what is the derivative of sinx- cosy = 0 in its simplest form? I got to the point y'=-consx/siny, but I wasn't too sure. This could probably simplified further. help!

You're supposed to treat y as a function of x and differentiate it implicitly. We have
d/dx{sinx- cosy)=d/dx{sinx) - d/dx{cosy)=
cos(x)-(-sin(y)*y')=0
You should be able to solve for y'.

I did, and I ended up with y'=-cos(x)/sin(y)

How do I simplify this further?

That's as far as you can go unless you know something specific about y. Without knowing something about y(x) there's nothing else to do here.

The derivative you obtained, y' = -cos(x)/sin(y), is the simplest form of the derivative of the equation sin(x) - cos(y) = 0. Given that you don't have any additional information about the relationship between x and y, you cannot simplify it further.

The form -cos(x)/sin(y) is already the most simplified expression for the derivative in this case. Remember that in calculus, simplifying further depends on the specific relationships between variables and any additional constraints or assumptions. Without those, you have reached the simplest form of the derivative.