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

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Answer 1 · 2016-08-22T15:36:06

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!

Answer 2 · 2018-04-05T01:48:58

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!