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$$\frac{d}{dx} \cos(y^2)=-2y \sin(y^2)$$

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.

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4 Answers

up vote 1 down vote accepted

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 $dy/dx = 0$....

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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: $$\dfrac{dy}{dx}=1$$

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Yes you are correct; $\cos(y^2)$ is a constant here.

The question would have been correct if it was $$\frac{d}{dy} \cos(y^2)=-2y \sin(y^2)$$

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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 $$ \frac{d}{dx}\cos(y^2) = -\sin(y^2)\cdot 2y\cdot 0 = 0. $$

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