Question

How do you find `f''(x)=csc x`?

Answer 1

(Question edited -- you can't use the keyboard's double quote for the second derivative, it interprets it to mean something else. You need do use two single quotes.)

I think you meant to ask how to find the second derivative of `f(x)=cscx`. (If not, please post another question.)

For `f(x)=cscx`, we have

`f'(x)=-cscx cotx = -[cscx cotx]`

To differentiate this, we'll need the product rule. (I use #(FS)'=F'S+FS')

`f''(x) = -[(-cscx cotx)cotx + (cscx)(-csc^2x)]`

`f''(x)=cscx cot^2x+csc^3x`