Find y'(x) when cos y - y^2 = 8
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?
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?
Nope, there was no typo, and the answer was 0. Thanks for the clarification!
To find y'(x), we need to differentiate both sides of the equation cos(y) - y^2 = 8 with respect to x using the chain rule.
Differentiating the left side with respect to x gives:
(-sin(y))(dy/dx) - (2y)(dy/dx) = 0
Since we want to find dy/dx or y'(x), we can isolate it on one side of the equation:
(dy/dx)(-sin(y) - 2y) = 0
dy/dx = 0 / (-sin(y) - 2y)
dy/dx = 0 (since the numerator is 0)
So we can see that the correct answer is indeed option C) 0. The derivative of y with respect to x is 0.