Strange form of Partial/Total Derivative

Question

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?

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 $r$, $$ \frac{\mathrm{d}x_i}{\mathrm{d}r} = \frac{\partial x_i}{\partial r}. $$