Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 58

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