Question

What is the derivative of `(cscx+cotx)^-1`?

Answer 1

Let's use chain rule here, by naming `u=(cscx+cotx)`

Thus, our function is written as `y=u^-1`.

The chain rule states that

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`(dy)/(du)=-u^-2`

`(du)/(dx)=-1*cscxcotx-1*csc^2x=-csc(cotx+cscx)`

Joining them both:

`(dy)(dx)=(cancel(-)u^-2)/(cancel(-)csc(cotx+cscx))`

Now, substituting `u`:

`(dy)/(dx)=cancel(cscx+cotx)/(cscxcancel(cscx+cotx))=1/cscx`