What is the second derivative of $f(x)=x/(x^2+1)$ ?

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Answer 1 · 2017-04-14T08:15:19

$f''(x)=(2x(x^2-3))/(x^2+1)^3$

$f(x)=x/(x^2+1)$

$f'(x)=(1 cdot (x^2+1)-x cdot 2x)/(x^2+1)^2 =(x^2+1-2x^2)/(x^2+1)^2 =(1-x^2)/(x^2+1)^2$

By Quotient Rule,

$f''(x)=(-2x cdot(x^2+1)^2-(1-x^2)cdot2(x^2+1)(2x))/(x^2+1)^4$

By factoring out $2x(x^2+1)$ from the numerator,

$=(2x(x^2+1)(-x^2-1-2+2x^2))/(x^2+1)^4$

By cancelling out $(x^2+1)$ ,

$=(2x(x^2-3))/(x^2+1)^3$

I hope that this was clear.

What is the second derivative of f(x)=x/(x^2+1) f(x)=x/(x^2+1) ?