Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 54

The derivative of y in terms of x is denoted by (dy)/(dx) or y’'
For the given problem: y = xarctan(2x) -1/4ln(1+4x^2) , we may apply the basic differentiation property:
d/(dx) (u-v) = d/(dx) (u) - d/(dx) (v)
Then the derivative of the function can be set-up as:
d/(dx)y =d/(dx)[ xarctan(2x) -1/4ln(1+4x^2)]
y ' = d/(dx) xarctan(2x) -d/(dx) 1/4ln(1+4x^2)

For the derivative of d/(dx)[ xarctan(2x) , we apply the Product Rule: d/(dx)(u*v) = u’*v =+u*v’ .
d/(dx)[ xarctan(2x)] = d/(dx)(x) *arctan(2x)+ x * d/(dx)arctan(2x) .
Let u=x then u' = 1
v=arctan(2x) then dv= 2/(4x^2+1)
Note: d/(dx)arctan(u)= (du)/(u^2+1)

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

For the derivative of d/(dx) 1/4ln(1+4x^2) , we apply the basic derivative property:
d/(dx) c*f(x) = c d/(dx) f(x) .
Then,
d/(dx) 1/4ln(1+4x^2)= 1/4 d/(dx) ln(1+4x^2)
Apply the basic derivative formula for natural logarithm function: d/(dx) ln(u)= (du)/u .
Let u =1+4x^2 then du = 8x
1/4d/(dx) ln(1+4x^2) = 1/4 *8x/(1+4x^2)
=(2x)/(1+4x^2)

Combining the results, we get:
y' = d/(dx)[ xarctan(2x)] -d/(dx)[ 1/4ln(1+4x^2)]
y ' = [arctan(2x) +(2x)/(4x^2+1)] - (2x)/(1+4x^2)
y ' = arctan(2x) +(2x)/(4x^2+1) - (2x)/(1+4x^2)
y ' = arctan(2x) +0
y'=arctan(2x)