# Another partial derivatives question

I think this is a common question in applied math but I find no occurrence of it in MSE.

$$u=u(x), v=v(x)$$

$$f=f(u(x),v(x))$$

1) $u$ and $v$ are both functions of the variable $x$. If $u$ varies, then that must be because of some variation in $x$, which in turn means that $v$ must also have varied. Is my logic correct thus far?

2) If (1) is correct, then there can be no variation in $u$ without a variation in $v$. A partial derivative of $f$, say $\frac{\partial f}{\partial u}$ would imply that $v$ is constant while $u$ varies. Isn't that a mathematical contradiction?

Thanks

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In 1), why can't $v$ be constant? In 2), I think you are getting confused about the difference between partial and total derivatives. When we talk about $\frac{\partial f}{\partial u}$ we are not thinking of $u$ as a function of $x$, we are just thinking of it as one of the inputs to $f$. We can talk about the inputs to $f$ varying and then later plug $(u(x), v(x))$ into $f$ and see what happens. – Qiaochu Yuan Sep 7 '12 at 1:33

It is the notation that is confusing. In this situation, you actually have two different functions of the same name $f$. To make the matter clearer, I'll give one of them a different name:

$$g(x) = f(u(x), v(x)).$$

Then you know that $f:\mathbb R^2 \to \mathbb R$ but $g:\mathbb R \to \mathbb R$. They are different functions.

Usually, the notation $\frac{\partial f}{\partial u}$ refers to the derivative of $f$ with respect to the first variable if you usually put $u$ in the first variable like this: $f(u, v)$. The notation $\frac{df}{dx}$ implies that $u(x)$ and $v(x)$ are provided, and it is simply $g'$.

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(1) is not correct - let $u(x) = x$, and $v(x) = 1$. Then $u$ changes any time $x$ does, but $v$ never changes. In general, $u$ changing implies nothing about how $v$ changes.

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