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I have no problem understanding the single variable derivative:

$$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

As this is merely a case of:

$$\text{rate of change} = \frac{\Delta y}{\Delta x}$$

Where $\Delta x$ masquerades as $h$ and becomes vanishingly small. The above formula is the only reason I understand why there's an $h$ in the denominator. Technically, $h$ tending to zero should make $f'(x)$ tend to infinity in my mind, which is a bit of shakiness in my head that I've never worked out but I can understand it still from the previous few sentences.

However, when then learning of the limit definition of a partial derivative, I lose the rate of change equation to justify the $h$ on the denominator.

$$\frac{\partial f}{\partial x} = \lim_{h\to 0}\frac{f(\mathbf x+h)-f(\mathbf x)}{h}$$

I'm perfectly fine with the numerator, but I now can no longer justify the $h$ in the denominator. It would make more sense to me if it was merely $$\frac{\partial f}{\partial x} = \lim_{h\to 0}h(f(\mathbf x+h)-f(\mathbf x))$$, or $$\frac{\partial f}{\partial x} = \lim_{h\to \infty}\frac{f(\mathbf x+h)-f(\mathbf x)}{h}$$.

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    $\begingroup$ Think of the denominator as $(x+h)-x$ that will make the h make more sense $\endgroup$ Commented Feb 13, 2018 at 21:02
  • $\begingroup$ That certainly makes sense to me with one variable, but with the vectors in the partial derivative one, I'm still having trouble picturing it. $\endgroup$ Commented Feb 13, 2018 at 21:04

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$$ \frac{\partial f}{\partial x} = \lim_{h\to 0}\frac{f(\mathbf x+h)-f(\mathbf x)} h $$ You shouldn't be "perfectly fine with the numerator" in this. Your notation needs to make clear that $h$ is added to just one of the independent variables, and it needs to say which one.

You have $$ \frac{\partial f}{\partial x} = \lim_{h\,\to\,0} \frac{f(s,t,u,\ldots, x+h,\ldots,z) - f(s,t,u,\ldots,x,\ldots, z)} h $$ i.e. you need to make it clear that the $h$ is added only to $x$ and not to any of the other variables upon which $f$ depends.

In slighlty more convnetional notation, $$ \frac{\partial f}{\partial x_k} = \lim_{h\,\to\,0} \frac{f(x_1,\ldots,x_{k-1},\,\, x_k + h,\,\, x_{k+1}, \ldots, x_n) - f(x_1,\ldots,x_k,\ldots, x_n)} h. $$ The "$h$" appears in both the numerator and the denominator because $h= \Delta x_k.$ It is the amount by which $x_k$ changes.

Your notation needs to make clear that $h$ is added to just one of the independent variables, and it needs to say which one.

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  • $\begingroup$ Oh, so it still is kind of like $h = dx$ in single variable calculus? So it can still represent $\text{rate of change} = \frac{\Delta f}{\Delta x}$ a little bit? $\endgroup$ Commented Feb 13, 2018 at 21:10
  • $\begingroup$ It exactly like one-variable calculus. When you take partial derivatives, you hold all the variables but one constant, and only one is "varying." $\endgroup$ Commented Feb 13, 2018 at 21:15
  • $\begingroup$ @sangstar : $h = \Delta x.$ Those are two different names for the same thing. One has $$ \frac{df}{dx} = \lim_{\Delta x\,\to\,0} \frac{\Delta f}{\Delta x}. $$ That is the usual definition of differentiation. $\endgroup$ Commented Feb 13, 2018 at 21:30
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It works exactly as for the functions of one variable $f(x): \mathbb{R}\to\mathbb{R}$, but now we are evaluating the limit of incremental ratio along one specific direction $\vec v$, thus for partial derivative we deal with vectors and, with $h$ number, the definition for directional derivative is

$$\frac{\partial f}{\partial \mathbf v} = \lim_{h\to 0}\frac{f(\mathbf x_0+ h \mathbf v)-f(\mathbf x_0)}{h}$$

and for partial derivatives

$$\frac{\partial f}{\partial x_i} = \lim_{h\to 0}\frac{f(\mathbf x_0+ h \mathbf e_i)-f(\mathbf x_0)}{h}$$

where $\mathbf e_i$ is the unitary vector relate to the $i^{th}$ variable $x_i$.

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  • $\begingroup$ This isn't getting at my question though, as to why there is an $h$ in the denominator. $\endgroup$ Commented Feb 13, 2018 at 21:06
  • $\begingroup$ Ah ok, I didn't undestood your doubt. It is just for the case of the function of one variable. We are evaluating the limit of incremental ratio along one direction as for the case of $f(x): \mathbb{R}\to\mathbb{R}$ $\endgroup$ Commented Feb 13, 2018 at 21:07
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The "standard" definition of derivate in $1$ variable calculus is: $$f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$ and the $$f'(a)=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$ is equivalent with it. But the $\Delta x$ represents the $x-a$. The partial derivate (in the first variable) is definied similar (without a lot of mathematical detail/restriction, but I can add it if you want):
$$f:\mathbb{R}^n\to \mathbb{R}\text{ , }a=(a_1,a_2,\dots,a_n)\text{ , }\phi(t):=f(t,a_2,\dots,a_n)$$Then: $$\partial_1f(a):=\phi'(a_1)=\lim_{t \to a_1} \frac{f(t,a_2,\dots,a_n)-f(a_1,a_2,\dots,a_n)}{t-a_1}$$ $$=\lim_{h \to 0} \frac{f(a_1+h,a_2,\dots,a_n)-f(a_1,a_2,\dots,a_n)}{h}$$ And it can be defined similary for all of the variables.

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