Answer
$y'=\dfrac{1-x^2}{x+x^3}$
Work Step by Step
In order to derivate this function you have to apply the chain rule
Let's make a «u» substitution to make it easier
$f(u) = $
$u = \dfrac{x}{1+x^2}$
Derivate the function:
$f'(u) = \dfrac{u'}{u}$
Now let's find u'
Note: Here you have to apply quotient rule
$u' = \dfrac{1-x^2}{(1+x^2)^2}$
Then undo the substitution, simplify and get the answer:
$f'(u) = \dfrac{1+x^2}{x} \times \dfrac{1-x^2}{(1+x^2)^2}$
$f'(u) = \dfrac{1-x^2}{x(1+x^2)}$
$f'(u)=\dfrac{1-x^2}{x+x^3}$
