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

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Answer 1 · 2018-06-19T15:25:59

The answer is #=(1-x^2)/(x^2+1)^2#

The derivative of a quotient is

#(u/v)'=(u'v-uv')/(v^2)#

Here, we have

#u=x#, #=>#, #u'=1#

#v=x^2+1#, #=>#, #v'=2x#

Therefore,

#(x/(x^2+1))=(1xx(x^2+1)-x xx2x)/(x^2+1)^2#

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

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

Answer 2 · 2018-06-19T15:28:17

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

After the Quotient rule we get

#f'(x)=(x^2+1-x*2x)/(x^2+1)^2#
which simplifies to

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