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

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) =...

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)`