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How do you find the derivative of #f(x) = 4x + piTan(x) #?

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Dec 15, 2017

#4+π(1+tan^2x)#

Explanation:

#f'(x)=(4x+πtan(x))' = 4(x)'+π(tanx)' = 4+π(1+tan^2x)#

because

#(tanx)'=(sinx/cosx)'##=((sinx)'cosx-sinx(cosx)')/cos^2x =#

#(cos^2x+sin^2x)/cos^2x = 1/cos^2x = 1+tan^2x#

& #sin^2x+cos^2x=1 <=> sin^2x/cos^2x+1=1/cos^2x <=> 1 + tan^2x = 1/cos^2x#

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