题目内容：

进行一次求导和二次求导 y=(x^2+2x)/(1-x) y=3x+In[(3x-4)/(x-1)]
y=(x^2+2x)/(1-x)
y=3x+In[(3x-4)/(x-1)]

优质解答

1、
y=(x^2+2x)/(1-x),
故
y'= [(x^2+2x)' *(1-x) - (x^2+2x)*(1-x)' ] / (1-x)^2,
显然(x^2+2x)'=2x+2,而(1-x)'= -1,
所以
y' = [(2x+2)(1-x)+(x^2+2x)] / (1-x)^2
= (-2x^2+2+x^2 +2x) / (1-x)^2
= (-x^2+2x+2) /(x-1)^2
= -1 + 3/(x-1)^2
而
y"= [3/(x-1)^2]'
= 3*(-2) /(x-1)^3
= -6/(x-1)^3
2、
y=3x+In[(3x-4)/(x-1)]
=3x + ln|3x-4| -ln|x-1|
故
y' = 3 + 3/(3x-4) - 1/(x-1)
而
y" = -9/(3x-4)^2 + 1/(x-1)^2