4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
4.6
383
Chain Rule II: Implicit Differentiation
In this section we will apply the chain rule in a setting of so-called implicit functions, which are
more general than the explicit functions we have dealt with so far. In later sections we will use
the chain rule in still a third context, related rates.
In this section we first consider a brief summary and re-examination of all differentiation rules,
including the chain rule. We then perform some simple differentiation problems which are of the
type typically encountered in a first study of implicit functions. We then proceed to the topic of
this section, which is finding the slope, dy/dx, for curves which are not necessarily functions in
the purest sense, but which are functions locally, and anyway for which it is reasonable to ask
about slope.
4.6.1
Review of Differentiation Rules
At this point we have many differentiation rules. If we look at all the rules so far, they can be
summed up as follows. First we had the very general rules, regardless of f (x) and g(x), and
fixed constants C ∈ R as long as the expressions on the right existed:
df (x)
d [Cf (x)]
=C·
dx
dx
d[f (x) + g(x)]
df (x) dg(x)
ˆ
=
+
dx
dx
dx
dg(x)
df (x)
d[f (x) · g(x)]
ˆ
= f (x) ·
+ g(x) ·
dx
dx
dx
ˆ
ˆ
g(x) ·
d f (x)
=
dx g(x)
ˆ
dC
=0
dx
df (x)
dx
− f (x) ·
[g(x)]2
dg(x)
dx
Then we had rules for specific functions, and their chain rule versions:50
ˆ
d
dx
[xn ] = n · xn−1
ˆ
d
dx
[un ] = n · un−1 ·
ˆ
d
dx
sin x = cos x
ˆ
d
dx
sin u = cos u ·
ˆ
d
dx
cos x = − sin x
ˆ
d
dx
cos u = − sin u ·
ˆ
d
dx
tan x = sec2 x
ˆ
d
dx
tan u = sec2 u ·
ˆ
d
dx
cot x = − csc2 x
ˆ
d
dx
cot u = − csc2 u ·
ˆ
d
dx
sec x = sec x tan x
ˆ
d
dx
sec u = sec u tan u ·
ˆ
d
dx
ˆ
d
dx
ˆ
d
dx
ˆ
d
dx
csc x = − csc x cot x
√
d
ˆ dx
x = 2√1 x
1
x
=
−1
x2
du
dx
du
dx
du
dx
du
dx
du
dx
du
dx
csc u = − csc u cot u ·
√
d
ˆ dx
u = 2√1 u · du
dx
1
u
=
−1
u2
·
du
dx
du
dx
(Note that the bottom four rules were just special cases of the power rules.)
While we will not venture here to reinvent the discussion of the previous section, we will at
least notice that when the variable of differentiation matches the input variable of the function,
50 The formulas in the right column assume that u is a function of x, but in fact if that is not the case—that
is if u is not a function of x—then both sides of those equations do not make sense. It is therefore customary to
proceed to write these formulas because a problem in the left-hand side of one of these equations will manifest in
the right-hand side of that equation as well.
384
CHAPTER 4. THE DERIVATIVE
we can use the simple derivative rules for that particular function, and when they do not, we
multiply by the derivative of that variable with respect to the variable of differentiation:
df (x)
= f ′ (x),
dx
du
df (u)
= f ′ (u) ·
.
dx
dx
(4.53)
(4.54)
For one example of (4.54), justified with Leibniz notation, we had f (x) = sin x and u = u(x)
yielding the following:
d sin u
d sin u du
du
=
·
= cos u ·
.
dx
du
dx
dx
The decomposition in the middle step allowed us to use the sine derivative formula because the
d
that appeared in the first term of
variable in the sine—u—matched the differential operator du
the Leibniz-style decomposition. Writing that step will become more burdensome if the input
of sine is more complicated, so we will skip that step in most future computations.
For a few slightly more complicated examples, consider the following, noting that the previous
rules still apply if we assume that y is a function of x (else Footnote 50, page 383 applies):
Example 4.6.1 Consider the following three derivative computations.
d 2
(y ) = 2y ·
dx
d √
1
ˆ
y= √
dx
2 y
ˆ
dy
.
dx
dy
.
·
dx
d(sin y)
d(y)
dy
dy
d
(y sin y) = y ·
+ sin y ·
= y · cos y ·
+ sin y ·
.
dx
dx
dx
dx
dx
Notice that the last example required first the product rule, and so the first step was to write—
exactly—the statement of the product rule, where the product is y · sin y. This is ultimately a
product of two functions of x, since we are assuming y is a function of x, and thus y sin y is also
ultimately a function of x. As with earlier derivatives, it is crucial to write out precisely what
the general rules dictate for the given function. This becomes even more critical in the following.
ˆ
Example 4.6.2 Consider the following derivative computations:
dy 2
dx
dy
dy
d
(xy 2 ) = x
+ y2
= x · 2y
+ y 2 (1) = 2xy
+ y2,
dx
dx
dx
dx
dx
dy
d
d 2
(x + y 2 )2 = 2(x2 + y 2 )1 (x2 + y 2 ) = 2 x2 + y 2 2x + 2y ·
ˆ
dx
dx
dx
dy
2
2
x+y
=4 x +y
.
dx
ˆ
Notice that we skipped an explicit writing of the “sum rule” step in the second computation
above:
d
d
dy
d
x2 + y 2 =
x2 +
y 2 = 2x + 2y ·
.
dx
dx
dx
dx
d
(x2 ) was a simple power rule because the variables matched.51 For the second
The first term, dx
dy 52
d
(y 2 ) = 2y dx
. Notice also
term they do not match, so we need the chain rule to compute dx
51 We
d(x2 )
can write dx = 2x · dx
but then, of course, “inner function’s” derivative is just
dx
usually does not continue to write, for example,
d ` 2´
d y 2 dy
dy
y =
·
= 2y ·
,
dx
dy dx
dx
52 One
dx
dx
= 1.
4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
385
that in the first computation we used the fact that dx
dx = 1, which is reasonable on its face, but
technically describes the fact that the function f (x) = x is a line with slope 1, i.e., f ′ (x) = 1,
regardless of x.53
Example 4.6.3 Consider the following derivative computation:
d
d(xy)
sin xy = cos xy ·
dx
dx
dx
dy
+y
= cos xy · x
dx
dx
dy
= cos xy · x
+y
dx
dy
= x cos xy
+ y cos xy.
dx
The above example used the chain rule first, and then the product rule “called” by the chain
rule. Note that the first step could have been written
d sin xy d(xy)
d sin xy
=
·
.
dx
d(xy)
dx
(4.55)
Of course it is important to note that, in the Leibniz notation, d(xy) is taken to be one, encapsulated quantity. It is not a multiplication of three quantities, so for instance d(xy)/dx 6= y.
Indeed, d(xy)/dx = x · (dy/dx) + y · (dx/dx) = x · (dy/dx) + y by the product rule. However
the chain rule does allow for the apparent division and multiplication by d(xy) in the derivative
notation in (4.55) above.
Two more points should be emphasized regarding the product rule computation above, which
we repeat here:
d(xy)
dy
dx
=x
+y
.
(4.56)
dx
dx
dx
First, note that (4.56) is exactly what the product rule says for this product xy. Thus it should
d
(y),
be established true from the explicit result of the product rule. Next, when computing dx
we get exactly what we would from the chain rule, or from the power rule. Thinking of y as a
function “raised to the power 1,” we might write:
d(y)
d(y) dy
dy
dy
=
·
=1·
=
.
dx
dy dx
dx
dx
(4.57)
Thus once again the derivative rules are self-consistent, as is Leibniz notation, properly interpreted.
4.6.2
Implicit Functions and Their Derivatives
In this subsection we discuss the context and concepts of implicit differentiation, following one
example through the discussion as an illustration. In subsequent subsections we will discuss
more numerous examples.
but skips the explicit decomposition step in the middle. We repeat it occasionally just as a reminder of the
computational reasonableness of the chain rule, best illustrated with Leibniz notation as above.
53 Of course dx = 1 also follows from the power rule, loosely interpreted: d (x1 ) = 1x0 , which seems to give 1,
dx
dx
though technically x0 = 1 only for x > 0 (for reasons we will see later in the text). Still, in this case the “formal
answer” x0 = 1, that is, 1, is in fact the correct answer for
d(x1 )
.
dx
386
CHAPTER 4. THE DERIVATIVE
5
4
3
2
1
−5 −4 −3 −2 −1
−1
1
2
3
4
5
−2
−3
−4
−5
Figure 4.20: The graph of x2 + y 2 = 25 gives y as a function of x locally except at (−5, 0)
and (5, 0). Some possible open rectangles centered at various (x0 , y0 ) on the graph are drawn,
within each of which (separately) y is a function of x.
In previous sections we were interested in functions y = f (x), i.e., where y was given explicitly
as a function of x. However, there are relationships and graphs of interest where y is related to
x implicitly, for instance, by an equation which contains both y and x as variables, except that y
is not solved for as an explicit function of x. Indeed, the graph of the equation (i.e., the graph of
all (x, y) which satisfy the equation) might not represent a function at all. However, it is quite
possible for y to be a function of x locally near a point (x0 , y0 ) on the graph, and to speak of
“slope” there, found through so-called implicit differentiation with the equation describing the
curve.
Definition 4.6.1 We will say y is locally (or implicitly) a function of x near (x0 , y0 ) if and
only if there exist δ, ε > 0 and an open rectangle
(x, y) (|x − x0 | < δ) ∧ (|y − y0 | < ε)
such that, within that open rectangle, the graph represents y as a function of x.
A simple example is a circle such as x2 + y 2 = 25. Except at (5, 0) and (−5, 0), we can find a
small enough, open rectangle around each point (x0 , y0 ) on the graph so that y is a function of
x inside that rectangle. Note that it is not necessary that there exists a vertical line touching
the graph inside the rectangle, but if one does touch the graph inside the rectangle it can only
do so at one point.
Now suppose we would like to find the slope of the graph of x2 + y 2 = 25, without solving
for y first.54 If we are at a point (x0 , y0 ) satisfying the equation x2 + y 2 = 25, and for which
we can (in principle) find an open rectangle centered at (x0 , y0 ) in which y = y(x), i.e., in which
y is a function of x, then inside such a rectangle, we have that y 2 = (y(x))2 is a (composite)
function of x, and x2 + y 2 is therefore also a function of x. Furthermore, 25 is a (rather trivial)
p
this one we can solve for y, almost: y = ± 25 − x2 , with the “plus” case being the upper semicircle and
the “minus” case being the lower semicircle. For many curves we cannot isolate y, or cannot do so easily. Even
when we can, it is sometimes easier to use the equation as it stands than in its “solved” form.
54 For
4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
387
function of x and, inside the rectangles, x2 + y 2 and 25 are in fact the same function of x:
=
25.
x2 + y 2
|{z}
| {z }
same function of x
function of x
in the rectangles
in each rectangle
Because these are the same functions of x in an open rectangle, they have the same derivatives
with respect to x there. Thus we can state
d
d
x2 + y 2 =
25 .
dx
dx
(4.58)
Now suppose that we restrict our analysis to just such a rectangle in which y is a function of x.
In the above we are taking the derivative with respect to x. The first term is a simple power
rule, but the second is a chain rule version of the power rule since the variables do not match,
while the other side of the equation is a constant. Thus, (4.58) becomes
2x + 2y
dy
= 0.
dx
The process above is called implicit differentiation, since we are differentiating both sides
of an equation which represents a curve in the xy-plane, on which y is locally a function of
x, given “implicitly” (not explicitly) by the equation. The differentiation process is simply a
manifestation of the chain rule, though the name “implicit differentiation” does give the process
dy
context.55 It is dx
that we want, i.e., the slope on the original curve. Fortunately we can now
solve for it:
dy
=0
dx
dy
= −2x
2y
dx
−2x
x
dy
=
=− .
dx
2y
y
2x + 2y
⇐⇒
⇐⇒
Some more notation is now needed for when we wish to evaluate this at particular points on the
dy
at such (x0 , y0 ) on the curve is written
curve. The value of dx
dy .
dx (x0 ,y0 )
dy
evaluated at the point (x0 , y0 ).” Looking back at the circle, we can
Out loud, this is said, “ dx
55 The term implicit differentiation unfortunately is sometimes used to refer to any computation of a derivative of
an expression with respect to one variable, where the expression contains other variables. For instance sometimes
the term is used when the variable of differentiation is “hidden” in the original equation, as is t in P V = C =⇒
d
d
(P V ) = dt
(C) =⇒ P dV
+ V dP
= 0 (assuming C is constant). In all cases it is implied that the other
dt
dt
dt
variables are functions of the variable of differentiation (or the differential operator with respect to that variable
cannot be meaningfully applied), though technically “implicit differentiation” should only refer to differentiation
on curves on which one variable is “implicitly” (i.e., not explicitly) given as a function (at least “locally”) of the
variable of differentiation.
In fact in all of these cases, we are simply applying the chain rule in different contexts.
388
CHAPTER 4. THE DERIVATIVE
find the slope at any (x0 , y0 ) which is on the graph:
x 3
dy =− =− ,
dx (3,4)
y (3,4)
4
dy x 0
=
−
= − = 0,
dx (0,5)
y (0,5)
5
4
x −4
dy = ,
=− =−
dx (−4,3)
y (−4,3)
3
3
etc.
A quick check of the graph in Figure 4.20, page 386 shows that these slope calculations
ring true.
dy In fact, it is interesting to notice what happens if we attempt to evaluate dx . If we naively
(5,0)
dy
insert (5, 0) into dx
= − yx , we would get − 50 , which is undefined. Thus the derivative there does
not exist, which is born out by the graph in the sense that there is no defined slope there. (Note
that in the Euclidean geometry sense the tangent to the circle is a vertical line there.) Similarly
for (−5, 0).
We might also be interested in equations
of tangent lines to this graph. For instance,
at (x, y) = (−4, 3), we have the slope
−x −(−4)
dy = 4/3,
=
=
dx (−4,3)
y (−4,3)
3
and so with the point (−4, 3) we have the line
shown at right, with equation
4
y = 3 + (x + 4).
3
5
4
3
2
1
−5 −4 −3 −2 −1
−1
1 2 3 4 5
−2
−3
−4
−5
Finding the slope at points on the circle above does not require the implicit differentiation
technique, that is finding dy/dx from an implicit curve where y is not an explicit function of x.
This is not an absolute requirement here because we can write y as an explicit function of x and
compute the derivative, √
near any (x0 , y0 ) on the curve except at (±5, 0). Indeed, on the upper
semi-circle we have y = 25 − x2 , and so
d 25 − x2
−x
d p
−2x
dy
1
2
= √
.
=
= √
25 − x = √
2
2
dx
dx
dx
2 25 − x
2 25 − x
25 − x2
Notice that (on the upper semicircle) this is the same as the derivative acquired through implicit
differentiation:
−x
−x
dy
=
= √
.
dx
y
25 − x2
A similar analysis shows that on the lower semicircle we also get two equivalent forms of the
derivative:
i
dy
d h p
−x
x
x
=
=
.
=
− 25 − x2 = · · · = √
2
dx
dx
−y
y
25 − x
4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
389
However it is arguably easier to compute the derivative with the implicit form of the curve as
we did originally.
Next we consider other examples where it is not clearly possible to solve for y as an explicit
function of x.
Example 4.6.4 Consider the curve (x2 + y 2 )3/2 = 2xy, which is graphed in Figure 4.21, page 390.
dy
by applying
(Much later in the text we will see how this graph came about.) First we find dx
d
the differential operator dx
to both sides:
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
Now we will
(x2 + y 2 )3/2 = 2xy
i
d
d h 2
(x + y 2 )3/2 =
[2xy]
dx
dx
3 2
d 2
dy
dx
2 1/2
2
(x + y ) ·
(x + y ) = 2 x
+y
2
dx
dx
dx
3 2
dy
dy
2 1/2
=2 x
(x + y ) · 2x + 2y
+y·1
2
dx
dx
dy
dy
3
3 2
= 2x
(x + y 2 )1/2 (2x) + (x2 + y 2 )1/2 2y
+ 2y
2
2
dx
dx
p
p
dy
dy
3x x2 + y 2 + 3y x2 + y 2 ·
= 2x
+ 2y
dx
dx
p
p
dy
dy
− 2x
= 2y − 3x x2 + y 2
3y x2 + y 2 ·
dx
dx
i dy
h p
p
= 2y − 3x x2 + y 2
3y x2 + y 2 − 2x
dx
p
dy
2y − 3x x2 + y 2
.
= p
dx
3y x2 + y 2 − 2x
√ compute the slope for two points on the curve, √12 , √12 and 43 , 43 , which
are labeled in the aforementioned figure. (The reader should verify that these points are actually
on the original curve.)
r 2
2
1
1
p
√
√
√1
√1
2
·
−
3
·
+
2
2
2
2
2
2
2y − 3x x + y dy r p
=
=
ˆ
“
”
2
2
2
2
dx “ √1 , √1 ”
3y x + y − 2x √1 , √1
2
2
√1
+ √12 − 2 · √12
3 · √12
2
2
2
√ √
√3 −
2
2 − √32 · 1
2
√ =− 3
√ = −1.
= 3
√ ·1−
√ −
2
2
2
2
This should seem reasonable
given the position of this point on the graph. For the com√ dy
3
3
putation of dx at 4 , 4 we will be more brief.
q
q
√
√
√
√
√
√
9
3
3
3
3
9
2 · 43 − 3 · 34 12
8
4 3−9 3
dy 16
2 − 4 ·
4
2√ − 4√· 2
· =
=
=
=
ˆ
√ q
√ q
dx “ 3 , √3 ”
9 − 12
3
3
3
3
3 · 43 · 23 − 23 8
−
2
·
−
3 · 43 12
3
·
4
4
16
4
4
4
2
√
−5 3
5√
=
=
3 ≈ 2.88675135.
−3
3
This should also seem reasonable from the graph.
390
CHAPTER 4. THE DERIVATIVE
1
−1
√1 , √1
2
2
√ 3
3
4, 4
1
−1
Figure 4.21: Graph of (x2 + y 2 )3/2 = 2xy. The slope dy/dx at
(x, y) on the
”
“ curve
“ each point
√ ”
is calculated in Example 4.6.4, page 390. The points (x, y) = √12 , √12 and (x, y) = 43 , 43
are plotted as well, for which slopes were computed in that example.
The argument in the proof of the power rule for rational powers of x, Theorem 4.4.2 on page
dy
without
359, used this implicit differentiation technique. As in that proof, we can compute dx
worrying about actually finding the open rectangles which give y locally as a function of x. The
rectangles are important in justifying the technique, but if something goes wrong, it will usually
show up in the final form of the computed derivative.
dy
Example 4.6.5 Find dx
for the graph 5x + x2 + y 2 + xy = tan y. This one will require two
2
chain rule calls, for the y and tan y terms, and a product rule for the xy term. As before, if we
are careful in writing out the product rule we are less likely to have errors.
d tan y
d 5x + x2 + y 2 + xy =
dx
dx
dy
dy
dy
dx
=⇒ 5 + 2x + 2y ·
= sec2 y ·
+ x
+y
dx
dx
dx
dx
dy
dy
dy
=⇒
5 + 2x + 2y ·
+x
+ y = sec2 y ·
dx
dx
dx
dy
dy
dy
− 2y
−x
=⇒
5 + 2x + y = sec2 y
dx
dx
dx
dy
=⇒
5 + 2x + y = sec2 y − 2y − x
dx
5 + 2x + y
dy
=⇒
=
.
sec2 y − 2y − x
dx
To find the tangent line through (0, 0), which is on the graph, we then compute
dy 5 + 2x + y 5+0+0
= 5,
=
=
dx (0,0)
sec2 y − 2y − x (0,0)
1−0−0
and so the tangent line through (0, 0) has equation y = 5x.
4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
391
Though we do not have the tools to prove it here, it is true that we will always be able to solve
dy
dy
for dx
in these problems. The basic idea of a proof is that dx
will always be a factor in the terms
in which it appears, and will only appear to the first degree, so we are basically solving a linear
dy
, albeit with nonconstant coefficients. So in fact it is no different
equation in the “variable” dx
fundamentally than solving Ax + By + C = Dx + Ey + F for y (or for x, for that matter). One
moves all terms containing that variable to one side, the other terms to the other side, factors
the variable from the side which then contains it, and divides by the other factor.
Of course our implicit differentiation technique begins with calculus steps and ends with
algebra steps. The entire process can be summarized by five steps:
(1) Apply
d
dx
to both sides of the equation describing the curve.
(2) complete all differentiation steps, flushing out all terms with a factor
(3) put all terms with
(4) factor the
dy
dx
dy
dx
dy
dx ,
factors on one side, other terms on the other side of the equation,
from the side which contains it, and finally,
(5) divide by the remaining factor, leaving
dy
dx
on one side by itself.
Example 4.6.6 Consider the equation x = sin y, pictured below and to the right.
The slope
dy
dx
x = sin y
is then computed as follows:
x = sin y
d
d
=⇒
[x] =
[sin y]
dx
dx
dy
=⇒
1 = cos y
dx
dy
1
=
.
=⇒
cos y
dx
So for instance the slope at 21 , π6 is given
√
dy
1
= 1/ cos π6 = √3/2
= 2/ 3, and the
by dx
equation of the tangent line there is
1
2
π
x−
.
y= +√
6
2
3
It is also worth noting what happens
5π
when cos y = 0, (y = ± π2 , ± 3π
2 ,± 2 ,···,
dy
, in light
x = ±1) and its implications for dx
of the graph, given on the right.
2π
π
−1
1
−π
−2π
The next example is quite long, but with moderate persistence is also quite do-able.
392
CHAPTER 4. THE DERIVATIVE
p
x2 + y 2 = cos 2x.
i d cos 2x
p
d h 3
y sec x2 + y 2 =
dx
dx
i
h p
3
p
d
d
y
d 2x
=⇒
y3
= − sin 2x ·
sec x2 + y 2 + sec x2 + y 2 ·
dx
dx
dx
p
p
p
d
3
x2 + y 2
=⇒
y sec x2 + y 2 tan x2 + y 2 ·
dx
p
dy
= − sin 2x · 2
+ sec x2 + y 2 · 3y 2 ·
dx
p
p
1
d 2
(x + y 2 )
=⇒ y 3 sec x2 + y 2 tan x2 + y 2 · p
·
2
2
dx
2 x +y
p
dy
+3y 2 sec x2 + y 2 ·
= −2 sin 2x
dx
p
p
y 3 sec x2 + y 2 tan x2 + y 2
dy
p
=⇒
· 2x + 2y
2
2
dx
2 x +y
p
dy
+3y 2 sec x2 + y 2 ·
= −2 sin 2x.
dx
Example 4.6.7 Find
dy
dx
on the graph of y 3 sec
dy
terms and get
Next we apply the distributive in the first term to flush out the dx
p
p
p
p
xy 3 sec x2 + y 2 tan x2 + y 2
y 4 sec x2 + y 2 tan x2 + y 2 dy
p
p
+
·
dx
x2 + y 2
x2 + y 2
p
dy
+ 3y 2 sec x2 + y 2 ·
= −2 sin 2x.
dx
dy
on one side (here, the left side), and the others on the
Now we put all terms with the factor dx
opposite side:
p
p
p
y 4 sec x2 + y 2 tan x2 + y 2 dy
dy
p
·
+ 3y 2 sec x2 + y 2 ·
2
2
dx
dx
x +y
p
p
xy 3 sec x2 + y 2 tan x2 + y 2
p
= −2 sin 2x −
x2 + y 2
Next we factor the
y 4 sec
dy
dx :
!
p
p
p
x2 + y 2 tan x2 + y 2 dy
dy
p
·
+ 3y 2 sec x2 + y 2
2
2
dx
dx
x +y
= −2 sin 2x −
Finally, we divide to solve for
dy
dx :
xy 3 sec
p
p
x2 + y 2 tan x2 + y 2
p
.
x2 + y 2
p
p
x2 + y 2 tan x2 + y 2
p
−2 sin 2x −
dy
x2 + y 2
p
p
.
= 4
p
dx
y sec x2 + y 2 tan x2 + y 2
2
2
2
p
+ 3y sec x + y
x2 + y 2
xy 3 sec
4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
393
p
If we would like, we can multiply the numerator and denominator by x2 + y 2 to get:
p
p
p
−2 x2 + y 2 sin 2x − xy 3 sec x2 + y 2 tan x2 + y 2
dy
p
p
p
p
.
=
dx
y 4 sec x2 + y 2 tan x2 + y 2 + 3y 2 x2 + y 2 sec x2 + y 2
Although this latest example may seem tedious, no particular step is conceptually difficult.
Success in such a project is nearly as much dependent upon our bookkeeping skills as upon
understanding of derivative rules.
The next example requires some interpretation of the derivative computation.
Example 4.6.8 Show that there are no horizontal tangent lines for the curve xy 5 + x5 y = 2.
d
Solution: As before, we apply dx
to both sides:
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
xy 5 + x5 y = 2
d 5
d
xy + x5 y =
(2)
dx
dx
dx
dy
dx5
dy 5
+ y5
+ x5
+y
=0
x
dx
dx
dx
dx
dy
dy
+ y 5 + x5
+ y · 5x4 = 0
x · 5y 4
dx
dx
dy
5xy 4 + x5
= −y 5 − 5x4 y
dx
y 5 + 5x4 y
dy
=−
dx
5xy 4 + x5
dy
y(y 4 + 5x4 )
=−
.
dx
x(5y 4 + x4 )
To have a horizontal tangent, we would need the numerator above to be zero, i.e., y(y 4 +5x4 ) = 0.
This would imply y = 0 or y 4 + 5x4 = 0.
1. Case y = 0: This never happens on the curve x5 y + xy 5 = 2, because it would imply 0 = 2,
a contradiction.
2. Case y 4 +5x4 = 0: This would require x, y = 0, since (y 4 ≥ 0)∧(5x4 ≥ 0) =⇒ y 4 +5x4 ≥ 0,
and we get equality only when both (nonnegative) terms in the sum are zero, thus requiring
x, y = 0, which is impossible again due to the fact that we require (x, y) to be on the curve
xy 5 + x5 y = 2.
The above example demonstrates the importance of applying the derived formula for dy/dx
to only points (x, y) which are on the original curve, though the formula otherwise “makes
sense” for nearly any pair (x, y). The actual curve is the setting in which the formula for dy/dx
is meaningful. The curve is given in Figure 4.22, page 394.
4.6.3
A Mistake to Avoid
Before finishing this section a remark is in order. It is tempting to try to simplify an algebraic
equation by taking derivatives of both sides, especially in the case of polynomial equations.
However, in such cases we are looking for points where one side’s expression equals the other,
which is very different from saying the two sides are the same functions of, say, x. Rephrased,
their outputs coinciding at a particular x-value is indeed different from saying the functions on
each side are the same functions (therefore having the same derivatives).
394
CHAPTER 4. THE DERIVATIVE
2
1
−2
−1
1
2
−1
−2
Figure 4.22: Graph of the curve xy 5 + x5 y = 2 from Example 4.6.8, page 393. The curve
dy
can have no horizontal tangents, though we can see that dx
→ 0 as x → ±∞. (We also see
˛ dy ˛
˛ ˛ → ∞ as x → 0. We did not prove this fact, though we can see that y must get large
dx
quickly as x → 0 if the equation xy 5 + x5 y = 2 is to hold, since x → 0 alone would shrink xy 5
and x5 y, and so y must grow to compensate if the sum is to be fixed at 2.)
16
12
f (x) = x2 − 2x + 1
8
4
−3
−2
−1
1
2
3
g(x) = 5 − 2x
Figure 4.23: The graphs of y = f (x) and y = g(x) intersect at x = ±2, which is the
solution to f (x) = g(x). However, it is clear from the picture that, while f (x) = g(x) at
x = ±2, the derivatives (slopes) f ′ (x) and g ′ (x) at those two points are not the same. In fact,
f ′ (x) = g ′ (x) at x = 0 only. A look at the graphs above indicates that the slopes of y = f (x)
and y = g(x) at x = 0 do appear to agree, but those at x = 2 and x = −2 do not. Thus
f (a) = g(a) =⇒
6
f ′ (a) = g ′ (a).
4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
395
Example 4.6.9 For a very simple case, consider the equation
x2 − 2x + 1 = 5 − 2x.
(4.59)
This succumbs easily to the earlier methods:
⇐⇒
⇐⇒
x2 − 2x + 1 = 5 − 2x
x2 = 4
x = −2, 2
so the solution is simply x = ±2. Now suppose instead we tried to take derivatives of both sides
of (4.59):
⇐⇒
⇐⇒
2x − 2 = −2
2x = 0
x = 0.
We see that we get the incorrect answer, if our goal was to solve the original equation x2 −2x+1 =
5 − 2x. If we let x = 2 this equation becomes 22 − 2 · 2 + 1 = 5 − 2 · 1, i.e., 1 = 3, which is clearly
false.
Thus if we are solving f (x) = g(x), it does not follow that f ′ (x) = g ′ (x). It is true if they
are the same functions, i.e., same heights everywhere (and we use this fact in our implicit differentiation process), then they have the same derivatives but when we solve algebraic equations
we are only interested in those points where the graphs of the two (usually different) functions
intersect. It is unlikely that they would share the same slopes there as well as the heights.
Hence it is important to use algebraic arguments where appropriate, and calculus arguments
where appropriate. See Figure 4.23, page 394.
4.6.4
An Application
Example 4.6.10 When two resistors with values R1 and R2 are connected in parallel, the
resulting resistance R is given by the equation
1
1
1
+
.
=
R
R1
R2
(4.60)
Suppose R2 = R1 + 4Ω.
1. Find an equation for R in terms of r = R1 .
2. Show that r2 = 2rR + 4R − 4r.
3. Find dR/dr in terms of r using the equation you found in part 1 directly.
4. Find dR/dr in terms of r by differentiating the equation in part 2. Use part 1 to show this
is the same as the answer in part 3.
Solution:
1. After making the substitution R2 = R1 + 4 we have
1
1
1
+
=
.
R
R1
R1 + 4
396
CHAPTER 4. THE DERIVATIVE
Using r = R1 this then becomes
1
1
1
= +
.
R
r
r+4
Taking reciprocals of both sides gives us
R=
=⇒ R =
1
1
1
r + r+4
2
·
r2 + 4r
r(r + 4)
=
r(r + 4)
(r + 4) + r
r + 4r
.
2r + 4
2. Multiplying the above equation by (2r + 4) gives us
2rR + 4R = r2 + 4r
2rR + 4R − 4r = r2
⇐⇒
⇐⇒ r2 = 2rR + 4R − 4r, q.e.d.
3. We find dR/dr using the quotient rule:
d
d
(2r + 4) dr
(r2 + 4r) − (r2 + 4r) dr
(2r + 4)
dR
=
dr
(2r + 4)2
(2r + 4)(2r + 4) − (r2 + 4r)(2)
=
[2(r + 2)]2
2
4r + 16r + 16 − 2r2 − 8r
=
4(r + 2)2
2
2r + 8r + 16
=
4(r + 2)2
2(r2 + 4r + 8)
=
4(r + 2)2
2
r + 4r + 8
dR
.
=
=⇒
dr
2(r + 2)2
4. Using part 2, we apply
d
dr
to both sides:
d 2
d
r =
[2rR + 4R − 4r]
dr
dr
d
dR
d
−4
=⇒
2r = 2r (R) + R (2r) + 4
dr
dr
dr
dR
dR
+ 2R + 4
−4
=⇒
2r = 2r
dr
dr
dR
=⇒ 2r + 4 − 2R = (2r + 4)
dr
2r + 4 − 2R
dR
=
.
=⇒
dr
2r + 4
Now, using R = (r2 + 4r)/(2r + 4) we get
2
+4r
2r + 4 − 2 · r2r+4
2r + 4
(2r + 4)2 − 2(r2 + 4r)
dR
,
=
·
=
dr
2r + 4
2r + 4
(2r + 4)2
4.6. CHAIN RULE II: IMPLICIT DIFFERENTIATION
397
which was one of our intermediate results in the computations in part 3 (specifically, the
second line there), and so continued simplification would give us the same as the final
simplification in part 3 as well, q.e.d.
Exercises
For Exercises 1–10 compute
plicit differentiation.
dy
dx
by im-
1. Ax + By = C, where A, B and C are
consants.
2. sin x + cos y = 12 .
(c) What can you say about the slope
dy
dx as y → 1?
(d) What can you say about the slope
dy
−
+
dx as x → 1 or x → −1 ?
12. Consider the curve xy = 1.
3. x2 − y 2 = 9. Find also the tangent line
at (5, −4).
(a) Use implicit differentiation to find
dy
dx .
4. x2 + 3x + y 2 = 16 + 7y.
(b) Use y =
5. x + tan x = y 3 + sec y.
(c) Show algebraically how these are
the same answers.
6. x2 + 2xy + y 2 = cos x.
(a) Define functions f (x) = x2 and
g(x) = 9 and graph them together
on one grid.
9. tan xy = 9.
10. sin(xy) + y = csc x − 11
11. Consider the curve graphed below
given by (x2 + y 2 )3 = (x2 − y 2 )2 .
dy
dx .
(b) Find the equation
of the tangent
√ !
1
3
.
,
line at
4 4
1
(b) What is its solution of the original
equation, i.e., of f (x) = g(x)?
(c) What is the graphical significance
of the solution of f (x) = g(x)?
(d) Now consider the equation
f ′ (x) = g ′ (x) for these two functions f and g.
What is the
solution of this new equation
f ′ (x) = g ′ (x)?
(e) Explain the significance of the sod
d
f (x) = dx
g(x) for this
lution of dx
particular example.
1
−1
dy
dx .
x2 = 9.
8. x sin y + y sin x = 2.
−1
to find
13. Consider the algebraic equation
7. sin x2 = cos y 3 .
(a) Find
1
x
(f) Explain why, if a particular x satisfies f (x) = g(x), we cannot expect that it also satisfies f ′ (x) =
g ′ (x).