Question

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

Answer 1

`g'(x)=-cotx,g''(x)=csc^2x`

Explanation:

According to the chain rule,

`d/dx(ln(u))=1/u*(du)/dx`

Thus,

`d/dx(ln(cscx))=1/cscx*d/dx(cscx)`

`=>1/cscx*(-cscxcotx)`

`=>-cotx=g'(x)`

To find the second derivative, know that `d/dx(cotx)=-csc^2x`.

Thus,

`d/dx(-cotx)=csc^2x=g''(x)`