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How do you find the derivatives of inverse function #f(x) = arctan(cos(3x))#?

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Jun 17, 2015

#d/(dx)[arctan(u(v(x)))] = 1/(1+u^2)*((du)/(dv))((dv)/(dx))#

#d/(dx)[cosu] = -sinu*((du)/(dx))#

With #u(v) = cosv# and #v(x) = 3x#:

#d/(dx)[arctan(cos3x)] = 1/(1+cos^2(3x))*-sin(3x)*3#

#= (-3sin3x)/(1+cos^2(3x))#

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