Question

高中数学例题5求解~

Answer 1

解：
显然，y>0，且各项（x-n）>0 ，其中n=1.....100
对原式两边去自然对数，则：
lny = ln(x-1)+ln(x-2)+ln(x-3)+....+ln(x-100)
对上式求x的导数：
d(lny)/dx = 1/(x-1) + 1/(x-2) + .....+ 1/(x-100)
y'/y = 1/(x-1) + 1/(x-2) + .....+ 1/(x-100)
因此：
y' = y[1/(x-1) + 1/(x-2) + .....+ 1/(x-100)]
=[(x-1)..(x-100)]·[1/(x-1) + 1/(x-2) + .....+ 1/(x-100)]

Answer 2

已知lnx对x求导为1/x
lny=ln(x-1)+ln(x-2)+....+ln(x-100)

lny对x求导(lny)'先对中间变量y求导，y再对x求导
即为y'/y
ln(x-1)+ln(x-2)+....+ln(x-100)对x求导
和的导数等于导数的和
[ln(x-1)+ln(x-2)+....+ln(x-100)]'=[ln(x-1)]'+[ln(x-2)]'+...+[ln(x-100)]'
分别把x-1，x-2，...，x-100看成中间变量，
先对中间变量求导，中间变量再对x求导
[ln(x-1)]'+[ln(x-2)]'+...+[ln(x-100)]'=1/(x-1)+1/(x-2)+...+1/(x-100)
所以
y'/y=1/(x-1)+1/(x-2)+...+1/(x-100)

y'=y[1/(x-1)+1/(x-2)+...+1/(x-100)]

y'=(x-1)(x-2)....(x-100)[1/(x-1)+1/(x-2)+...+1/(x-100)]