · y ′ ) = 4xy 2 + 4x2 y · y ′
We now expand the parentheses and solve this equation for y ′ :
3x5 + 3x4 y · y ′ + 6x3 y 2 + 6x2 y 3 · y ′ + 3xy 4 + 3y 5 · y ′ = 4xy 2 + 4x2 y · y ′
so
3x4 y · y ′ + 6x2 y 3 · y ′ + 3y 5 · y ′ − 4x2 y · y ′ = −3x5 − 6x3 y 2 − 3xy 4 + 4xy 2
hence
y ′ (3x4 y + 6x2 y 3 + 3y 5 − 4x2 y) = −3x5 − 6x3 y 2 − 3xy 4 + 4xy 2
therefore
y′ =
−3x5 − 6x3 y 2 − 3xy 4 + 4xy 2
3x5 + 6x3 y 2 + 3y 4 x − 4y 2 x
=
−
3x4 y + 6x2 y 3 + 3y 5 − 4x2 y
3y 5 + 6y 3 x2 + 3x4 y − 4x2 y
8
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
EXAMPLE: Find y ′ if sin(x + y) = y 2 cos x.
Solution: We have
sin(x + y) = y 2 cos x
=⇒
(sin(x + y))′ = (y 2 cos x)′
therefore
cos(x + y) · (x + y)′ = (y 2 )′ cos x + y 2 (cos x)′
hence
cos(x + y) · (1 + y ′ ) = 2y · y ′ cos x − y 2 sin x
We now solve this equation for y ′ :
cos(x + y) + cos(x + y) · y ′ = 2y · y ′ cos x − y 2 sin x
so
cos(x + y) · y ′ − 2y · y ′ cos x = − cos(x + y) − y 2 sin x
therefore
y ′ (cos(x + y) − 2y cos x) = − cos(x + y) − y 2 sin x
Dividing both sides by cos(x + y) − 2y cos x, we get
y′ =
cos(x + y) + y 2 sin x
− cos(x + y) − y 2 sin x
=
cos(x + y) − 2y cos x
2y cos x − cos(x + y)
9
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
Appendix
EXAMPLE: If (x2 + y 2 − 2x)2 = 4(x2 + y 2 ), find
Solution: To find
dy
.
dx
dy
we differentiate both sides:
dx
(x2 + y 2 − 2x)2 = 4(x2 + y 2 )
=⇒
[(x2 + y 2 − 2x)2 ]′ = [4(x2 + y 2 )]′
hence
2(x2 + y 2 − 2x)2−1 · (x2 + y 2 − 2x)′ = 4(x2 + y 2 )′
(7)
(x2 + y 2 − 2x)′ = (x2 )′ + (y 2 )′ − (2x)′ = 2x + 2y · y ′ − 2 = 2(x + y · y ′ − 1)
(8)
(x2 + y 2 )′ = (x2 )′ + (y 2 )′ = 2x + 2y · y ′
(9)
Note that
and
Substituting (8) and (9) into (7), we obtain
2(x2 + y 2 − 2x) · 2(x + y · y ′ − 1) = 4(2x + 2y · y ′ )
(x2 + y 2 − 2x)(x + y · y ′ − 1) = 2x + 2y · y ′
We now expand the parentheses and solve this equation for y ′ :
x3 + x2 y · y ′ − x2 + xy 2 + y 3 · y ′ − y 2 − 2x2 − 2xy · y ′ + 2x = 2x + 2y · y ′
so
x2 y · y ′ + y 3 · y ′ − 2xy · y ′ − 2y · y ′ = −x3 − xy 2 + 3x2 + y 2
hence
y ′ (x2 y + y 3 − 2xy − 2y) = −x3 − xy 2 + 3x2 + y 2
therefore
y′ =
x3 + xy 2 − 3x2 − y 2
−x3 − xy 2 + 3x2 + y 2
=
−
x2 y + y 3 − 2xy − 2y
y 3 + x2 y − 2xy − 2y
10
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
EXAMPLE: Sketch the curve r = 1 + sin θ, 0 ≤ θ ≤ 2π (cardioid).
Solution: We have
r=1+sinHthetaL, theta=Pi6
-1.5
-1.0
r=1+sinHthetaL, theta=2 Pi6
2.0
2.0
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
r=1+sinHthetaL, theta=8 Pi6
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
r=1+sinHthetaL, theta=11 Pi6
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
11
1.5
r=1+sinHthetaL, theta=12 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=9 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=6 Pi6
1.5
-0.5
1.0
-0.5
2.0
r=1+sinHthetaL, theta=10 Pi6
-1.0
-1.0
2.0
-0.5
-1.5
-1.5
2.0
r=1+sinHthetaL, theta=7 Pi6
-1.0
1.5
r=1+sinHthetaL, theta=5 Pi6
-0.5
-1.5
1.0
-0.5
r=1+sinHthetaL, theta=4 Pi6
-1.0
0.5
-0.5
-0.5
-1.5
r=1+sinHthetaL, theta=3 Pi6
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
1.0
1.5
Section 2.6 Implicit Differentiation
2010 Kiryl Tsishchanka
EXAMPLE: Sketch the curve r = 1 − cos θ, 0 ≤ θ ≤ 2π (cardioid).
Solution: We have
r=1-cosHthetaL, theta=Pi6
-2.0
-1.5
-1.0
r=1-cosHthetaL, theta=2 Pi6
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-1.5
-1.0
-1.5
-1.0
-1.5
-1.0
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=5 Pi6
r=1-cosHthetaL, theta=6 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=8 Pi6
r=1-cosHthetaL, theta=9 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=10 Pi6
-2.0
-1.5
-0.5
r=1-cosHthetaL, theta=7 Pi6
-2.0
-2.0
-0.5
r=1-cosHthetaL, theta=4 Pi6
-2.0
r=1-cosHthetaL, theta=3 Pi6
1.5
r=1-cosHthetaL, theta=11 Pi6
r=1-cosHthetaL, theta=12 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
12