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=Pi6
-1.5
-1.0
r=1+sinHthetaL, theta=2 Pi6
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 Pi6
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 Pi6
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 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=9 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=6 Pi6
1.5
-0.5
1.0
-0.5
2.0
r=1+sinHthetaL, theta=10 Pi6
-1.0
-1.0
2.0
-0.5
-1.5
-1.5
2.0
r=1+sinHthetaL, theta=7 Pi6
-1.0
1.5
r=1+sinHthetaL, theta=5 Pi6
-0.5
-1.5
1.0
-0.5
r=1+sinHthetaL, theta=4 Pi6
-1.0
0.5
-0.5
-0.5
-1.5
r=1+sinHthetaL, theta=3 Pi6
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=Pi6
-2.0
-1.5
-1.0
r=1-cosHthetaL, theta=2 Pi6 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 Pi6
r=1-cosHthetaL, theta=6 Pi6
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 Pi6
r=1-cosHthetaL, theta=9 Pi6
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 Pi6
-2.0
-1.5
-0.5
r=1-cosHthetaL, theta=7 Pi6
-2.0
-2.0
-0.5
r=1-cosHthetaL, theta=4 Pi6
-2.0
r=1-cosHthetaL, theta=3 Pi6
1.5
r=1-cosHthetaL, theta=11 Pi6
r=1-cosHthetaL, theta=12 Pi6
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
