how do you determine the derivative is true or false? $$\frac{d}{dx} \cos(y^2)=-2y \sin(y^2)$$
I need to find out if this statement is true or false. My answer is false, but I'm not sure.It is false because the variables disagree.
 A: You are correct. To salvage the statement, it should be written as:
$$
\frac{d}{dx} \cos(y^2)=-2y \sin(y^2) \frac{dy}{dx}
$$
where $y$ is assumed to be some function of $x$.

Note that that the statement could only have been true if $y=x+C$ for some constant $C$ so that: $$\dfrac{dy}{dx}=1$$
A: Yes you are correct; $\cos(y^2)$ is a constant here.
The question would have been correct if it was $$\frac{d}{dy} \cos(y^2)=-2y \sin(y^2)$$
A: More formally, use the Chain Rule:
$$
\frac{d\cos(y^2)}{dx} = -\sin(y^2) \frac{d y^2}{dx} = -2y \sin(y^2) \frac{dy}{dx}.
$$
If you are not told anything about the relationship of $y$ to $x$, they are assumed independent, so $y$ is constant with respect to $x$ and $dy/dx = 0$....
A: Yes, that statement is false. We can see that this is wrong by applying the chain rule:
$$
\frac{d}{dx}\cos(y^2) = \frac{d}{dy}\left[\cos(y^2)\right]\cdot\frac{dy}{dx}
$$
is the correct formula, but $\frac{dy}{dx} = 0$ here, as there is no information indicating that $y$ depends on $x$. So we would have
$$
\frac{d}{dx}\cos(y^2) = -\sin(y^2)\cdot 2y\cdot 0 = 0.
$$
