For a function $f(x)$ that is differentiable at $x$, the derivative of $f$ with respect to $x$ at the point $x$ is usually given as $$\frac{df(x)}{dx} = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}.$$

I think a much more natural (and perfectly rigorous) definition is one that more similarly resembles the usual $\frac{df(x)}{dx}$, namely $$\frac{df(x)}{dx} = \lim_{dx\to0} \frac{df(x)}{dx},$$ where $df = f(x + dx) - f(x).$

This is essentially the same thing, but to me it carries more meaning.

To me, not only does this more clearly show what the derivative means (the gradient of $f$ at $x$, as opposed to between two points), but it also allows $dx$ -- and likewise $df$ -- to be understood as variables like any other, rather than some sort of new notation (which is what I and every one of my classmates thought when first taught differentiation in school). It justifies the algebraic manipulation of $dx$ as a variable like any other, for example in Leibniz's notation for the chain rule: $\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}$. It would no longer require teachers to answer the question, "but... can we do that?" with, "yes, just trust me." (a teacher would likely need to remind students that $dx$ approaches zero; algebraic operations are merely applied before the limit is evaluated)

It also lends to easier understanding of the integral by making concrete the idea that $dx$ is just another variable: $$\int_a^b f(x)\ dx$$ is just the sum of all values of f(x) times $dx$ in the interval $[a,b]$. Since multiplication distributes over addition, it allows the above integral to be (perfectly rigorously) written as $$dx \int_a^b f(x),$$ provided it is understood that this represents the sum as $dx$ approaches zero.

It also reveals why differentiation and integration are inverse operations: the derivative of $f$ is $f(x)$ divided by $dx$, while the integral of $f$ is $f(x)$ multiplied by $dx$ (roughly speaking). If the integral of a function's derivative is taken: $$\int \frac{df}{dx} \, dx,$$ the $dx$s evaluate to $1$, which leaves $$\int df,$$ the sum of infinitely many infinitesimal changes in $f$, which is just $f(x)$.

Anyway, I was wondering why we don't use $dx$ as the limiting variable in the derivative definition -- I see no reason why the limiting variable and the variable of differentiation can't be the same thing.

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    $\begingroup$ that is not perfectly rigorous $\endgroup$
    – janmarqz
    Apr 26, 2016 at 17:16
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    $\begingroup$ The widths of the intervals $\Delta x$ are not assumed to be uniform in the definition of the definite integral. Even if they were, you can't assume that this represents a sum, it doesn't. Even if this was allowed, $dx$ doesn't "approach zero". There is a problem in your understanding in the previous paragraph as well. The definite integral is not "just the sum of all values of f(x) times $dx$". I agree that the concepts being applied are delicate. But the definition of the definite integral requires a two sided convergence to a single value (sometimes called the "pinching theorem"). $\endgroup$ Apr 26, 2016 at 17:35
  • $\begingroup$ In the Riemann sum for integrals, the height of the rectangles is $f(a + k \Delta x)$ where $k$ is the iterator of the sum, and the width of the rectangles is $\Delta x$ -- this is what I meant by "all the values of f(x) times $dx$" When you say the definite integral doesn't represent a sum, what does it represent? i thought that the antiderivative and the Riemann sum were the same thing (or at least evaluated to the same thing) $\endgroup$
    – Nubtom
    Apr 26, 2016 at 17:40
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    $\begingroup$ My guess is that $dx$ already means infinitesimal so there is no point to take its limit $\endgroup$
    – High GPA
    Jun 30, 2020 at 21:32
  • $\begingroup$ @Nubtom as previously said the width may not be uniform and again it is not a sum nor the derivative is the quotient which you are thinking $\endgroup$ Feb 18, 2021 at 13:31

1 Answer 1


It's common to write $\Delta x$ instead of $h$. It would be bad to use $dx$, because then $dx$ would mean two different things.

  • $\begingroup$ Which two different things? $\endgroup$
    – Nubtom
    Apr 26, 2016 at 17:43
  • $\begingroup$ Well, we'd be using it to mean both dx in the usual sense and also $\Delta x$. $\endgroup$ Apr 26, 2016 at 21:29
  • $\begingroup$ But what's wrong with that? To me, if $x$ is the variable being differentiated with respect to, then a small (infinitesimal) change in $x$ -- $dx$ -- is what should be approaching zero. $\endgroup$
    – Nubtom
    Apr 27, 2016 at 7:25
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    $\begingroup$ @Nubtom It is very bad to have a definition of the form $a:=\Phi(a)$. $\endgroup$ May 6, 2016 at 17:11
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    $\begingroup$ @bof why would you want to use the same notation for two distinct concepts, namely $\Delta x$ or $h$ which is a real number vs $dx$ which in differential geometry is a 1-form (and even if you want to avoid terminology like "1-form", this is really what's going on). Also, if you were really taught $dx=\Delta x$ then that is just very poor form. It's like saying an apple is the same thing as a tomato, and calling them both by the term "apple" just because they're both red. It's just all sorts of confusing, misleading and wrong. $\endgroup$
    – user580918
    Feb 18, 2021 at 3:46

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