Question

What are the first and second derivatives of `g(x) = lnsinx`?

Answer 1

Using chain rule, we can rename `u=sinx` and solve it following chain rule definition: `(dy)/(dx)=(dy)/(du)(du)/(dx)`

Explanation:

`(dy)/(dx)=(1/u)*cosx`

`(dy)/(dx)=cosx/sinx`

This is the same as `tan^(-1)x` or `1/tanx`, which is the same as `cotx`

`(dy^2)/(d^2x)=-csc^2x`