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.