How do you differentiate #y=ln(sin^(2) x)#?

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Answer 1 · 2018-04-09T00:36:31

#y'=2cotx#

#d/dxlnu=1/udu#

by applying this to the function:

#y'=1/sin^2x(d/dxsin^2x)#

#d/dxsin^2x=2sinxcosx# power rule

so,
#y'=1/sin^2x2sinxcosx#

by simplification

#y'=2cosx/sinx#=#2cotx#

You could also simplify the function to look like this:
#y=2lnsinx#
and it will be way easier that way.