Question

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

Answer 1

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

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!