How do you find the derivative of #(x/(x^2+1))#?

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Answer 1 · 2016-09-10T03:08:06

#-((x^2-1))/(x^2+1)^2#

The quotient rule states that:

#d/dx(f(x)/g(x)) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2#

In this example; #f(x) = x# and #g(x) = (x^2+1)#

Hence, using the quotient rule:

#d/dx(x/(x^2+1)) = (1* (x^2+1) - x* (2x+0))/(x^2+1)^2#

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

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