f'(x) = 0

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

assume that:

f(x) = u/v such that:

u = x ==> u'=1

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

f'(x) = (u'v- uv')/v^2

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

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

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

f'(x) = 0

==> 1-x^2 = 0

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

==> x1= 1

==> x2= -1

First, we'll have to differentiate the given function.

We notice that the function is a ratio and we'll calculate it's derivative using quotient rule:

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

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

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

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

We have, at numerator, a difference of squares:

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

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

Because the denominator of f'(x) is always positive, for any value of x, only the numerator could be zero.

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

We'll set each factor as zero.

1-x=0, x=1

1+x=0, x=-1

**The solutions of f'(x) = 0 are : {-1 ; 1}.**

f(x) = x/(1+x^2) to find f'(0),

Solution:

We use {u(x)/(v(x) }= {u'(x) v(x)- u(x)v'(x)}{v(x) }^2

f '(x) = { x/(1+x^2)^2} .

= {(x)'*(1+x^2) - x(1+x^2)'}/(1+x^2)^2

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

f'(x) = (1-x^2)/(1+x^2)^2. Put x= 0.

f'(0) = (1-0)/(1+0^2)^2

f'(0) = 1/1 =1