Determine the first derivative of the function below:
y=ln sin2x
g=sin2x ⇒ y=ln g
u=sin x ; u'=cos x⇒ g=u2 ⇒ g'=2·u·u'=2·sin x·cos x
y'=1g·g'
y'=1sin2x·2·sinx·cosx
y'=2cosxsinx
y'=2ctgx
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