What's the derivative of $arctan(x)/(1+x^2)$ ?

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Answer 1 · 2017-10-06T00:09:04

$d/dx ( arctanx/(1+x^2) ) = ( 1- 2xarctanx )/(1+x^2)^2$

$d/dx (f/g) = (g * (df)/dx - f * (dg)/dx)/g^2$

we have:

$d/dx ( arctanx/(1+x^2) ) = ( (1+x^2) d/dx arctanx - arctanx d/dx (1+x^2))/(1+x^2)^2$

$d/dx ( arctanx/(1+x^2) ) = ( (1+x^2)(1/(1+x^2)) -2x arctanx )/(1+x^2)^2$

$d/dx ( arctanx/(1+x^2) ) = ( 1- 2xarctanx )/(1+x^2)^2$

What's the derivative of arctan(x)/(1+x^2)arctan(x)/(1+x^2)?