Page 123 - Foundations Of Differential Calculus

106 6. On the Differentiation of Transcendental Functions
m 1 m
VI. If y = x ln x − x , and we differentiate each part, we see that
m
m
d.x ln x = mx m−1 dx ln x + x m−1 dx
1
and d. x m = x m−1 dx, so that
m
m−1
dy = mx dx ln x.
n
m
VII. If y = x (ln x) , then
m−1 n m−1 n−1
dy = mx dx (ln x) + nx dx (ln x) .

VIII. If logarithms of logarithms occur, so that y =ln ln x,welet p =ln x.
dp dx
Then y =ln p and dy = ; but dp = , so that
p x
dx
dy = .
x ln x
IX. If y = lnlnln x and we let p =ln x, then y =ln ln p, and by the
preceding example
dp
dy = .
pdp
But dp = dx/x, so that by substitution we have

dx
dy = .
x ln x ln ln x

186. Now that we have discussed the differentiation of logarithms, we
move on to exponential quantities, that is, powers of the sort where the
exponent is a variable. The differentials of this kind of function of x can
be found by differentiating their logarithms, as follows. If we want the
x
x
differential of a ,welet y = a and take the logarithm of each: ln y =
x ln a. When we take the differentials we have dy/y = dx ln a, so that dy =
x
x
ydx ln a. Since y = a ,wehave dy = a dx ln a, which is the differential
x
p
of a . In a similar way, if p is any function of x, the differential of a is
p
a dp ln a.
187. This differential could also be found immediately from the nature of
p
exponential quantities discussed in Introduction. Let a be given where p
is any function of x. When we substitute x + dx for x we obtain p + dp.
p
Hence, if we let y = a and x becomes x+dx,wehave y +dy = a p+dp , and
so
p

dy = a p+dp − a = a p a dp − 1 .