How do you find the derivatives of #y=lnabs(cosx)#?

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Jan 16, 2017

Answer:

#-tan x, x ne # an odd multiple of #pi/2#.

Explanation:

To make y real, #x ne# an odd multiple of #pi/2#'

Applying function of function of function rule,

#y'=1/|cos x| |cos x|'#

#=1/|cos x|(1) (cos x)'=-sin x/|cos x|=-sinx/cosx=-tan x#,

when #cos x >0 to x in open Q_1 uuu Q_4# and

#=1/|cos x|(-1) (cos x)'=sin x/|cos x|=sinx/(-cos x)=tan x#,

when #cos x <0 to x in open Q_2 uuu Q_3#.

In brief,

#y'=-tan x, x ne# an odd multiple of #pi/2#.

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Jan 16, 2017

Answer:

#d/(dx) (ln abs(cosx)) = -tanx#

Explanation:

We note that the function:

#f(x)= ln abs (cosx)#

is defined in the intervals:

#x=(-pi/2+kpi, pi/2+kpi)# with #k in ZZ#

where #cosx !=0#

In the central interval #x in (-pi/2,pi/2)# we have #cosx >0#, so that:

#ln abs (cosx) = ln(cosx)#

#d/(dx) ln abs (cosx) = d/(dx) ln(cosx) = -sinx/cosx = -tanx#

In any other interval we have:

#x in (-pi/2+kpi, pi/2+kpi) => y = (x -kpi) in (-pi/2,pi/2)#

but:

#abs(cosx) = abs (cos(y+kpi)) = abs ((-1)^kcosy) = abs(cosy) #

so that #f(x)# is periodic of period #pi# and #f'(x)# must also be necessarily periodic of period #pi#

In conclusion:

#d/(dx) (ln abs(cosx)) = -tanx#

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