Solve for x, f'(x)=0 if f(x)=x/(x^2+1).

The given function is a ratio and we'll determine it's
derivative using the quotient rule:

(u/v)'=
(u'*v-u*v')/v^2

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

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

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

We've
noticed that, at numerator, we have a difference of
squares:

a^2-b^2=(a-b)(a+b)

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

So,
the solution of the equation f'(x)=0
are

(1-x)(1+x)=0

We'll set
each factor as zero:

1-x=0,
x=1

1+x=0, x=-1