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

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Answer 1 · 2015-06-06T21:24:45

You can do this two ways.

a)
Quotient Rule:
#d/(dx)[(f(x))/(g(x))] = (g(x)(df(x))/(dx) - f(x)(dg(x))/(dx))/((g(x))^2#

#d/(dx)[x/(1+x^2)] = [(1+x^2)(1) - (x)(2x)]/(1+x^2)^2#

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

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

#d/(dx)[f(x)*g(x)] = f(x)(dg(x))/(dx) + g(x)(df(x))/(dx)#

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

#= [x][-(1+x^2)^(-2)*(2x)] + [1/(1+x^2)][1]#

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

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

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