What is the derivative of #ln(sqrt(sin(2x)))#?

1 Answer
Gió
Mar 19, 2016

I found: #f'(x)=cos(2x)/sin(2x)#

Explanation:

This is quite good!

We can recognize 4 functions nested one into the other!!!

We can use the Chain Rule deriving (from the outher to the inner):
#ln# then #sqrt()# then #sin# and finally #2x#:

we get:

#f´(x)=1/sqrt(sin(2x))1/(2sqrt(sin(2x)))cos(2x)*2=#

#=cos(2x)/sqrt(sin(2x))*1/sqrt(sin(2x))=cos(2x)/sin(2x)#

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