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of x, its differential dy will have the form dy = pdx, where p can always
be found from the laws we have set down. Furthermore, the function p
is also an algebraic function of x, since in determining the differential no
other operations were used except the usual ones for algebraic functions.
For this reason if y is an algebraic function of x, then dy/dx is also an
algebraic function of x. Furthermore, if z is an algebraic function of x, such
that dz = qdx, since q is an algebraic function of x, we also know that
dz/dx is an algebraic function of x, and indeed so is dz/dy an algebraic
function of x which is equal to p/q. Hence, if the formula dz/dy is part of
some algebraic expression, this does not prevent the whole expression from
being algebraic, provided only that y and z are algebraic functions.
176. We can extend this line of reasoning to second- and higher-order
differentials. If y is an algebraic function of x, dy = pdx and dp = qdx,
then with dx remaining constant, we have d2y = qdx2, as we have already
seen. Since for the reasons already given q is also an algebraic function of
x, it follows that d2y/dx2 is not only a finite quantity but also an algebraic
function of x, provided only that y is such a function. In a similar way we
see that
d3y d4y
, , . . .
dx3 dx4
5. On the Differentiation of Algebraic Functions of One Variable 97
are algebraic functions of x, provided that y is such a function. Furthermore,
if z is also an algebraic function of x, all finite expressions made up of
differentials of any order of y, z, and dx, such as
d2y d3y dx d4y
, , ,
d2z dz d2y dy3d2z
are all likewise algebraic functions of x.
177. Since the first differential of any algebraic function of x can now
be found by the method given, using the same method we can investigate
the second- and higher-order differentials. If y is any algebraic function of
x, from differentiation we have dy = pdx, and we note the value of p. If
we differentiate again and obtain dp = qdx, then d2y = qdx2, supposing
that dx is constant. In this way we have defined the second differential.
When we differentiate q, so that dq = rdx, we have the third differential
d3y = rdx3. In this way we investigate the differentials of higher order,
and since the quantities p, q, r, . . . are all algebraic functions of x, the
given laws for differentiation are sufficient. Therefore, we have continuous
differentiation. If we omit the dx in the differentiation of y, we obtain the
value dy/dx = p, which is again differentiated and divided by dx to obtain
q = d2y/dx2. Each time we divide by dx, since everywhere the differential
dx is omitted. In a similar way we obtain r = d3y/dx3, and so forth.
a2
I. Let y = ; find the first- and higher-order differentials.
a2 + x2
First we differentiate and divide by dx to obtain
dy -2a2x
=
dx
(a2 + x2)2
and then
d2y -2a4 +6a2x2
= ,
dx2 (a2 + x2)3
d3y 24a4x - 24a2x3
= ,
dx3 (a2 + x2)4
d4y 24a6 - 240a4x2 + 120a2x4
= ,
dx4
(a2 + x2)5
d5y -720a6x + 2400a4x3 - 720a2x5
= ,
dx5
(a2 + x2)6
and so forth.
98 5. On the Differentiation of Algebraic Functions of One Variable
1
II. Let y = " ; find the first- and higher-order differentials.
1 - x2
dy x
= ,
dx
(1 - x2)3/2
d2y 1+2x2
= ,
dx2 (1 - x2)5/2
d3y 9x +6x3
= ,
dx3 (1 - x2)7/2
d4 9+72x2 +24x4
= ,
dx4 (1 - x2)9/2
d5y 225x + 600x3 + 120x5
= ,
dx5
(1 - x2)11/2
d6y 225 + 4050x2 + 5400x4 + 720x6
= ,
dx6
(1 - x2)13/2
and so forth. These differentials can easily be continued, but the law
by which the terms proceed may not be immediately obvious. The
coefficient of the highest power of x is the product of the natural num-
bers from 1 to the order of the differential. Meanwhile, if we to
"wish
continue further our investigation, we will find that if y =1/ 1 - x2,
generally we have
dny 1 2 3 n
=
1
dxn (1 - x2)n+ 2
1 n (n - 1)
xn + xn-2
2 1 2
1 3 n (n - 1) (n - 2) (n - 3)
+ xn-4
2 4 1 2 3 4
1 3 5 n (n - 1) (n - 5)
+ xn-6
2 4 6 1 2 6
1 3 5 7 n (n - 1) (n - 7)
+ xn-8 + .
2 4 6 8 1 2 8
Examples of this kind are useful not only for acquiring a habit of differ-
entiating, but they also provide rules that are observed in differentials of
all orders, which are very much worth noticing and can lead to further
discoveries.
6
On the Differentiation of
Transcendental Functions
178. Besides the infinite class of transcendental, or nonalgebraic, quanti-
ties that integral calculus supplies in abundance, in Introduction to Analysis
of the Infinite we were able to gain some knowledge of more usual quan-
tities of this kind, namely, logarithms and circular arcs. In that work we
explained the nature of these quantities so clearly that they could be used
in calculation with almost the same facility as algebraic quantities. In this
chapter we will investigate the differentials of these quantities in order that
their character and properties can be even more clearly understood. With
this understanding, a portal will be opened up into integral calculus, which
is the principal source of these transcendental quantities.
179. We begin with logarithmic quantities, that is, functions of x that,
besides algebraic expressions, also involve logarithms of x or any functions
of logarithms of x. Since algebraic quantities no longer are a problem, the
whole difficulty in finding differentials of these quantities lies in discovering
the differential of any logarithm itself. There are many kinds of logarithms,
which differ from each other only by a constant multiple. Here we will
consider in particular the hyperbolic, or natural, logarithm, since the others
can easily be found from this one. If the natural logarithm of the function
p is signified by ln p, then the logarithm with a different base of the same
function p will be m ln p where m is a number that relates logarithms with
this base to the hyperbolic logarithms. For this reason ln p will always
indicate the hyperbolic logarithm of p.
100 6. On the Differentiation of Transcendental Functions
180. We are investigating the differential of the hyperbolic logarithm of x
and we let y =ln x, so that we have to define the value of dy. We substitute
x + dx for x so that y is transformed into yI = y + dy. From this we have
dx
y + dy =ln (x + dx) , dy =ln (x + dx) - ln x =ln 1+ .
x
But we have seen before1 that the hyperbolic logarithm of this kind of
expression 1 + z can be expressed in an infinite series as follows:
z2 z3 z4
ln (1 + z) =z - + - + .
2 3 4
When we substitute dx/x for z we obtain
dx dx2 dx3
dy = - + - .
x 2x2 3x3
Since all of the terms of this series vanish in the presence of the first term,
we have
dx
d ln x = dy = .
x
It follows that the differential of any logarithm whatsoever that has the
ratio to the hyperbolic logarithm of n : 1, has the form n dx/x.
181. Therefore, if ln p for any function p of x is given, by the same argu-
ment, we see that its differential will be dp/p. Hence, in order to find the
differential of any logarithm we have the following rule:
For any quantity p whose logarithm is proposed, we take the differential
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