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.

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