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I need somebody to explain me why

$$\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$$

I don't get it. I'd say the left hand side is 0, because there is no change in $y^2$ when x changes slightly.

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Actually, the answer depends on what $y$ is. If $y$ is just a free variable, then you're right and the derivative is $0$. But if $y$ is a function of $x$ (i.e., $y=y(x)$) then the identity you wrote is an example of the chain rule in action.

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    The identity that Andrew wrote is correct either way - but if $y$ is a free variable, then ${dy\over dx}$ is $0$ and so of course $2y{dy\over dx}$ is identically $0$.2011-11-14
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    @StevenStadnicki True, but $2y\frac{dy}{dx}$ is a pretty verbose way to write $0$. :-)2011-11-15
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You are considering that $y$ is some function of $x$. An example would be if $y=x^3$. Then $y^2=x^6, \frac{d}{dx}(y^2)=2y\frac{dy}{dx}=2x^3(3x^2)=6x^5=\frac{d}{dx}x^6$. This is an example of the chain rule in action.

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    One question, how you went from $\frac{d}{dx}(y^2) \ to \ 2y\frac{dy}{dx}$, I can understand $y^2=2y$ but where does that $\frac{dy}{dx}$(next to 2y) come from?2011-11-14
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    @Andrew: $y^2 \neq 2y$... but the $dy/dx$ comes from the chain rule. Since we are looking for $\frac{d}{dx}y(x)$ we need to use the chain rule, that is, $\frac{d}{dx}y(x) = y'(x) = \frac{dy}{dx}$. I think the difficulty comes in when you're composing $y(x)$ with other functions.2011-11-14
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    OK, I got it totally. I'll tell you what was the difficulty. I am used to having the chain rule as $\frac{dy}{dx}$ here, the y was used as the variable. I changed it to u and still kept the dy/dx as in the original form and everything is clear now. :) Thank you a lot.2011-11-14