Question

How do you find the derivative of `ln(csc 5x)`?

Answer 1

Use the chain rule and

use `d/dx(lnu) = 1/u (du)/dx`.

We'll also need `d/dx(cscu) = -cscucotu (du)/dx`

And also #d/dx(5x) = 5

`d/dx(ln(csc(5x)))=1/csc(5x) d/dx(csc(5x))`

`color(white)"sssssssssssssssss"` `=1/csc(5x) *(-csc(5x)cot(5x)) d/dx(5x)`

`color(white)"sssssssssssssssss"` `=1/csc(5x) *(-csc(5x)cot(5x)) (5)`

We are finished with the calculus, but we can rewrite the answer:

`d/dx(ln(csc(5x))) = -5cot(5x)`