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:

  1. 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.
  2. 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.


2 Answers 2


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 $h$: $\;h\longmapsto 2x h$.

  • $\begingroup$ I feel that you should explicitly acknowledge that this is only one of many interpretations of $d$. $\endgroup$
    – Arthur
    May 30, 2021 at 12:01
  • $\begingroup$ This is Cartan's classic definition. Of course, one may want to stick to non-standard analysis. $\endgroup$
    – Bernard
    May 30, 2021 at 12:08
  • $\begingroup$ Or measure theory. Or differential geometry. And I'm sure there are others. $\endgroup$
    – Arthur
    May 30, 2021 at 12:17
  • 1
    $\begingroup$ As a geometer, I agree with @Bernard, but I would be more precise: the differential at $p$ of the function $f$ is a linear map $\mathrm{d}f(p)$. The differential of $f$ is the function $p \mapsto \mathrm{d}f(p)$, which takes value in some set formed of linear functions (usually, $L(E,F)$ for fector spaces or in a space of sections of a vector bundle.) $\endgroup$
    – Didier
    May 30, 2021 at 19:52
  • 1
    $\begingroup$ @Didier: that's precisely what I denoted $\mathrm d_x y$ or $\mathrm d_x f$. $\endgroup$
    – Bernard
    May 30, 2021 at 20:01

The Leibniz notation of differential, also called the derivative at x defines it as $$\frac{dy}{dx}=\lim_{h\rightarrow 0}{\frac{y(x+h)-y(x)}{h}}$$ Therefore $\frac{dy}{dx}$ is the differential and represents the slope of the tangent to the graph of the function at point (x,y).

Further unifying Leibniz with Newton notation, $$\frac{dy}{dx}=y\prime(x)$$

Therefore $dy$ is an infinitesimal difference of y.

When $y\prime$ exists it makes sense to write $$dy=y\prime \cdot dx$$

In formal notation $\Delta$ represents a measurable difference resulted from evaluating a definite integral. To understand this better think in terms of $$\text{average speed}=\frac{\text{total distance}}{\text{total time}}\Leftrightarrow \frac{1}{\Delta t}\int_{t_1}^{t_2}{v dt}=\frac{\Delta x}{\Delta t}$$

However, the symbol is used with the general meaning of “difference” for clarity in algebraic expressions. This includes the Riemann sum which despite being defined as limit sum with an infinite number of terms, is numerically evaluated by a finite number of terms, where again $\Delta$ regains its meaning of measurable difference.


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