求y=√([(x-1)(x-2)]/[(x-3)(x-4)])的导数吧,用对数求导法
两边求对数
lny=1/2[ln(x-1)+ln(x-2)-ln(x-3)-ln(x-4)]
两边对x求导,记住y不是自变量（看作中间变量）
y'/y=1/2[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)]
∴y'=√[(x-1)(x-2)]/[(x-3)(x-4)]*[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)]/2