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

1 Answer

Apr 1, 2016

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

Explanation:

By the chain rule:

#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)#

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