How does $f(x+\delta x)=(1+\delta x)f(x)\implies \frac{d f(x)}{dx}=f(x)$ generalizes to multivariables?

Question

I am looking for two generalizations:

Multivariable (with scalars)

$$ f(x+\delta x,y+ \delta y)=(1+\delta x + \delta y)f(x,y)\implies \frac{f(x+\delta x,y+\delta y)-f(x,y)}{(\delta x +\delta y)}=f(x,y) $$

Is this a derivative in some sense?

Multivariable (with vectors)

$$ f(x+\delta x,y+ \delta y)=(1+ \hat{\mathbf{x}} \delta x + \hat{\mathbf{y}} \delta y)f(x,y)\implies \frac{f(x+\delta x,y+\delta y)-f(x,y)}{(\hat{\mathbf{x}} \delta x + \hat{\mathbf{y}} \delta y)}=f(x,y) $$

Do any of these generalizations connect to known derivatives?

Presumably, you want $f(x,y)$ on the right side of the equalities. — Thomas Andrews, Jun 30 '21 at 20:01
@ThomasAndrews The notation $\hat x$ often denotes a unit vector along the $x$ direction. — Mark Viola, Jun 30 '21 at 20:06
But you can't divide by vectors, @MarkViola, as the OP clearly does. — Thomas Andrews, Jun 30 '21 at 20:09
@anon21 What are you trying to do? For $x\in \mathbb{R}$ and $f\in C^2$, the equation $f'(x)=f(x)$ can be written $f(x+\Delta x)-f(x)=f'(x)\Delta x+o(\Delta x)=f(x)\Delta x+o(\Delta x)$ or $f(x+\Delta x)=(1+\Delta x)f(x)+o(\Delta x)$. — Mark Viola, Jun 30 '21 at 20:10
@ThomasAndrews Yes. I thought that you were asking about the notation $\hat x$. — Mark Viola, Jun 30 '21 at 20:15
A good sign that you're going to be in trouble is that this is undefined when $\delta x + \delta y = 0$. If you changed to something like $\sqrt{(\delta x)^2 + (\delta y)^2}$, then it would resemble a directional derivative. — Ted Shifrin, Jun 30 '21 at 20:32

score 0 · Answer 1 · answered Jun 30 '21 at 22:55

For the first one, notice that: $$\begin{align} \phantom{=}&f(x+\delta x,y+\delta y)-f(x,y)\\ =&\delta x\underbrace{\frac{f(x+\delta x,y+\delta y)-f(x,y+\delta y)}{\delta x}}_{\approx\left.\frac{\partial f}{\partial x}\right|_{(x,y+\delta y)}}+\delta y\underbrace{\frac{f(x,y+\delta y)-f(x,y)}{\delta y}}_{\approx\left.\frac{\partial f}{\partial y}\right|_{(x,y)}} \end{align}$$ so it looks like the total derivative, which is: $$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$

How does $f(x+\delta x)=(1+\delta x)f(x)\implies \frac{d f(x)}{dx}=f(x)$ generalizes to multivariables?

1 Answers1