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

Answer 1

William Abdalla

#y'=2cotx#

#d/dxlnu=1/udu#

Differentiating Logarithmic Functions with Base e

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.

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Answer 2

William Abdalla

The derivative of ( y = \ln(\sin^2(x)) ) with respect to ( x ) is ( \frac{2\sin(x)\cos(x)}{\sin^2(x)} ) or ( \frac{2\cos(x)}{\sin(x)} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.