differetiation

y=xex
dy/dx=xex+ex=(x+1)ex
dy/dx=(x+1)ex+ex=(x+2)ex
......
So, we can observe there exists a pattern that dny/dxn=(x+n)ex
∴d100y/dx100=(x+100)ex

[(d^100)(xe^x)]/(dx^100)

100C0* (d^100/dx^100) x * e^x + 100C1* (d^99/dx^99) x * (d/dx)e^x
+ 100C2 * (d^100/dx^100) x * (d^2/dx^2)e^x + ......
+ 100C99 * (d/dx) x * (d^99/dx^99)e^x + 100C100 * x * (d^100/dx^100)e^x

Since (d/dx) e^x = e^x , ( d^n/dx^n) e^x = e^x

(d/dx) x = 1
(d^2/dx^2) x = 0

So,

100C0* (d^100/dx^100) x * e^x + 100C1* (d^99/dx^99) x * (d/dx)e^x
+ 100C2 * (d^100/dx^100) x * (d^2/dx^2)e^x + ......
+ 100C99 * (d/dx) x * (d^99/dx^99)e^x + 100C100 * x * (d^100/dx^100)e^x

= 0+0+0+...+ 100C99 * (d/dx) x * (d^99/dx^99)e^x + 100C100 * x * (d^100/dx^100)e^x

= 100 e^x + x e^x
= e^x (100+x)

2008-07-02 19:33:51 補充：
My method is 參考ed from binomial theorem ...

(x+y)^n = x^n + nC1 x^n-1 y + ... + nCn-1 x y ^n-1 + y&n

d(xe^x)/dx = xe^x + e^x.
d2y/d2x = d(xe^x + e^x)/dx = d(xe^x)/dx + d(e^x)/dx = xe^x + e^x + e^x = xe^x + 2e^x.
When the process continue,
d^100y/dx^100 = xe^x + 100e^x.