Page 95 - Foundations Of Differential Calculus

78 5. On the Differentiation of Algebraic Functions of One Variable
n−2
(n − 1) x dx. When this is multiplied by ndx, we have the second dif-
ferential
2
2
n
d .x = n (n − 1) x n−2 dx .
In a similar way, if the differential of x n−2 , which is equal to (n − 2) x n−3 dx,
2
is multiplied by n (n − 1) dx , we have the third differential
3 n n−3 3
d .x = n (n − 1) (n − 2) x dx .
Furthermore, the fourth differential will be
4 n n−4 4
d .x = n (n − 1) (n − 2) (n − 3) x dx ,
and the fifth differential is
5
5
n
d .x = n (n − 1) (n − 2) (n − 3) (n − 4) x n−5 dx .
The form of the following differentials is easily understood.
154. As long as n is a positive integer, eventually the higher differentials
will vanish; these are equal to 0, because differentials of higher powers of
dx vanish. We should note a few of the simpler cases:
2 3
d.x = dx, d .x =0, d .x =0, ...,
2 2 2 2 3 2 4 2
d.x =2x dx, d .x =2dx , d .x =0, d .x =0, ...,
3 2 2 3 2 3 3 3 4 3
d.x =3x dx, d .x =6xdx , d .x =6dx , d .x =0, ...,
2
4
3
4
4
4
4
3
4
2
3
2
d.x =4x dx, d .x =12x dx ,d .x =24xdx ,d .x =24dx ,
5 4
d .x =0 ...,
5 4 2 5 3 2 3 5 2 3
d.x =5x dx, d .x =20x dx ,d .x =60x dx ,
4 5 4
d .x = 120xdx ,
5 5 5 6 5
d .x = 120dx ,d .x =0, ....
It is clear that if n is a positive integer, then the differential of order n of
n
n
x will be a constant, that is, it will be equal to 1·2·3 ··· ndx . The result
is that all differentials of higher order will be equal to 0.
155. If n is a negative integer, differentials of x with such negative powers
can be taken, such as
1 1 1
, , , ...,
x x 2 x 3
since
1 −1 1 −2
= x , = x ,
x x 2