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How do I find the derivative of this
g(x)= arctanx + arccotx
asked Jan 11, 2013 in CALCULUS by rockstar Apprentice
reshown Jan 11, 2013 by bradely

1 Answer

+3 votes
Given g(x)= arctanx + arccotx

Derivative  each side with respect to " x"

 d / dx ( g(x) ) = d / dx (arctanx) +d / dx( arccotx)

 g'(x) = 1 / 1+x^2  + (-1) / 1+x^2

g'(x) = 1 / 1+x^2  -1 / 1+x^2

g'(x) =  0

derivative of  g(x)= arctanx + arccotx  = 0
answered Jan 11, 2013 by ricky Pupil

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