Mark M. answered 10/06/23
Retired math prof. Very extensive Precalculus tutoring experience.
f(z) = ln l cos(5z) l
f'(z) = (cos(5z))' / cos(5z) = -5sin(5z) / cos(5z) = -5tan(5z)
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y = ln√tanx = ln(tanx)1/2 = (1/2)ln(tanx)
y' = (1/2)[ (tanx)' / tanx ] = (1/2)sec2x / tanx = sec2x / (2tanx)