3.8: Implicit Di↵erentiation
Problem 1 On the graph below, sketch the tangent lines at x = 0. Then, explain why both the x-coordinate
and the y-coordinate are generally needed to find the slope of the tangent line at a point on the graph of an
equation of the form F (x, y) = 0
Solution: Given the equation F (x, y) = 0, its graph may not pass the “vertical line test”. As illustrated
in the figure below, fixing a value for x (in this case x = 0) will not typically specify a unique point on the
graph because there may be more than one corresponding y-value. That is why you need to specify both the
x and the y coordinate, to make sure that you are giving the coordinates for a unique point on the graph.
In this case, there are two tangent lines to the curve at x = 0. One passes through the point (0, 1) and the
other through (0, 1).
1
Problem 2 Consider the equation x2 +4xy+9y 2 = 9. Note: This equation is equivalent to x2 +4xy+9y 2 9 =
0. Therefore it has a form F (x, y) = 0
(a) Find
dy
.
dx
Solution:
d 2
d
(x + 4xy + 9y 2 ) =
(9)
dx
dx
2
✓
◆
dy
dy
2x + 4y + 4x
+ 18y
=0
dx
dx
dy
dy
+ 18y
= 2x 4y
dx
dx
dy
(4x + 18y)
= 2x 4y
dx
dy
2x 4y
=
dx
4x + 18y
4x
provided that 4x + 18y 6= 0.
(b) Find the equation(s) of the tangent line(s) when x = 0. Draw the tangent line(s) on the above picture.
Solution:
Plugging into the original equation, we see that when x = 0 we have that
02 + 4(0)y + 9y 2 = 9
Then,


y2 = 1
=)
dy
dx
dy
dx
=
(0,1)
=
(0, 1)
4
=
18
4
=
18
=)
y = ±1
2
9
2
9
So there are two tangent lines to the graph of the given equation when x = 0. The two points are
2
(0, 1) and (0, 1), and the tangent lines at both points have slope
. Thus, the equations of the two
9
tangent lines are
2
2
y 1=
(x 0)
=)
y=
x+1
9
9
2
2
y+1=
(x 0)
=)
y=
x 1
9
9
3
(c) Find the point(s) where the tangent line is horizontal. Draw the point(s) and line(s) on the above
picture.
Solution: A line is horizontal if and only if its slope is 0. So we are looking for the points (x0 , y0 )
dy
such that
= 0. So we solve:
dx (x0 ,y0 )
2x 4y
=0
4x + 18y
=)
2x
4y = 0
=)
But we also need the point to lie on the the given curve. So, letting x =
x=
2y
2y, we solve:
x2 + 4xy + 9y 2 = 9
( 2y)2 + 4( 2y)y + 9y 2 = 9
4y 2
8y 2 + 9y 2 = 9
5y 2 = 9
9
y2 =
5
3
y = ±p
5
So there exists two solutions
to the
where the tangent line to the graph has 0 slope.
✓
◆ given
✓ equation ◆
6
3
6
3
p ,p
Those two points are
and p , p .
5
5
5
5
Problem 3 A part of the curve with equation cos(⇡xy) + x + y = 1 is sketched below.
4
(a) Use the implicit di↵erentiation to find the derivative dy/dx.
Solution:
Implicitly di↵erentiate equation:
d
d
(cos(⇡xy) + x + y) =
(1) =)
dx
dx
Solve for
sin(⇡xy) · (⇡y + ⇡x
dy
dy
)+1+
=0
dx
dx
dy
:
dx
sin(⇡xy) · (⇡y + ⇡x
dy
dy
)+1+
= 0 =)
dx
dx
dy
dy
sin(⇡xy) + 1 +
=0
dx
dx
dy
=) 1 ⇡x sin(⇡xy)
= ⇡y sin(⇡xy) 1
dx
dy
⇡y sin(⇡xy) 1
=)
=
if 1 ⇡x sin(⇡xy) 6= 0
dx
1 ⇡x sin(⇡xy)
⇡y sin(⇡xy)
⇡x
(b) Consider the point (1, 1). Show (algebraically) that this point lies on the curve.
Solution:
(1, 1) lies on this curve:
cos(⇡ · 1 · 1) + 1 + 1 = cos(⇡) + 2
=
1+2=1
(c) Find the equation of the line tangent to the curve at (1, 1). Draw this line in the figure above.
Solution:
Slope of tangent line:

dy
dx
=
(1,1)
⇡ · 1 · sin(⇡1 · 1) 1
1 ⇡ · 1 · sin(⇡1 · 1)
⇡ sin(⇡) 1
1 ⇡ sin(⇡)
1
=
= 1
1
=
5
Equation of tangent line:
y
1=
1(x
1) =) y =
x+2
Graph of tangent line:
Problem 4 For each of the curves given by the following equations, find a formula for the slope of the tangent
line at a point (x, y).
(a) ex
2 3
y
5x + 7y = 36
Solution:
◆
dy
dy
e
2xy + x (3y )
5+7
=0
dx
dx
2 3
2 3 dy
dy
2xy 3 ex y + 3x2 y 2 ex y
5+7
=0
dx
dx
2 3 dy
2 3
dy
3x2 y 2 ex y
+7
= 2xy 3 ex y + 5
dx
dx
⇣
⌘ dy
2 3
2 3
3x2 y 2 ex y + 7
= 2xy 3 ex y + 5
dx
2 3
dy
2xy 3 ex y + 5
=
dx
3x2 y 2 ex2 y3 + 7
x2 y 3
✓
3
2
2
6
(b) 7 = 22 tan(y) +
4
x
7
y
Solution:
0 = 22 sec2 (y)
22 sec2 (y)
dy
dx
4
7 dy
+ 2
x2
y dx
dy
7 dy
4
+
= 2
dx y 2 dx
x
4
dy
x2
=
dx
22 sec2 (y) +
(c) cos(xy)
7
y2
x3 = 5y 3
Solution:
✓
dy
sin(xy) · y + x
dx
y sin(xy)
x sin(xy)
x sin(xy)
◆
dy
dx
3x2 = 15y 2
dy
dx
3x2 = 15y 2
dy
dy
+ 15y 2
= y sin(xy)
dx
dx
dy
y sin(xy) 3x2
=
dx
x sin(xy) + 15y 2
dy
dx
3x2
Provided that x sin(xy)+15y 2 6= 0. It is worth pointing out that equivalent condition was not necessary
for parts (a) and (b) because the denominator of those solutions cannot be 0.
Problem 5 The volume of a doughnut with an inner radius of a and an outer radius of b is
V = ⇡2
(b + a)(b
4
a)2
.
Find db/da if the volume of a doughnut is 64⇡ 2 and does not change.
Solution:
We have to (implicitly) di↵erentiate
64⇡ 2 = ⇡ 2
(b + a)(b
4
a)2
.
Before doing this we’ll perform a bit of algebra to simplify our calculuations:
64⇡ 2 = ⇡ 2
(b + a)(b
4
a)2
=) 256 = (b + a)(b
a)2 .
Now we’ll di↵erentiate with respect to a:
256 = (b + a)(b
a)
2
✓
◆
✓
db
db
2
=) 0 =
+ 1 (b a) + (b + a)2 · (b a) ·
da
da
db
db
=) 0 =
(b a)2 + (b a)2 + 2(b2 a2 )
2(b2
da
da
7
◆
1
a2 ).
To finish we solve for
db
:
da
db
db
(b a)2 + (b a)2 + 2(b2 a2 )
2(b2 a2 )
da ✓
da
◆
db
db
=)
(b a)2 + 2(b2 a2 )
= (b a)2 2(b2
da
da
db
(b a)2 2(b2 a2 )
=)
=
.
da
((b a)2 + 2(b2 a2 ))
0=
Problem 6 Find
Solution:
d2 y 1/3
. x + y 2/3 = 2.
dx2
Take the first derivative:
1
x
3
Solving for
2/3
2
+ y
3
1/3
·
dy
=0
dx
dy
:
dx
1
3x2/3
2
3y 1/3
dy
=
dx
Take the second derivative:
We need to subsitute in for
1
2/3
3y
d2 y
=
dx2
·
dy
dx
=
y 1/3
2x2/3
· 2x2/3 + y 1/3 · 43 x
4x4/3
1/3
dy
we have:
dx
d2 y
=
dx2
1
2/3
3y
·
y 1/3
2x2/3
8
· 2x2/3 + y 1/3 · 43 x
4x4/3
1/3
a2 )