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What is the derivative of #(x/(x^2+1))#?

1 Answer

Jun 4, 2015

Using quotient rule, which states that for #y=f(x)/g(x)#,

#y'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#,

We can proceed to derivate this with no big problems.

#(dy)/(dx)=(1*(x^2+1)-x*2x)/(x^2+1)^2#

#(dy)/(dx)=(x^2+1-2x^2)/(x^2+1)^2#

#color(green)((dy)/(dx)=(-x^2+1)/(x^2+1)^2)#

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