A professor of mine was talking about the difference between derivative with respect to a slot and a derivative with respect to a variable. Apparently there is some notational issue there. I didn't really got what he meant there and at the moment I am not able to ask him.

Let $x=(x^1, x^2, \dots, x^d)$. We have notationally speaking

$$ \partial_{x^{j}} f(x) := (\partial_{j} f)(x) $$

where the left side the derivative in respect to the variable and on the right the derivative with respect to the slot.

Now the argument was that this might lead to confusion in the case of the chain rule, for example:

$$ \partial_{t} f(tx) = \sum_{j=1}^{d} x^{j}(\partial_{j}f)(tx) $$

(the evaluation here depends on some real $t$ variable so we don't evaluate at $x$ anymore which confuses me somehow) and also that the fundamental theorem of calculus does not apply to slot derivatives.

Could anyone elaborate on that? An example would be awesome to see the difference and what the notational problems are in this cases. Especially the chain rule example I don't understand. Thank you!

  • $\begingroup$ If $x$ is a vector this just means taking the derivative with respect to the j-th entry. $\endgroup$ Jul 12 '21 at 17:29
  • $\begingroup$ I'm not sure if this answer of mine is directly relevant to your question, but it may be helpful. $\endgroup$
    – peek-a-boo
    Jul 12 '21 at 18:03

Here's how you'd write everything precisely using the chain rule. There are actually two functions involved here. First is $f:\Bbb{R}^n\to\Bbb{R}$ and second is the map $\mu:\Bbb{R}\times\Bbb{R}^n\to\Bbb{R}^n$ defined as \begin{align} \mu(t,x)&:=tx. \end{align} So, $\mu$ is the "scalar-multiplication map". Now, we can consider the composite function $f\circ\mu:\Bbb{R}\times\Bbb{R}^n\to\Bbb{R}$ and ask what is its first partial derivative. The answer is given by the chain rule: for all $(t,x)\in\Bbb{R}\times\Bbb{R}^n$ \begin{align} [\partial_1(f\circ \mu)](t,x)&=\sum_{j=1}^n(\partial_jf)(\mu(t,x))\cdot (\partial_1\mu^j)(t,x)\\ &=\sum_{j=1}^n(\partial_jf)(tx)\cdot x^j, \end{align} where the last line is simply because $\mu^j(t,x)=tx^j$.

Note that here, you shouldn't pay attention to the individual letters $t$ and $x$. They are just arbitrary curved symbols. Sure, it may be tradition to use these letters, but mathematically, one is not obligated to use them. I could just as well write: for all $(\ddot{\smile},@)\in\Bbb{R}\times\Bbb{R}^n$,

\begin{align} [\partial_1(f\circ \mu)](\ddot{\smile},@)&=\sum_{\#=1}^n(\partial_{\#}f)(\ddot{\smile}@)\cdot @^j \end{align}

As this crazy rewriting shows, logically speaking, one should not use notation such as $\partial_{x_j}f$ or $\frac{\partial f}{\partial x^j}$ in place of $\partial_jf$, simply because the symbol $x$ has no inherent mathematical meaning.

As for the remark about the FTC not being applicable, I'm not sure what your Professor meant, because the FTC is certainly applicable (assuming $f$ is $C^1$ for example): for all $x\in\Bbb{R}^n$, \begin{align} f(x)&=f(0)+\int_0^1\sum_{j=1}^n(\partial_jf)(tx)\cdot x^j\,dt \end{align}

  • $\begingroup$ Ok although I think I get what you mean I am not sure if I get the point the professor is making there. Also $\partial_{1}$ means that we derivate in respect to the first argument right - which is $t$ in the example? And also if $x$ is a vector then $\mu$ would be $(tx^{1}, tx^{2}, \dots, tx^{n})$ I guess. But then I am not sure again - is the slot $t$ or $x$ or is it one of the vector elements... $\endgroup$
    – craaaft
    Jul 13 '21 at 7:09
  • $\begingroup$ And as for the FTC he wrote that $\int_{0}^{1} \partial_{t} f(tx) dt = f(x) - f(0) \neq \int_{0}^{1} f'(tx) dt$ the last term being a slot derivative. $\endgroup$
    – craaaft
    Jul 13 '21 at 7:43
  • $\begingroup$ @craaaft yes, $\partial_1$ here is what you've written $\partial_t$. For your second question, I'm not sure what you mean $\mu(t,x)=tx=(tx^1,\dots, tx^n)$, so $(\partial_1\mu^j)(t,x)=x^j$... or clasically we'd write this as $\frac{\partial \mu^j}{\partial t}=\frac{\partial}{\partial t}(tx^j)=x^j$. For your last query, what does $f'(tx)$ mean? Do you mean the $n\times 1$ matrix of partial derivatives? In that case it is obviously true that the two things are not equal. $f(x)-f(0)$ is an element of $\Bbb{R}$, while $\int_0^1f'(tx)\,dt$ is an integral of a row vector, and hence a row vector. $\endgroup$
    – peek-a-boo
    Jul 13 '21 at 8:05
  • $\begingroup$ anyway, even in single variable calculus, that equality is not true. The FTC implies that if $f:\Bbb{R}\to\Bbb{R}$ then $f(x)-f(0)=\int_0^1\frac{d}{dt}[f(tx)]\,dt=\int_0^1f'(tx)\cdot x\,dt\neq \int_0^1f'(tx)\,dt$. Another way of saying it is that you're fixing a value of $x$ and then defining the function $g(t)=f(tx)$. Then, $f(x)-f(0)=g(1)-g(0)=\int_0^1g'(t)\,dt=\int_0^1f'(tx)\cdot x\,dt$. $\endgroup$
    – peek-a-boo
    Jul 13 '21 at 8:08
  • $\begingroup$ If you're confused about this last step it just means you're assigning too much importance to the letter $x$, because I'm sure that if I write $g(t)=f(c\cdot t)$, where $c$ is a fixed number, then you'd (hopefully) immediately say that by the chain rule, $g'(t)=f'(ct)\cdot c$ (for example, differentiating $g(t)=e^{ct}$ yields $g'(t)=ce^{ct}$). $\endgroup$
    – peek-a-boo
    Jul 13 '21 at 8:14

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