﻿830 Dr. I. J. Schwatt on the 



By Leibnitz's theorem for the derivative of a product we 

 have 



l.(-"'(JHj,C)£?^£S-»- • m I 



In order to prove (7), it is only necessary to show that 

 it holds for u=xp, when it becomes 



U=o y =i V ; 0\y\ (n-fyl (n-y)\(p0-n + y)\(p—y+l)\ 



which is evidently independent of gc. But when x is a 

 constant, n is also a constant, and therefore (7) holds for all 

 values of u. So that: 



V ^fo V ^ W dx dx» a %^ H « J U dx'^ U 



and therefore 



^t+iy ^_»+i 1/z; 1 n a//A a ^+V-« <fy 



The following examples will illustrate the theorem. 

 To calculate 



d K x m 



(i.) 



^*(i— # n y 



By Leibnitz's theorem for the derivative of a product of 

 two functions, 



d K x™ « /«\ J*-« d a .. , n 



dx< a^y = *X*)dx^« xm d* a{1 - x) 



Hm\ x m ~ K , £ / m \x m ~ K+a d a p -i 



^O^^^l^-aj-^dx-^ 1 ^^ } 



Let x n = u, and (1— u)~ p =y, then 



A _ f. 1 %\ n v /A _ * w S -,, Kf+i)(p+2)..(y+/3-i) 



