Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 57 doc

17.3 Implicit Differentiation 541
6. Suppose y =
f(x)
g(x)
, where f(x)and g(x) are positive for all x. Use logarithmic differ-
entiation to find
dy
dx
. Verify that this is the same result you would get had you used the
Quotient Rule.
7. If you felt so inclined, you could come up with a “rule” for taking the derivative of
functions of the form f(x)
g(x)
where f(x) is positive. You might call it “the Tower
Rule” since you have a tower of functions, or you might think of a more descriptive
name. In any case, what would this rule be?
17.3 IMPLICIT DIFFERENTIATION
When using the process of logarithmic differentiation, we differentiate an equation in which
y is not explicitly expressed as a function of x. Logarithmic differentiation is a special case
of the broader concept of implicit differentiation, a concept with far-reaching implications
and applications. The basic idea is that we can find
dy
dx
even when y is not given explicitly
as a function of x. We differentiate both sides of the equation that relates x and y, applying
the Chain Rule to differentiate terms involving y because y varies with x.
Implicit differentiation is an important concept; we’ll begin with a very straightforward
example to illustrate what is going on.
◆
EXAMPLE 17.4 Consider the circle of radius 2 centered at the origin.
5

It is given by x
2
+ y
2
= 4. Find the
slope of the line tangent to the circle at the following points.
(a) (1,
√
3)
(b) (1, −
√
3)
y
x
–11
1
2
2
(1, –√3)
(1, √3)
Figure 17.1
SOLUTION Although y is not a function of x, it can be expressed as two different functions of x.
y =

4 − x
2
(the top semicircle) and y =−

4 − x
2

(the bottom semicircle)
Each of these functions gives y explicitly as a function of x. One approach is to differentiate
the former expression to get information about the point (1,
√
3) and the latter to deal with
5
This circle is the set of all points a distance 2 from the origin. If (x, y) is a point on this circle, then the distance formula
tells us that

(x −0)
2
+ (y − 0)
2
= 2. And conversely, if (x, y) satisfies the equation

x
2
+ y
2
= 2, then (x, y) is a point on the
circle. Therefore, x
2
+ y
2
= 4 is the equation of the circle.
542 CHAPTER 17 Implicit Differentiation and its Applications
(1, −
√
3). Do this on your own and compare your answers with those below. We’ll take a
more efficient approach.

We are looking for
dy
dx
, the rate of change of y with respect to x, at a point. The equation
x
2
+ y
2
= 4 implicitly gives a relationship between x and y. Because
dy
dx
is a local measure
of a rate of change, we are interested in a bug’s-eye view rather than a bird’s-eye view of the
curve. Let’s peer at the point (1,
√
3) through a magnifying glass. In the immediate vicinity
of (1,
√
3) the curve looks like the graph of a differentiable function, despite the fact that a
bird’s-eye view tells us it is not.
6
Because a derivative is a local measure of a rate of change,
this is all that is really important. We differentiate both sides of the equation treating y as if
it were a function of x. We differentiate each side with respect to x because we are looking
for
dy
dx
, the derivative of y with respect to x.
d
dx

[x
2
+y
2
] =
d
dx
[4]
Because we are treating y as if it were a function of x, when we differentiate y
2
we get
2y
dy
dx
. Just as we needed to use the Chain Rule to find
d
dx
[ln y] when we did logarithmic
differentiation in Example 17.2, we need to use the Chain Rule here to evaluate
d
dx
y
2
.
Because y depends on x, what we are really trying to find here is
d
dx
(mess)
2
. The Chain

Rule tells us that the derivative of (mess)
2
is 2(mess) · (mess)

. Therefore,
2x + 2y
dy
dx
=0 Now solve for
dy
dx
.
2y
dy
dx
=−2x
dy
dx
=
−x
y
.
You may initially be startled that the formula for
dy
dx
involves not only x,butyalso. On
second thought, this should not be too surprising, because y is not a function of x.Agiven
x-value may correspond to more than one y-value and therefore may have more than one
slope associated with it. We need to know both coordinates of a point on the curve, not just
the x-value, to determine the slope of the tangent line. For example, x =1 at points (1,

√
3)
and (1, −
√
3) on the circle; we need to specify a value for y to know which one is meant.
It makes sense that our formula for
dy
dx
should involve both x and y.
At (1,
√
3), the slope of the tangent line is
−1
√
3
, while at (1, −
√
3) the slope of the
tangent line is
1
√
3
.
We’ve solved the original problem, but let’s look back at it one more time. We found
dy
dx
=
−x
y
. If we are looking at a point on the top semicircle where y =

√
4 − x
2
, we can write
dy
dx
=
−x
√
4−x
2
; if we were looking at a point on the bottom semicircle where y =−
√
4 − x
2
,
we can write
dy
dx
=
−x
−
√
4−x
2
.
dy
dx
is undefined when y = 0. This corresponds to the points (2, 0) and (−2, 0) on
the circle. Get out your magnifying glass again and take a good look at the curve in the

immediate vicinity of each of these points. No matter how much the curve is magnified,
6
The graph of the circle does not pass the vertical line test. Any bird with enough height can see that.
17.3 Implicit Differentiation 543
it does not look like the graph of a function. Intuitively speaking,
dy
dx
is defined at a point
P , and we can find its value using implicit differentiation only if, under magnification, the
curve around P looks like the graph of a differentiable function.
◆
◆
EXAMPLE 17.5 The curve shown below is called the folium of Descartes. It is the set of all points satisfying
the equation x
3
+ y
3
= 6xy. The point (3, 3) lies on this curve; when we substitute x = 3
and y = 3 into the equation, both sides are equal. Find the slope of a tangent line to
x
3
+ y
3
= 6xy, first in general, and then at the point (3, 3).
y
x
(3, 3)
x
3
+ y

3
= 6xy
Folium of Descartes
Figure 17.2
SOLUTION We need to determine
dy
dx
. Notice that we do not have an explicit formula for y in terms of
x. In fact, a glance at the graph shows that y is not a function of x. While it is possible to
solve for y explicitly in terms of x using several functions, it is very difficult. However, if
we magnify the curve right around the point (3, 3), it does look like the graph of a function,
so we can use implicit differentiation treating y as if it were a function of x and applying
the Chain Rule where necessary.
d
dx
(x
3
+ y
3
) =
d
dx
(6xy) Differentiate each side with respect to x,
applying the Chain Rule.
3x
2
+ 3y
2
dy
dx

= 6y + 6x
dy
dx
In differentiating the 6xy on the right side of the equation, we have used the Chain Rule
in combination with the Product Rule. Now we need to solve for
dy
dx
. The equation we have
is linear in
dy
dx
, so we use the standard strategy for solving linear equations.
3x
2
+ 3y
2
dy
dx
= 6y + 6x
dy
dx
Bring all terms involving
dy
dx
to one side.
3y
2
dy
dx
− 6x

dy
dx
= 6y − 3x
2
Factor out a
dy
dx
.
dy
dx
(3y
2
− 6x) = 6y − 3x
2
Divide through to isolate
dy
dx
.
dy
dx
=
6y − 3x
2
3y
2
− 6x
dy
dx
=
2y − x

2
y
2
− 2x
544 CHAPTER 17 Implicit Differentiation and its Applications
We now have a formula for
dy
dx
; this expression makes sense, provided y
2
= 2x.
7
To fi nd the slope of the line tangent to the folium of Descartes at the point (3, 3),
substitute x = 3 and y = 3 into the expression for
dy
dx
, obtaining
dy
dx
=−1. This looks quite
reasonable considering the symmetry of the graph around the line y = x.
◆
◆
EXAMPLE 17.6 Find the slope of the tangent line to the curve 3y
4
+ 4xy
2
− 2x
2
= 9 at the point (2, 1).

SOLUTION We take the derivative of both sides with respect to x, thinking of y as if it were a function
of x.
d
dx
(3y
4
+ 4xy
2
− 2x
2
) =
d
dx
(9)
12y
3
dy
dx
+ 4y
2
+ 8xy
dy
dx
− 4x = 0
Although we could solve for
dy
dx
and then substitute in x = 2 and y = 1, because a general
formula is not called for, it is algebraically much cleaner to substitute in the x- and y-values
immediately after differentiation.

8
12(1)
3
dy
dx
+ 4(1)
2
+ 8(2)(1)
dy
dx
− 4(2) = 0
28
dy
dx
− 4 = 0
dy
dx
=
1
7
Thus, the slope at the point (2, 1) is
1
7
.
◆
◆
EXAMPLE 17.7 The equation 2x
2
+ 4xy + 3y
2

= 6 describes an ellipse
9
centered at the origin and rotated
as shown below. Find the maximum and minimum values of y on this curve.
y
x
√2
√3
–√2
–√3
Figure 17.3
SOLUTION We need to find the points at which the line tangent to the curve is horizontal, i.e., where
dy
dx
= 0. To find
dy
dx
we use implicit differentiation.
7
At the points on the folium of Descartes where y
2
= 2x the derivative is undefined. y does not look like a differentiable
function of x, no matter how much you magnify the area around the point in question. For example, look at the point (0, 0).
8
Don’t substitute in values of x and y before differentiating; you would just be taking the derivative of constants.
9
For information on conic sections, see Appendix E, Conic Sections.
17.3 Implicit Differentiation 545
d
dx

(2x
2
+4xy + 3y
2
) =
d
dx
(6)
4x +4y +4x
dy
dx
+6y
dy
dx
=0
dy
dx
(4x +6y) =−4x−4y
dy
dx
=
−2x −2y
2x + 3y
dy
dx
=0where the numerator of the fraction is zero and the denominator is not simultaneously
zero. Simplifying −2x − 2y = 0givesy=−x.
Wemust check that the denominator is not zero at the same time. Substituting y =−x
into the equation 2x + 3y = 0givesx=0.Thus, when x and y are both zero,
dy

dx
is not
defined. But the point (0, 0) is not on the ellipse, so we need not worry about it.
y =−xis an entire line of points; we need to find out which of these points are also on
the ellipse. We find the points of intersection of the ellipse and the line by solving the two
equations simultaneously.

2x
2
+ 4xy +3y
2
= 6
y =−x
The simplest way to do this is to substitute the second equation into the first to eliminate
the variable y and then solve for x.
2x
2
+ 4xy +3y
2
= 6
2x
2
+ 4x(−x) +3(−x)
2
=6
2x
2
−4x
2
+3x

2
=6
x
2
=6
x =
√
6orx=−
√
6
These are the two values of x at which the tangent line to the ellipse is horizontal. Our
original goal is to find the maximum and minimum y-values, so we can substitute these x-
values back into the original equation. Or, because we know that they will lie on the line
y =−x,and this is a much simpler equation, we can say that at x =
√
6, the minimum y-
value of −
√
6 is attained and that at x =−
√
6, the maximum y-value of
√
6 is attained.
◆
The Process of Implicit Differentiation
1. Decide which variable you want to differentiate with respect to (x if you want
dy
dx
, t if
you want

dy
dt
, etc.).
2. Differentiate both sides of the equation with respect to that variable. Remember the
Chain Rule!
Suppose you are differentiating with respect to t. Distinguish between quantities that
vary with t (treating them as functions of t ) and those that are independent of t (treating
them as constants). In particular, when looking for
dy
dt
, think of y as a function of t.
546 CHAPTER 17 Implicit Differentiation and its Applications
3. If necessary, solve to find a formula for the desired derivative. If you only want to
know the derivative at a specific point, substitute in the coordinates of that point before
solving for the derivative you’re trying to find.
A Brief Recap of Differentiating “with respect to” a Particular Variable
In practice, because a functional relationship between variables is not explicitly spelled out
when we use implicit differentiation, it is necessary to think about what quantities change
with which other variables. Suppose, for instance, that we will differentiate both sides of an
equation with respect to the variable time t; we must establish which quantities vary with
respect to time and which do not. We’ll clarify this by means of an example.
◆
EXAMPLE 17.8 Let P be the pressure under which a gas is kept, V be the volume of the gas, and T be
temperature measured on the absolute (or Kelvin) scale. Then the combined gas law tells
us that
PV
T
= K, where K is a constant.
10
(a) Suppose temperature is kept constant. Express the rate of change of volume with respect

to pressure.
(b) Suppose volume is kept constant. Express the rate of change of temperature with respect
to pressure.
11
(c) Suppose volume is kept constant but pressure and temperature change with time. What
is the relationship between the change in pressure with respect to time and the change
in temperature with respect to time?
(d) Suppose pressure, temperature, and volume all change with time. How are these rates
of change related?
SOLUTION (a) Here we are thinking of P as the independent variable and looking for
dV
dP
. V varies
with P , so we treat it as a function of P . T is treated as a constant in this part of the
problem. K is also a constant. We can emphasize this by writing
P
T
V(P)= K.
Wewant to find
dV
dP
. One option is to solve for V in terms of P to get
V(P)=
KT
P
= KT P
−1
.
To fi nd out how V varies with P we differentiate with respect to P , obtaining
d

dP
[V ] =
d
dP
[KT P
−1
]
10
Here’s a little history for the chemists among you. In 1660 Robert Boyle published his gas law stating that if temperature
is kept constant, then the product of the pressure and the volume is a constant. Later Jacques Charles looked at the relationship
between volume and temperature when pressure is kept constant. The absolute (Kelvin) temperature scale allows Charles’slaw
to be written V/T = C
1
, where T is the temperature on the absolute scale and the constant C
1
depends on the pressure and the
mass of gas present. In 1802 Joseph Guy-Lussac’s investigation of gases yielded the result that P/T = C
2
,where C
2
is a constant.
Combining Boyle’s, Charles’s, and Guy-Lussac’s laws gives the more general gas law, PV/T = C
3
,where the value of the constant
C
3
depends on the amount of gas present and T is the absolute (or Kelvin) temperature. I found this law handy on Christmas day
1996 when baking a cake for a potluck dinner in the Ecuadoran Andes at an altitude of 2530 meters (about 1.5 miles above sea
level).
11

This was the goal in the cake baking mentioned in the previous footnote. The pressure had dropped due to the high altitude.
I was trying to figure out how the baking temperature should change in order to counterbalance this.
17.3 Implicit Differentiation 547
dV
dP
= KT (−1)P
−2
=
−KT
P
2
.
12
Another option is to differentiate implicitly.
d
dP

P
T
V(P)

=
d
dP
K,so
1
T

P
dV

dP
+ V(P)

=0
P
T
dV
dP
=
−V(P)
T
.
Solving for
dV
dP
gives
dV
dP
=
−V(P)
P
. This is equivalent to the previous answer, because
V(P)=
KT
P
.
(b) Here we are thinking of P as the independent variable and looking for
dT
dP
.NowV

is treated as a constant and T as a function of P . We can emphasize this by writing
PV
T(P)
=K.The simplest strategy here is to solve for T explicitly and then differentiate
with respect to P .
T(P)=
PV
K
=
V
K
P , where V and K are both constant. Therefore,
dT
dP
=
V
K
.
(c) Here we are thinking of time t as the independent variable; T and P vary with time but
we consider both V and K as constants. We differentiate implicitly.
It may be helpful to write
P(t)· V
T(t)
=K
to emphasize that P and T vary with time.
We will differentiate the equation with respect to t. Our job will be easier if we
rewrite the equation as
V · P(t)=KT(t)
because we won’t need the Product and the Quotient Rules.
Differentiating with respect to t gives

V
dP
dt
= K
dT
dt
or
dP
dt
=
K
V
dT
dt
.
If we like, we can eliminate V entirely, because V = K
T
P
.
12
Some remarks about notation:
d
dP
is an operator; it is a symbol that represents the operation of differentiating with respect to P .
dV
dP
is an expression, in
the same way that P + V is an expression.
dV
dP

= lim
P →0
V
P
, where V = V(P +P ) − V(P).The notation
dV
dP
reminds us that we are differentiating V with
respect to P . Here it reminds us that we are looking at the change in volume induced by a change in pressure.
548 CHAPTER 17 Implicit Differentiation and its Applications
dP
dt
= K ·
P
KT
·
dT
dt
or
dP
dt
=
P
T
dT
dt
.
(d) Again we think of time t as the independent variable. Here P , V , and T vary with time.
It may be helpful to write
V(t)·P(t)= K · T(t).

We’ ll differentiate with respect to t,sowe’ll need to use the Product Rule on the left-
hand side.
V(t)·
dP
dt
+
dV
dt
· P(t)= K ·
dT
dt
We can eliminate K completely, replacing it by
PV
T
.
V(t)·
dP
dt
+
dV
dt
· P(t)=
PV
T
·
dT
dt
.
Observe that if we know P ,V , and T at a certain instant, then knowing two of the
rates of change at that moment allows us to determine the third.

REMARK Notice that the notation V

can be ambiguous. Is it
dV
dP
or
dV
dt
? Weuse it only when
it is clear from context what is meant. The notation V

can also be ambiguous. The second
derivative of V with respect to P ,
d
dP

dV
dP

can be written as
d
2
V
dP
2
. The second derivative
of V with respect to t,
d
dt


dV
dt

can be written as
d
2
V
dt
2
.
◆
PROBLEMS FOR SECTION 17.3
1. The equation x
2
+ y
2
= 169 describes a circle with radius 13 centered at the origin.
(a) Solve explicitly for y in terms of x.Isyafunction of x?
(b) Differentiate your expression(s) from part (a) to find
dy
dx
.
(c) Now use implicit differentiation on the original equation to find
dy
dx
.
(d) Which method of differentiation (that used in part(a), or that used in part (b)) was
easier? Why?
(e) What is the slope of the tangent line to the circle at the point (5, 12)?At(5, −12)?
2. Which is larger, the slope of the tangent line to x

2
+ y
2
= 25 at the point (4, −3) or
the slope of the tangent line to x
2
+ 4y
2
= 25 at the point (4, −3/2)? You can answer
this analytically (find the two slopes), or you can answer this by looking at the graphs
of the ellipse and the circle and sketching the tangent to each at the designated points.
(To sketch the ellipse, look at the x- and y-intercepts.)
3. Consider the equation x
2
y + xy
2
+ x = 1. Find
dy
dx
at all points where x = 1.
4. Find the slopes of the tangent lines to (x − 2)
2
+ (y − 3)
2
= 25 at the two points where
x = 6.
17.3 Implicit Differentiation 549
5. Find the equation of the line tangent to the curve x
3
+ y

3
− 3x
2
y
2
+ 1 = 0 at the point
(1, 1).
6. Find
dy
dx
for the curve x
3
+ 3y + y
2
= 6. What is the slope at (2, −1)? At what points
is the slope zero?
7. At what points (if any) is the tangent line to the curve 3x
2
+ 6xy + 8y
2
= 8 vertical?
8. (a) Graph the ellipse 4(x − 1)
2
+ 9(y − 3)
2
= 36. (Sketch 4x
2
+ 9y
2
= 36 by looking

for x- and y-intercepts and then shift the graph horizontally and vertically as
appropriate.)
(b) Find the slope of the line tangent to the ellipse at the point (1, 5) in two ways. First
solve for y explicitly (using the appropriate half of the ellipse) and find
dy
dx
. Then
do the problem by differentiating implicitly.
9. Consider the circle given by x
2
+ y
2
= 4.
(a) Show that
dy
dx
=−
x
y
.
(b) Show that
d
2
y
dx
2
=
d
dx
(−

x
y
) =−
4
y
3
. Explain your reasoning completely.
10. The equation 2(x
2
+ y
2
)
2
= 25(x
2
− y
2
) gives a curve that is known as a lemniscate.
Find the slope of the tangent line to the lemniscate at the point (−3, 1).
x
y
Lemniscate: 2(x
2
+ y
2
)
2
= 25(x
2
– y

2
)
11. Find an equation for the tangent line to y
2
= x
3
(3 − x) at the point (1, 2). What can
you say about the tangent line to this curve at the point (3, 0)?
x
y
y
2
= x
3
(3 – x)
12. Find
dy
dx
.
(a) 3x
2
+ 6y
2
+ 3xy = 10 (b) (x − 2)
3
· (y − 2)
3
= 1 (c) xy
2
+ 2y = x

2
y + 1
(d) (x
2
y
3
+ y)
2
= 3x (e) e
xy
= y
2
(f) x ln(xy
3
) = y
2
(g) ln(xy) = xy
2
550 CHAPTER 17 Implicit Differentiation and its Applications
13. (a) Sketch a graph of (x − 2)
2
− (y − 3)
2
= 25. (Sketch x
2
− y
2
= 25 by looking
for x- and y-intercepts and then shift your graph vertically and horizontally as
appropriate.)

(b) Find the equation of the tangent line to the graph at each of the following points.
i. (15, −9) ii. (−3, 3)
14. Find an equation for the tangent line to x
2/3
+ y
2/3
= 5 at the point (8, 1).
x
y
x + y = 5: astroid
5
3
2
/
5
3
2
/
2
3
/
2
3
/
15. Match the equations with the appropriate graphs. (These are graphs of conic sections.
To learn more about conic sections, refer to Appendix E.)
(a)
x
2
5

+
y
2
5
= 8 (b) 3x
2
− 3y
2
= 27 (c) 3x
2
+ y
2
= 12
(d) x
2
+ 3y
2
= 12 (e) −x
2
+ 3y
2
= 12
y
x
A
y
x
C
y
x

B
y
x
E
y
x
D
17.4 IMPLICIT DIFFERENTIATION IN CONTEXT:
RELATED RATES OF CHANGE
In Sections 17.2 and 17.3 we introduced logarithmic and implicit differentiation. The former
is particularly useful when we want to differentiate something of the form [f(x)]
g(x)
and
the latter when it is either unappealing or impossible to solve for the variable we wish
to differentiate. In fact, we can get a great deal more mileage from the two fundamental
principles we used.
Principle (i): If two expressions are equal, their derivatives are equal.
Principle (ii): Whether y is given implicitly or explicitly in terms of x, we can differ-
entiate an expression in y with respect to x by treating y the same way we would treat
g(x) and using the Chain Rule.