Question

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

Answer 1

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