Can partial derivatives be expressed as a fraction?

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.

• It won't always hold that $\frac{\partial a}{\partial c} = a/c$. For example, try with $a = bc^{2}$. – Minus One-Twelfth Feb 18 at 1:55
• Not true in general.. this is just because $a$ happens to be a linear function of $c$. – GReyes Feb 18 at 1:55

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.
• 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). – Allure Feb 18 at 3:22
• 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. – siegehalver Feb 19 at 3:36
• Doesn't this only assume that $b$ is independent of $c$ (since it's a partial derivative not full derivative)? – Allure Feb 19 at 3:43
• But if its independent of $c$, then its not changing at all in the context of this problem, and is therefore constant correct? – siegehalver Feb 19 at 3:46
• 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 $c$ – siegehalver Feb 19 at 4:01