When looking through this document, I was confused by Equation $(4)$. It is stated there that a function $f(x,y)$, with $x,y\in\mathbb{R}^n$ and $x=x(r),y=y(r)$ for some $r\in\mathbb{R}$, has the derivative

$$ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x_i}\frac{\partial x_i}{\partial r} + \frac{\partial f}{\partial y_j}\frac{\partial y_j}{\partial r}$$ using Einstein's notation for summation. But to me, this makes no sense - wouldn't this reduce to $$ \frac{\partial f}{\partial r} = n\cdot\frac{\partial f}{\partial r} + n\cdot\frac{\partial f}{\partial r}$$ via the chain rule? Assuming that the original author made a mistake, would the total derivate $$ \frac{df}{dr} = \frac{\partial f}{\partial x_i}\frac{dx_i}{dr} + \frac{\partial f}{\partial y_j}\frac{dy_j}{dr}$$

be a reasonable fix for the notation? Or am I misunderstanding something?

  • $\begingroup$ I think you are giving far too much weight to the notation. We are allowed to use $\partial$ for derivatives even though there is only one parameter. So what the author writes is perfectly OK. Life would be hell in multivariate calculus if every time we had to stop and say "oh wait, is $n=1$, shouldn't I write $d$ and not $\partial$ here?" $\endgroup$ Jul 1, 2021 at 8:36

1 Answer 1


The chain rule for partial derivatives is not $$ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x_i} \frac{\partial x_i}{\partial r} $$ (where I mean a fixed $i$, not Einstein summation convention). In other words, they don’t “cancel” like normal single-variable differentials do. Hence, the step from the first to the second equation in your post is wrong.

Indeed, the first equation is the chain rule for partial derivatives, i.e. with the sum over all variables that Einstein summation gives. (It is slightly confusing here because you have $x$ and $y$, both of which represent several variables themselves.)

To help you remember this, many authors will use the last formula in the post where one is not tempted to cancel the differentials. But the first formula is fine, too: For a function $x_i$ of one variable $r$, $$ \frac{\mathrm{d}x_i}{\mathrm{d}r} = \frac{\partial x_i}{\partial r}. $$

  • $\begingroup$ They kind of can’t cancel: $\frac{\partial f}{\partial x_i}$ only tells you what happens to $f$ if you vary $x_i$ and keep the other variables fixed. But if $r$ changes, the other variables change, too (i.e., they’re not fixed), and $\frac{\partial f}{\partial x_i}$ and $\frac{\partial x_i}{\partial r}$ together do not contain all the information needed to compute $\frac{\partial f}{\partial r}$. You need information about how $f$ reacts to changes in all the variables, i.e. all partial derivatives. The multi-variable chain-rule then tells you how you can put this information together. $\endgroup$ Jul 1, 2021 at 9:18
  • $\begingroup$ Thanks, that clarifies the issue for me. I guess the part $\frac{dx_i}{dr}=\frac{\partial x_i}{\partial r}$ is what I overlooked in my post $\endgroup$
    – mto_19
    Jul 1, 2021 at 10:00

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