Page 125 - Foundations Of Differential Calculus

108 6. On the Differentiation of Transcendental Functions
so that

yq dp q q−1
dy = ydq ln p + = p dq ln p + qp dp,
p
q
since y = p . Hence this differential consists of two members, the first
q
q
of which, p dq ln p, would arise if in the proposed quantity p the p were
constant and only the exponent q were variable; the other member would
q
arise if in the proposed quantity p the exponent q were constant and only
the quantity p were variable. This differential could have been found by the
general rule given above in paragraph 170.
q
190. The differential of this same expression p can also be found from
q
the nature of an exponential quantity as follows. Let y = p and let x be
q+dq
replaced by x + dx so that y + dy =(p + dp) . This expression, when it
is expressed in the usual way by a series, gives
y + dy = p q+dq +(q + dq) p q+dq−1 dp

(q + dq)(q + dq − 1) q+dq−2 2
+ p dp + ··· ,
1 · 2
so that

q
dy = p q+dq − p +(q + dq) p q+dq−1 dp.
The following terms, which involve higher powers of dp vanish in the pres-
q+dq−1
ence of (q + dq) p dp. But
q
p q+dq − p = p q p dq − 1
2 2
dq (ln p) q
q
= p 1+ dq ln p + + ··· − 1 = p dq ln p.
2
p
In the second term (q + dq) p q+dq q+dq−1 dp, if we write q instead of q + dq
we obtain qp q−1 dp so that the differential is as was found before: dy =
q
p dq ln p + qp q−1 dp.
191. This same differential can more easily be investigated from the nature
of exponential quantities in the following way. Since we have taken the
number e for the number whose hyperbolic logarithm is equal to 1, we
q
let p = e q ln p , because the logarithm of each is the same, q ln p. Hence
y = e q ln p . It follows, since now the quantity e that is raised to a power is
constant, we have

qdp
q ln p
dy = e dq ln p + ,
p