Question

y= (x+1)(x+2)(x+3)…(x+n),求y'(0)注意,是y'(0)

Answer 1

你是不是少乘了一个x?
如果是的话:
y=
x
设g(x)=(x+1)(x+2)(x+3)…(x+n)
∴y=x*g(x)
对y求导
y'=(x)'*g(x)+
x*g'(x)=g(x)+
x*g'(x)
y'(0)=g(0)+0*g'(0)=g(0)=1*2*3…n=n!
那就是两边同时求对数:
lny=ln[(x+1)(x+2)(x+3)…(x+n)]
lny=ln(x+1)+ln(x+2)+…+ln(x+n)
再同时求导:(注意是对x求导)
[lny]'=[ln(x+1)]'+[ln(x+2)]'+…+[ln(x+n)]'
(1/y)*y'=1/(x+1)+1/(x+2)+…+1/(x+n)
y'=[1/(x+1)+1/(x+2)+…+1/(x+n)]*y
∵y=(x+1)(x+2)(x+3)…(x+n)
分别与中括号中的各项相乘得到
y'=(x+2)(x+3)…(x+n)+(x+1)(x+3)…(x+n)+…+(x+1)(x+2)…(x+n-1)
∴y'(0)=n!/1+n!/2+…+n!/n
=n!*(1+1/2+1/3+…+1/n)