What is the derivative of #(cos x) / (1 + tan x)#?

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Answer 1 · 2015-04-14T13:43:01

The answer is: #y'=(-sinxcosx-sin^2x-1)/(cosx(1+tanx)^2)#.

Remembering the division rule, that is:

#y=f(x)/g(x)rArry'=(f'(x)*g(x)-f(x)*g'(x))/[g(x)]^2#,

than:

#y'=(-sinx(1+tanx)-cosx*1/cos^2x)/(1+tanx)^2=#

#=(-sinx-sinx*sinx/cosx-1/cosx)/(1+tanx)^2=#

#=(-sinxcosx-sin^2x-1)/(cosx(1+tanx)^2)#.