f(x)=x(x-1)(x-2)(x-3).(x-n),则f(x)的n+1阶求导

问题描述:

f(x)=x(x-1)(x-2)(x-3).(x-n),则f(x)的n+1阶求导

(n+1)! 通过观察可知道F(X)展开后最高次为N+1次,所有低于N+1次的求N+1次导都为0,X^n+1 求N+1次导为(N+1)!

f(x)为n+1阶多项式,所以n+1阶求导后只会剩下x的n+1次方的导数,为n+1的阶乘