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

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

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