`y = arctan(x/2) - 1/(2(x^2+4))` Find the derivative of the function

The derivative of y with respect to is denoted as` y'` or `(dy)/(dx)` .

 For the given equation: `y = arctan(x/2) -1/(2(x^2+4))` ,

we may apply the basic property of derivative:

`d/(dx) (u-v) =d/(dx) (u) - d/(dx)(v)`

Then the derivative of y will be:

`y' = d/(dx)(arctan(x/2) -1/(2(x^2+4)))`

`y' =d/(dx)(arctan(x/2)) - d/(dx)( 1/(2(x^2+4)))`

To find the derivative of the first term: `d/(dx)(arctan(x/2))` , recall the basic derivative formula for inverse tangent as:

`d/(dx) (arctan(u)) = ((du)/(dx))/(1+u^2)`

With `u = x/2` and `du=(1/2) dx ` or `(du)/(dx) =1/2` , we will have:

`d/(dx)(arctan(x/2)) = (1/2) /(1+(x/2)^2)`

                           `= (1/2) /(1+(x^2/4))`

Express the bottom as one fraction:

`d/(dx)(arctan(x/2)) = (1/2) /((x^2+4)/4)`

Flip the bottom to proceed to multiplication:

`d/(dx)(arctan(x/2)) = 1/2*4/(x^2+4)`

                         ` = 4/(2(x^2+4))`

                         ` =2/(x^2+4)`

For the derivative of the second term:` d/(dx)(1/(2(x^2+4)))` , we can rewrite it using the basic property of derivative: `d/(dx) (c*f(x)) = c* d/(dx) f(x)` where c is constant.

`d/(dx)(1/(2(x^2+4))) = (1/2) d/(dx)(1/(x^2+4))`

Then apply the Quotient Rule for derivative: `d/(dx) (u/v)= (u' * v- v'*u)/v^2` on  `d/(dx)(1/(2(x^2+4)))` .

We let:

`u = 1` then   ` u' = 0`

`v = x^2+4` then` v'=2x`

`v^2= (x^2+4)^2`

Applying the Quotient rule, we get:

`d/(dx)(1/(2(x^2+4))), = (0*(x^2+4)-(1)(2x))/(x^2+4)^2`

                         `=(0-2x)/ (x^2+4)^2`

                        ` =(-2x)/ (x^2+4)^2`

Then `(1/2) * d/(dx)(1/(x^2+4)) =(1/2) * (-2x)/ (x^2+4)^2`

                                              `= -x/ (x^2+4)^2`

 For the complete problem: 

`y' =d/(dx)(arctan(x/2)) - d/(dx)( 1/(2(x^2+4)))`

`y' =2/(x^2+4) + x/ (x^2+4)^2`