Question

Is the second derivative of
\[y = \frac{ 2x }{ 1+x ^{2} }\]\[y'' = \frac{ 4x(x ^{2}-3) }{ (1-x ^{2})^{3} }\]or did I mess up somewhere?

Answer 1

let's see
\[\large y'=\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}=
\frac{2+2x^2-4x^2}{(1+x^2)^2}=\frac{2-2x^2}{(1+x^2)^2} \]
right?

Answer 2

and the second derivative
\[\large y''=\frac{-4x(1+x^2)^2-(2-2x^2)2(1+x^2)(2x)}{(1+x^2)^4} \]
\[\large =\frac{(-4x)(1+x^2)[(1+x^2)+(2-2x^2)]}{(1+x^2)^4} \]
\[\large =\frac{(-4x)(3-x^2)}{(1+x^2)^3}=\frac{4x(x^2-3)}{(1+x^2)^3} \]

Answer 3

okay, thank you :)
so in conclusion, my answer was correct. :P

Answer 4

thank you everyone for your time! <3