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What is the derivative of #y = ln(cscx)#?

Answer 1

Charles Anctil

Alternatively, using logarithm laws, we can say:

#y = ln(1/sinx) = ln(1) - ln(sinx) = 0 - ln(sinx) = -ln(sinx)#

It follows by the chain rule that

#y' = cosx * -1/sinx = -cosx/sinx = -cotx#

Hopefully this helps!

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Answer 2

Charles Arocha

We will use the chain rule to differentiate. Let #y = lnu# and #u = cscx#

The derivative of #cscx#, by the quotient rule, is #(cscx)' = (-cosx)/sin^2x = -csc^2xcosx = -cotxcscx#

The derivative of #lnu# is #1/u#.

#dy/dx= 1/u xx -cotxcscx = 1/cscx xx -cotxcscx = sinx xx (-cotx xx 1/sinx)=- cotx#

Hopefully this helps!

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Answer 3

Luke Alvarez

The derivative of ( y = \ln(\csc x) ) is:

[ \frac{d}{dx} \ln(\csc x) = -\cot x \csc 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.