Question

How do you find the derivative of `y= x arctan 2x - (1/4) ln (1+4x^2)`?

Answer 1

The answer is: `y'==arctan2x`.

Using the product rule and the chain rule:

`y'=1*arctan2x+x*1/(1+(2x)^2)*2-1/4*1/(1+4x^2)*8x=`

`=arctan2x+(2x)/(1+4x^2)-(2x)/(1+4x^2)=arctan2x`.

I suppose that this is a joke. Someone calculate the integral of `y=arctan2x`, and than asked its derivative!