How do you find the derivative of $y = x arctan 2 x - (\frac{1}{4}) ln (1 + 4 x^{2})$ $y= x arctan 2x - (1/4) ln (1+4x^2)$ ?

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Answer 1 · 2015-03-01T15:57:18

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

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

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

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

How do you find the derivative of y= x arctan 2x - (1/4) ln (1+4x^2)y=xarctan2x−(14)ln(1+4x2)y= x arctan 2x - (1/4) ln (1+4x^2)?