Question

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

Answer 1

`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`

Answer 2

`-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`.