What is the derivative of $arctan (8/x^2)$ ?

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Answer 1 · 2016-04-01T11:51:03

$(-16x)/(x^4+64)$

$y=arctanu, u = 8/x^2$

$dy/(du) = 1/(1+u^2), (du)/dx = -16/x^3$

$dy/dx=dy/(du)*(du)/dx$

$= 1/(1+(8/x^2)^2)*-16/x^3$

$= -16/(x^3+64/x) = (-16x)/(x^4+64)$

What is the derivative of arctan (8/x^2)arctan (8/x^2)?