Consider: \begin{multline} f = x^2 + y^2 \end{multline} \begin{multline} \frac{\partial f}{\partial x} = 2x \end{multline} and suppose: \begin{multline} y = x + u \end{multline} Then: \begin{multline} f = x^2 + (x+u)^2 \end{multline} \begin{multline} \frac{\partial f}{\partial x} = 2x + 2(x+u) \end{multline} Oops!

Of course, it's not a very deep paradox, because it all depends on what is being held constant.

It is interesting, though, how the notation is letting us down.

So how should we write things instead?

We could make things a bit more explicit, like this: \begin{multline} \frac{\partial f(x,y)}{\partial x} = 2x \end{multline} \begin{multline} \frac{\partial f(x,u)}{\partial x} = 2x + 2(x+u) \end{multline} Let's see how this notation copes with a bit of chain rule. \begin{multline} \frac{\partial f(x,y)}{\partial x} = \frac{\partial f(x,y(x,u))}{\partial x} = \frac{\partial f(x,y)}{\partial x} + \frac{\partial f(x,y)}{\partial y} \frac{\partial y}{\partial x} = 2x + 2y \end{multline} Hmm, what a muddle!

I'm not even sure what it means!

It seems like, at the very least, we need two \( f \) s. Or maybe more formality about what a partial derivative is. \begin{multline} \partial :: (R \rightarrow R \rightarrow R) \rightarrow (R \rightarrow R \rightarrow R) \end{multline} Or maybe, for safety, go back to first principles, with things like this: \begin{multline} f(x + \delta x, y) = f(x + \delta x, y(x + \delta x, u)) \end{multline}