# $dx=\frac {dx}{dt}dt$. Why is this equality true and what does it mean? [duplicate]

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$dx=\frac {dx}{dt}dt$. I know that this deduction is obvious from the chain rule, given that we treat our dx and dt as just numbers. But I find it quite unsatisfactory to think of it in that sense. Is there a better / more "calculus-inclined" way of thinking about this equality. Can you please explain both the LHS and RHS individually.

## marked as duplicate by Emily, Johanna, Chappers, Mark Fantini, N. F. TaussigApr 11 '15 at 0:40

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• Differentials should never be treated as numbers. There are examples where you simply get the wrong answer. – Matt Samuel Apr 10 '15 at 21:59
• @MattSamuel From a mathematician's point-of-view you are absolutely correct. But since calculus is aimed towards engineering and physics students, this way of thinking of derivatives as infinitesimal quantities is quite useful. It gives you a lot of motivation to set up the right equation. – Nicolas Bourbaki Apr 10 '15 at 22:33
• It would be helpful to know how you define $dx$ and $dt$. – A.P. Apr 10 '15 at 22:52
• @JustinThong Please let me know how I can improve my answer. It addressed precisely the rigorous definition of a differential. I don't understand the disconnect. – Mark Viola Apr 10 '15 at 23:02

## 3 Answers

Here's how I would think about it. Imagine you are at the blue dot in the graph below and you are moving along the curve one unit to the right (ΔX = 1). Where will this leave you? You can find out where you will be by taking the slope at the blue point (ΔT/ΔX) and multiplying by ΔT • I think this is misleading. While that's where the notation comes from historically, you shouldn't treat $\frac{dx}{dt}$ as a ratio... – A.P. Apr 10 '15 at 22:48

Statements like $dx = \frac{dx}{dt} dt$ are best viewed as mnemonic devices, and nothing more (until way down the mathematical road, potentially, if at all, maybe).

They are helpful for remembering techniques when it comes to separable differential equations, for example, or the first derivative of a parametric equation:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$

But, do not buy into this notion too seriously. This analogy will break down. For example,

$$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \neq \frac{d^2y/dt^2}{dx^2/dt^2}.$$

• "until way down the mathematical road, potentially, if at all, maybe" that's a very safe way to go about it :) – hjhjhj57 Apr 10 '15 at 22:50
• @hjhjhj57 My thoughts exactly! I've heard talk of these "differential forms", but have not made it there myself; I was indeed alluding to your answer there. – pjs36 Apr 10 '15 at 22:52
• My first approach to differential forms was with David Bachman's book. It's quite accessible and very geometrically oriented if I remember correctly. – hjhjhj57 Apr 10 '15 at 22:54

There are two good ways to think about this.

First - integration (which is usually what this type of notation is shorthand for).

If $x=x(t)$, then $$dx=\frac{dx}{dt}dt=x'(t)\;dt$$ is shorthand for $$x(T)-x(0)=\int_{0}^{T}x'(t)\;dt,$$ which is something you can readily verify in a fully rigorous way. Let $\Pi=\{0=t_{0},t_{1},\ldots,t_{n-1},t_{n}=T\}$ be some partition of the interval $[0,T].$ Then, \begin{align*} x(T)-x(0)&=\sum_{k=1}^{n}x(t_{k})-x(t_{k-1})\\ &=\sum_{k=1}^{n}x'(t_{k}^{*})(t_{k}-t_{k-1})\;\;\;\;(t_{k}^{*}\in(t_{k-1},t_{k}))\\ &\to\int_{0}^{T}x'(t)\;dt\;\text{as}\;n\to\infty. \end{align*} (Note how the first expression is unaffected by the limit $n\to\infty$.)

Intuitively, you can get $\Delta_{[0,T]} x$ by adding up a bunch of $dx$'s or a bunch of $\frac{dx}{dt}dt$'s.

In the second line we use the mean value theorem (or linear version of Taylor's theorem), which says $$x(t+\delta t)-x(t)=x'(t^{*})\delta t$$ for some $t^{*}_{k}\in(t,t+\delta t)$.

This leads us into the second interpretation, which is a fully computable/rigorous approximation to the expression using only finite quantities/concepts you are already familiar with.

If we switch to Leibniz notation and evaluate $dx/dt$ at $t^{*}$ then we have $$\delta x=\frac{dx}{dt}\delta t,$$ and you can see that this is basically the expression you are trying to verify the validity of. However, we do not typically know $t^{*}$ and anyway, the expression is suggesting we can evaluate $dx/dt$ at $t$. But if we do that we only get something slightly different: $$\delta x=\frac{dx}{dt}\delta t+o(|\delta t|).$$ The expression above is computable (since we know all of the quantities involved), except the error term which obeys the bound $\leq C(\delta t)^{2}.$

Essentially, your expression $dx=\frac{dx}{dt}dt$ is the above when $\delta t\to0$ (note that the error term goes to $0$ and the remaining expression should be true). Literally though, both sides ends up going to $0$ in the limit (this is obvious). What is actually important is how the ratio behaves, and that is what the expression is really trying to communicate. Indeed, dividing by $\delta t$ we get $$\frac{\delta x}{\delta t}=\frac{dx}{dt}+\frac{o(|\delta t|)}{\delta t}.$$

The error term still goes to $0$ as $\delta t\to0$, and the result is $$\frac{dx}{dt}=\frac{dx}{dt}.$$

• $\delta x = x'(t) \delta t$ isn't the same as $dx = x'(t)dt$. You need to formally define what $d(\cdot)$ means. – hjhjhj57 Apr 10 '15 at 22:40
• I don't follow you - did you read the entire response? – Sargera Apr 10 '15 at 22:42
• Yes, but $\lim_{\delta t \to 0} \delta t$ = 0. – hjhjhj57 Apr 10 '15 at 22:44
• Correct, so you didn't read it apparently..... – Sargera Apr 10 '15 at 22:44
• You're right, didn't read it close enough. Let me delete my previous comments and elaborate a more precise comment. – hjhjhj57 Apr 10 '15 at 22:46