Page 124 - Foundations Of Differential Calculus

6. On the Differentiation of Transcendental Functions 107
3 z
We have shown that an exponential quantity a can be expressed by a
series as follows:
2
3
z (ln a) 2 z (ln a) 3
1+ z ln a + + + ··· .
2 6
It follows that
2 2
dp (ln a)
dp
a =1 + dp ln a + + ···
2
and a dp − 1= dp ln a, since the following terms vanish in the presence of
dp ln a. It follows that
p
p
dy = d.a = a dp ln a.
p
Therefore, the differential of an exponential quantity a is the product of
the exponential quantity itself, the differential of the exponent p, and the
logarithm of the constant quantity a that is raised to the variable exponent.

188. If e is the number whose hyperbolic logarithm is equal to 1, so that
x
x
ln e = 1, then the differential of the quantity e is equal to e dx.If dx is
taken to be constant, then the differential of this differential is equal to
2
x
x
e dx , which is the second differential of e . In a similar way the third
3
nx
x
differential is equal to e dx . It follows that if y = e , then dy/dx = ne nx
2 2 2 nx
and d y/dx = n e . Furthermore,
4
3
d y = n e , d y = n e , ....
4 nx
3 nx
dx 3 dx 4
Hence it is clear that the first, second, and following differentials of e nx
form a geometric progression, and it follows that the differential of order
m nx
m
nx
m
m of y = e , namely, d y, equals n e dx . Therefore,
m
d y
ydx m
m
is the constant quantity n .
189. If the quantity that is raised to a power is itself a variable, its differ-
ential can be investigated in a similar way. Let p and q be any functions of
q
x, and we consider the exponential quantity y = p . We take the logarithm
so that ln y = q ln p. When we differentiate these we have
dy qdp
= dq ln p + ,
y p
3 Introduction, Book I, Chapter VII; see also note on page 1.