y = arctanx + x/(1+x^2) Find the derivative of the function

Recall that the derivative of y with respect  to is denoted as y' or (dy)/(dx) .
 For the given equation: y = arctan(x) +x/(1+x^2) ,
we may apply the basic property of derivative:
d/(dx) (u+v) =d/(dx) (u) + d/(dx)(v)
 where we take the derivative of each term separately.
Then the derivative of y will be:
y' = d/(dx)(arctan(x) +x/(1+x^2))
y' =d/(dx)(arctan(x)) +d/(dx)(x/(1+x^2))
To find the derivative of the first term:d/(dx)(arctan(x)) , recall the basic derivative formula for inverse tangent as:
d/(dx) (arctan(u)) = ((du)/(dx))/1+u^2
With u = x and du=dx or (du)/(dx) =1 , we will have:
d/(dx)(arctan(x)) =1 /(1+x^2)
 
For the derivative of the second term:d/(dx)(x/(1+x^2)) , we apply the
Quotient Rule for derivative: d/(dx) (u/v)= (u' * v- v'*u)/v^2 .
Based fromd/(dx)(x/(1+x^2)) , we let:
u = x then u' = 1
v = 1+x^2 then v'=2x
v^2= (1+x^2)^2
Applying the Quotient rule,we get:
d/(dx)(x/(1+x^2)) = (1*(1+x^2)-(x)(2x))/(1+x^2)^2
d/(dx)(x/(1+x^2)) =(1+x^2-2x^2)/(1+x^2)^2
Combining like terms at the top:
d/(dx)(x/(1+x^2))= (1-x^2)/(1+x^2)^2
For the complete problem: 
y' =d/(dx)(arctan(x)) +d/(dx)(x/(1+x^2))
y' =1/(1+x^2) +(1-x^2)/(1+x^2)^2