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How can you take the derivative

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

I don't understand how the chain rule applies here. Someone told me that the chain rule applies here because $y$ can be expressed in some type of $x$ expression, but I still don't get it.

So, I do understand how to apply the chain rule, I just don't understand why it applies in this situation.

I checked this video.

Aaron Maroja
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user1534664
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4 Answers4

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The chain rule always applies. You derive a function of $y$ (namely $y^2$) with respect to $x$, which is the same as deriving $y^2$ with respect to $y$ and multiplying by the derivative of $y$ with respect to $x$.

$$\frac{dy^2}{dx}=\frac{dy^2}{dy}\cdot\frac{dy}{dx}=2y\frac{dy}{dx}.$$

Whether there is a dependency between $y$ and $x$ doesn't matter, the rule remains true.

If there is no dependency,

$$\frac{dy^2}{dx}=2y\frac{dy}{dx}=2y\cdot0.$$

Mark Bennet
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Notice that by the Chain Rule you have $g(f(x)) = g'(f(x))f'(x)$. Now assuming $y(x)$ is a function of $x$ we have that

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

Take $g(y) = y^2$ and $y = f(x)$.

Aaron Maroja
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  • I'll have a look at the video. The rest of the answer seems to state the same as in the video I linked. – user1534664 Feb 04 '15 at 19:49
  • He keeps on some examples, it's good to get the intuition. – Aaron Maroja Feb 04 '15 at 19:49
  • I studied the video, and it's doing the exact same trick at 7:18. Somehow it's starting to make sense intuitively, but I can't explain it. I'm going to follow the rest of the series on implicit differentiation and see if it works out. I'll come back if I can't figure it out :p Thanks for your help. – user1534664 Feb 04 '15 at 20:27
  • Great that you are catching it. I'm glad I could help! – Aaron Maroja Feb 04 '15 at 20:29
  • "Now assuming y(x) is a function of x" made me understand it properly now, because I tried thinking of how to differentiate the equation if y was any random x expression. I feel dumb now that I see how simple it is. Love it :) – user1534664 Feb 04 '15 at 22:31
  • Don't feel like that, it's part of the learning process. Keep up the good job! – Aaron Maroja Feb 05 '15 at 01:31
  • One more thing: I understand how to get the derivative of $\frac{dy^2}{dx}$, but how did we know we must assume $y$ is a function of $x$? If it wasn't a function of x, then $\frac{dy}{dx}$ would be $0$ (which would give a different result) according to Yves Daoust's answer (if I interpreted that correctly). So how can I tell by context that I have to assume this? – user1534664 Feb 05 '15 at 10:32
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    Yes you are right, if $y$ wasn't a function of $x$ then the derivative would be zero. This can be told when you are dealing with an implicit equation, implicit function, and context (exercise or theorem) will give you that. – Aaron Maroja Feb 05 '15 at 12:29
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$$ u(y) = y^2\\ y = f(x). $$ Thus we have by chain rule $$ \frac{du}{dx} = \frac{dy}{dx}\frac{du}{dy} = \frac{dy}{dx}\dfrac{d}{dy}y^2 = 2y'y $$ So we have that $$ \frac{du}{dx} =\dfrac{d}{dx}y^2 =2y'y $$

Chinny84
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I see this as differentiating with respect to $y$ and making up for it with a scale factor $\frac{dy}{dx}$ $$\frac{d}{dx}\{\}=\frac{dy}{dx}\frac{d}{dy}\{\}$$

Hence:$$\frac{d}{dx}\{y^2\}=\frac{dy}{dx}\times 2y$$

Karl
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