If $y = x^2$, then what is $dy$? (What's the definition of differential?) Suppose $y = f(x) = x^2$. We know that $dy = 2x dx$.
In high school calculus, I overlooked a lot of details and mechanically calculated $dy$. Now, I am unsure how the definition of differential is developed in elementary (calculus or analysis) language. Of course, if we let $F(x,y) = y - f(x) \equiv 0$ and take exterior derivative, then $dF = dy - d(x^2) = dy - 2x dx = 0$, so we have the desired result. However, this exterior derivative is a purely algebraic concept and I feel that a lot of elementary calculus perspective is missing. Here are a few explanations that I remember in high school calculus:


*Naive approach: $\frac{dy}{dx} = 2x$, so "multiply" by $dx$ yields $dy = 2x dx$. I am trying to stay away from this approach as much as possible.

*Better alternative: if $\Delta x, \Delta y$ are infinitesimal changes, then $\Delta x$ and $\Delta y$ are related by: $\Delta y \approx f' \Delta x$. Taking Riemann sum,$
\int_{[a,b]} f \approx \sum\limits_{\|\Delta x\| \rightarrow 0} f'(x) \Delta x = \sum\limits_{\Delta y} \Delta y$, where $\Delta y = f'(x) \Delta x$. This naturally implies the relation $dy =  f'(x) dx$.

I am satisfied with 1) but it nevertheless fails to address the question: "what is" differential? This being said, how should I understand what differential is?
Thank you in advance.

The differential $\mathrm df$ of a function $f$ is simply the linear function that realises the best linear approximation of the function at a point, in a precise sense given by asymptotic analysis:
$$f(x+h,y+k)=f(x,y)+\mathrm df_{(x,y)}(h,k)+o\bigl(\|(h,k)\|\bigr).$$
This definition can be generalised to normed vector spaces, whether finite dimensional or not.
Considering the example of your title, $\mathrm d_xy$ is the linear map of $h$:  $\;h\longmapsto 2x h$.
