An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

62

Successive differentiation of explicit functions.

Bk. I. ch VII.

If y be the product of two functions:
y= T(x). ~(x) = (p * then dy_ =, + *
d-, 2 =y,, d3 y _,, +
In general, if we denote -k by p(k) and the binominal coefficients
dx k
(~ 27) by nk, (n -= 1), we have the Rule:
dn
d-y - y(n) 4 + n -(n-1) 2(1) + yc,(-2) V4(2) +
dx
k=n
+- nk py(n)(-k) (.+... *p p(n) == ' (n-k) -p(k).
k=O
For, if we assume this formula proved for any value of n, differentiating,
dn+l kyn
d n~1 znk [%p(f+l-7c) p(c) + q(n-k)ip(k+1)]
k=O
k=n k-=n
= lk ( (n+rl-k)?(k) + nk q(n —k) (k+1)
k-=u k=0O
Writing apart the first term of the first sum and the last of the second:
jn+7-l k= —n k=n —1
dn == p(n+1) + (p (n+-)+Zi) f k ((n+-l-k) i(k)+ 7 nk(p(n-k)i (Ak1)
k —1 k —
we can evidently combine each pair of terms of the two sums so that
dnxl- - t(n+l)4/ + (n + n0) t(n) T )(1) + (n2 + l) (n-1)2) +
+ (nk + nk-1) 9)('+z- )(k) +.... p q(n+l).
But it is a property of the binominal coefficients that
4k + nk-1 = (n + 1)k,
thus this sum can be written according to the above notation:
dln+l ky=n
dx+~ - (n + ~-k)
k=o
which proves that if the assumed law holds for any n it remains valid
for the following number, and therefore for all that follow; but its
validity is directly seen for n =- 2 and n = 3.
According to this Rule we obtain the following exposition for:
dnr
4) y = tan x. If we put y.cos x == sin x, then, y(n) denoting dx-
y' cos x + y cos(x - + 4) = sin (x + — r)
y" os x - 2 y' os (x + -1 z) + y os (x -+ -) )== sin (x + — r)
y" cos x + - y +' os (x( ) + ) c + t 3 y'osx - CO (x + 3 )
= sin (x + i)........ o................................
s(n) cos x + n y (l) cos (x + - +ny-y ) cos (x Z). -
-+ 1, y(n) cos (x 4- k 7r) + * * y cos (x + -2 n ) = sin (x + -.1- n).
+ 91k 2 2 Y~~~~~~~~~~~av'/ r ""\