`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 from`d/(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`