41
Implicit differentiation
READING
Read Section 3.8 of Rogawski
Reading
So far all of the equations we have been interested in have been of the form
y = (a function in variable x),
such as
y = x2 − 3.
Such equations are called explicit equations since we can write y = f (x) with f (x) = x2 −3
and then the y-coordinate is given explicitly in term of x.
However, many interesting curves can not be written in this way. For example, the unit
circle is given by the equation
x2 + y 2 = 1,
where it is understood that a point (a, b) is on the unit circle if an only if a2 + b2 = 1.
Another example is the curve
which has the equation x3 + y 3 = 3xy.
Equations of this type are called implicit equations. Notice that implicit equations can
not necessarily be described in the form y = f (x).
It turns out that we can still differentiate implicit equations (and therefore find equations
of tangent lines and so on) using a technique called implicit differentiation.
Implicit
Step 1. Regard y as a function in x.
differentiation Step 2. Differentiate both sides of the implicit equation
(regarding y as a function in x).
Step 3. Solve for y 0 .
Example 1. Let y be a function in x. Calculate
Solution.
1
d
(y
dx
+ x),
d 2 d
y , dx xy 2 ,
dx
and
d (y+x)
e
,
dx
-
Example 2. Find the tangent line to the curve defined by x3 + y 3 = 3xy at the point ( 32 , 43 ).
Solution. We first find the gradient of the tangent by finding the derivative y 0 and evaluating
it at the point ( 23 , 43 ).
We use implicit differentiation to find the derivative y 0
x3 + y 3 = 3xy
so
d
d
x3 + y 3 =
(3xy)
dx
dx
so
d
d
d
x3 +
y3 =
(3xy)
dx
dx
dx
Remembering that y is a function in x, evaluating the derivative gives the equation
3x2 + 3y 2 y 0 = 3y + 3xy 0 .
Then
y0 =
y − x2
3y − 3x2
=
3y 2 − 3x
y2 − x
The gradient of the tangent line at ( 23 , 43 ) is then
0
y |( 2 , 4 )
3 3
4
3
− ( 23 )2
4
= 4 2 2 =
5
(3) − 3
The equation of the tangent line is then
4
y = x + c.
5
It passes through the point ( 23 , 43 ) so
4
42
=
+c
3
53
giving
4
c= .
5
3
3
The equation of the tangent line to x + y = 3xy at the point ( 23 , 34 ) is then
4
4
y = x+ .
5
5
2
Example 3. Let y 4 − y = x3 + x. Determine y 0 .
Solution. We use implicit differentiation.
y 4 − y = x3 + x
so
d
d
y4 − y =
x3 + x .
dx
dx
Then
d
d
y4 −
(y) = 3x2 + 1.
dx
dx
Then
= 3x2 + 1.
-
Solving for y 0 gives
y0 =
3x2 + 1
4y 3 − 1
Example 4. Let sin(x + y) = x + cos(y). Show y 0 =
1−cos(x+y)
.
cos(x+y)+sin(y)
-
Solution.
3
Example 5. Find the (x, y)-coordinates of all of the points on the graph of 3x2 +4y 2 +3xy =
24 where the tangent line is horizontal.
Solution. The tangent line at a point (x, y) is horizontal if and only if y 0 = 0. So we need
to determine all points (x, y)where y 0 = 0.
Implicit differentiation gives
d
d
3x2 + 4y 2 + 3xy =
24,
dx
dx
giving
6x + 8yy 0 + 3xy 0 + 3y = 0
Now,
y0 = 0
⇐⇒
6x + 3y = 0
⇐⇒
y = −2x.
So the tangent line at a point (x, y) to the curve 3x2 + 4y 2 + 3xy = 24 is horizontal only if
(x, y) = (x, −2x)
To find the points on 3x2 + 4y 2 + 3xy = 24 of the form (x, −2x) we substitute this point
into the equation:
(x, −2x) is on 3x2 +4y 2 +3xy = 24
r
⇐⇒
x=±
24
13
⇐⇒
3x2 +4(−2x)2 +3x(−2x) = 24
⇐⇒
13x2 = 24
r
r !
r
r !
24
24
24
24
x=
, −2
or
, −2
13
13
13
13
⇐⇒
HOMEWORK Rowgowski Section 3.8: Q 1, 9, 5, 13, 25, 35, 37, 48
4
Homework