Find y'(x) when cos y - y^2 = 8

Question

a) -siny - 2y
b) - (1/sin y - 2y)
c) 0
d) - (8/cos y - y)

So if I implicitly differentiate:
y'(x) = -sin(y) - 2y y' = 0

Am I right to say that there's nothing more you can do to this and pick C?
y' = 0 / (-siny - 2y)
So y'(x) = 0?

Answer 1

Since y is not a function of x in the problem you posted, the change of y with a change in x is inevitably zero.
Are you sure there is no typo in the problem statement?

Answer 2

Nope, there was no typo, and the answer was 0. Thanks for the clarification!