f" = [(-1)(x+1)^3 - 3(1-x)(x+1)^2]/(x+1)^6
= [-(x+1)+3(x-1)]/(x+1)^4 ( I don't understand how you got this!!)
Looks like division of top and bottom by (x+1)^2 and also -(1-x) = + (x-1)
I don't get how you get this part:
f' = (1-x)/(1+x)^3
f" = [(-1)(x+1)^3 - 3(1-x)(x+1)^2]/(x+1)^6
= [-(x+1)+3(x-1)]/(x+1)^4 ( I don't understand how you got this!!)
= (-x-1+3x-3)/(x+1)^4
= (2x-4)/(x+1)^4
3 answers
Thank you, Damon!
You are welcome.