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Let $a = bc$. Then $b = a/c$. From the first equation, we also have $\frac{\partial a}{\partial c} = b$. Equating, $\frac{\partial a}{\partial c} = b = a/c$, or $\frac{\partial a}{\partial c} = a/c$, and any partial derivative can be expressed a fraction of its constituents.

Is there something wrong with this proof? If so, what? This looks like such a simplification that I'm convinced it must be wrong, but I can't find a counterexample.

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  • $\begingroup$ It won't always hold that $\frac{\partial a}{\partial c} = a/c$. For example, try with $a = bc^{2}$. $\endgroup$ – Minus One-Twelfth Feb 18 at 1:55
  • $\begingroup$ Not true in general.. this is just because $a$ happens to be a linear function of $c$. $\endgroup$ – GReyes Feb 18 at 1:55
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You've assumed that $a$ is a scalar multiple of $c$ and therefore (assuming both are differentiable) $a'$ is going to be the same scalar multiple of $c'$ by elementary properties of derivatives. It's not a powerful result, because you put some pretty extreme restrictions on $a$ and $c$, namely that they are scalar multiples.

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  • $\begingroup$ Are you saying that there's an assumption that $b$ is a scalar in the proof? If so, where'd that enter the proof? It seems to me that just setting $a=bc$, there's no constraint on all three of $a$, $b$ and $c$ (other than the relation). $\endgroup$ – Allure Feb 18 at 3:22
  • $\begingroup$ Well you took a partial with respect to $c$ right? So $b$ is a constant, otherwise you're statement that $\frac{\partial a}{\partial c}=b$ is false, right there you have assumed $b$ is a constant. $\endgroup$ – siegehalver Feb 19 at 3:36
  • $\begingroup$ Doesn't this only assume that $b$ is independent of $c$ (since it's a partial derivative not full derivative)? $\endgroup$ – Allure Feb 19 at 3:43
  • $\begingroup$ But if its independent of $c$, then its not changing at all in the context of this problem, and is therefore constant correct? $\endgroup$ – siegehalver Feb 19 at 3:46
  • $\begingroup$ Basically, you should have done product rule when taking the derivative of the product, and assuming you did, the only way you got what you did is if $\frac{\partial b}{\partial c}=0$, namely that it is a constant with respect to $c$ $\endgroup$ – siegehalver Feb 19 at 4:01

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