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I have encountered a situation where I need to find the result of the following construction:

$$ \lim_{dx\to 0,dy\to 0} \frac{F(x+dx,y+dy)-F(x,y)}{dxdy} $$

Is this a valid derivative construction? What is the result of such a thing?

In fact, my actual situation has 4 terms:

$$ \lim_{dx\to 0,dy\to 0, dz\to 0, dt\to 0} \frac{F(t+dt, x+dx,y+dy z+dz)-F(t,x,y,z)}{dtdxdydz} $$

I assume the 2d cases generalizes easily to the 4d case.

Anon21
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  • Why not use https://en.wikipedia.org/wiki/Total_derivative ? – zkutch Jun 29 '21 at 23:07
  • @zkutch I don't have a choice; this construction is what I get. – Anon21 Jun 29 '21 at 23:08
  • Similar general construction is https://en.wikipedia.org/wiki/Gateaux_derivative , but your denominator makes me doubt. – zkutch Jun 29 '21 at 23:13
  • Suppose that $F: U \to V$ is a function of real vector spaces and $n \in \mathbb N_0$ let $h: {1,\ldots, n} \to U$ and $\tau: {1,\ldots, n} \to \mathbb R^*$ and $p \in U$. Then a good candidate for the difference quotient for the $n^\text{th}$ derivative of $F$ at point $p$ in the directions given by $h$ would be $\frac{1}{\prod_{i = 1}^n \tau_i} \sum_{X \subseteq {1,\ldots, n}} (-1)^{n - |X|}F(p + \sum_{x \in X}\tau_xh_x)$. The limit needed to be taken to get the required derivative is the one where all component $\tau_i$ converge uniformly to $0$ for all $ i \in {1,\ldots, n}$. – user2628206 Jun 29 '21 at 23:32
  • Are you sure that instead of $dxdy$ it should be $\sqrt{(dx)^2+(dy)^2}$ ? – jimjim Jun 30 '21 at 02:53
  • @jimjim ohhh maybe! I'll have to look into this. – Anon21 Jun 30 '21 at 13:05

1 Answers1

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This looks to be divergent for any differentiable function. For instance look at $$F(x,y) = x+y$$ Then $$\frac{F(x+h,y+k) -F(x,y)}{hk} = \frac{h+k}{hk} = \frac1h +\frac1k$$ Which clearly does not converge as $(h,k) \to 0$

In general if $F$ has continous partial derivatives $$\frac{F(x+h,y+k) -F(x,y)}{hk}$$ $$=\frac{F(x+h,y+k) -F(x+h,y)+F(x+h,y)-F(x,y)}{hk}$$ $$=\frac1k\frac{F(x+h,y+k) -F(x+h,y)}{h}+\frac1h\frac{F(x+h,y)-F(x,y)}{k}$$ $$=\frac1kD_2F(x+h,\zeta)+\frac1hD_1F(\xi,y)$$ where $\zeta \in (y,y+k)$ and $\xi \in (x,x+h)$. This obviously has no limit as $(h,k) \to 0$

Ben Martin
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