Question

find where the slope of the curve is defined?
x= cos y
I've already found the implicit differentiation -1/siny but I don't know what to do from here

Answer 1

x=cosy
the slope is \[\frac{dx}{dy}=-\sin y \] this defined for any value of y
so why do you need implicit derivative
unless you seeking dy/dx

Answer 2

if you are lookinf ro dy/dx you are esentially looking for the derivative of y=arccos(x)
\[x=\cos y \Longrightarrow 1=\frac{dy}{dx}(-\sin y) \]
which is what you got \[\frac{dy}{dx}=-\frac{1}{\sin y}\]
this defined for \[y=0, \pi, 2\pi... \]
if we generlize \[y' \text{is defined for the set of values y that satisfy the following:}\]
\[\ \{y: y\ne 2n\pi , ~~n\in \mathbb{Z} \}\ \]