Let $f, g \in \mathbb R^n \rightarrow \mathbb R$. Is there some way to make sense of the expression:

$$\frac{\partial f(x_1, x_2, \dots, x_n)}{\partial g(x_1, x_2, \dots, x_n)}$$

I want some way to measure "how much $f$ changes along $g$" --- I'm not sure what a reasonable definition of this. Here is one that came to mind.

Let us define $\frac{\partial f}{\partial g}\big(t \big)$: At each point $t \in \mathbb R^n$, we first compute $g'(t) \in \mathbb R^n$. Now, we compute the directional derivative of $f$ along $g'(t)$:

$$ \frac{\partial f(x_1, x_2, \dots x_n)}{g(x_1, x_2, \dots x_n)} (t) : \mathbb R^n \rightarrow \mathbb R\equiv (\nabla_{g'(t)} f)(t) = \lim_{h \rightarrow 0} \frac{f(t + hg'(t)) - f(t)}{h}$$

  • Is this a well-known idea? If so, what is it called?
  • My problem with this is that it returns a scalar at each point: what I am actually interested in is to find a new function which tells me "how to move $f$ infinitesimally so we can make it closer to $g$.

The given definition above clearly generalizes to any manifold: all I need is a directional derivative, which I do possess on a manifold: Can we say something interesting about this in a larger setting?


The construction you described doesn't actually work on manifolds - $g'(t)$ (or $dg$ as we would normally write) is not a tangent vector, but a cotangent vector; so the directional derivative $\nabla_{dg} f$ doesn't make sense without a metric (or some other additional structure) to identify $TM$ with $T^* M$.

It also doesn't seem to have the right behaviour to be casually called "$\partial f / \partial g$" - for example, $\nabla_{dg} f$ would double if you doubled $g$, which is the opposite scaling behaviour to what the notation would suggest.

Remember that partial derivatives are only defined in terms of a whole coordinate system - if you're given the two coordinate systems $(x,y)$ and $(x,z=y-x),$ the expression $\partial f/\partial x$ will have different values depending on whether you're holding $y$ or $z$ fixed! Thus I'm unsure how to interpret the intended spirit of "$\partial f/\partial g$".

Is this a well-known idea? If so, what is it called?

Not commonly enough to have a name; but it's a simple enough expression that you'll find it cropping up in equations here and there in vector calculus (as $\nabla g \cdot \nabla f$) and Riemannian geometry (as $\nabla_{\operatorname{grad}g}f$). The interpretation isn't really close to what you're looking for, however - it's just "the rate of change of $f$ in the direction $\nabla g$ (or vice versa).

what I am actually interested in is to find a new function which tells me "how to move f infinitesimally so we can make it closer to g".

I'm unsure how to interpret this - to me, the way to "move" a scalar function is to deform it by another scalar function (e.g. deform $f$ to $f + \epsilon \phi$ for a parameter $\epsilon$), in which case what you're describing would just be the scalar $g - f$? If you could formalize what you're asking for here (or at least give more of a geometric description) there might be something more useful to say.

  • $\begingroup$ You're right; I'll make my question more precise and ask again --- thanks a lot for pointing out that it doesn't really behave very well. If you could expand your answer on where on might see $\nabla_{grad~ g} f$, I'd be very interested! $\endgroup$ – Siddharth Bhat Feb 26 '20 at 9:51

I think $\frac{\partial f(x_1, x_2, \dots, x_n)}{\partial g(x_1, x_2, \dots, x_n)}$ should be able to predict small changes in $f$, given small changes in $g$. Now, if all we're given is a small change $dg$, then that's just $ \nabla g\cdot \vec{dr} $. Even knowing $\nabla{g}$ at the point, we can't invert this dot product to get back $\vec{dr}$. Without knowing $\vec{dr}$, we can't predict the change in $f$, which is $\nabla f\cdot \vec{dr}$. So in this sense, the derivative is not define-able.


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