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Answer 1 · 2017-12-04T18:21:40

#1-tanx #

Assumung #log# is in base #e#, #logbeta equiv lnbeta # for the basis of this quesion:

#log(e^x cosx )# = #log e^x + log(cosx)#

#=x+log(cosx)#
#d/dx ( x) = 1#
#d/dx ( log(cosx) ) #;

We know #d/dx ( log(gamma(x))) = (gamma'(x))/(gamma(x) #

So #d/dx ( log(cosx)) = 1-sinx / cosx = 1-tanx #

This answer differs for #log_10# but is simple for base #e#