Juxtaposing the limit definition of the partial derivative and single variable derivative

Question

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

Answer 1

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

Answer 2

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 $x_i$.

Answer 3

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.