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

2 Answers
Aug 22, 2016

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!

Apr 5, 2018

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!