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

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Answer 1 · 2016-07-19T16:22:33

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

Two straightforward ways.

#color(blue)("Method One")#

Rewrite as #(1+x^2)^(-1)# and use the power and chain rules:

#h'(x) = -(1+x^2)^(-2)*2x = -(2x)/(1+x^2)^2#

#color(blue)("Method Two")#

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

#h'(x) = (0 - 2x)/(1+x^2)^2 = -(2x)/(1+x^2)^2#