I'll start with an example.

In physics, $x(t)$ represents the $x$-position of a particle, and $v(t)$ its ($x$-)velocity. To determine the total displacement of a particle on the interval $[a, b]$, we can use the formula

$$\Delta x = \int_a^b{v(t)~dt}$$

To me, this makes sense, because $v = \frac {dx}{dt}\\$, so the above equation is equivalent to:

$$\Delta x = \int_a^b{v(t)~dt} = \int_a^b{\frac {dx}{dt}~dt} = \int_a^b{dx} = x(b) - x(a) = \Delta x$$

However, I've been told that you can't just "cancel" the $dt$ differential because it's not "proper."

Another example uses parametric arc length:

$$\ell = \int_a^b{\sqrt{\left ( \frac {dx}{dt} \right )^2 + \left ( \frac {dy}{dt} \right )^2} dt}$$

Now take a standard function, $y = f(x)$. We can define $x = t$, and then we have $y(t) = f(t)$, $x(t) = t$, and $\frac{dx}{dt} = 1$. Then, $\frac {dy}{dt} = \frac {{dy}~/~{dx}}{{dx}~/~{dt}} = \frac {{dy}~/~{dx}} 1 = \frac {dy}{dx}$, and our formula simplifies to

$$\ell = \int_a^b{\sqrt{1 + \left ( \frac {dy}{dx} \right )^2} dt}$$

This is indeed the correct formula for arc length of a function (also derivable with Pythagorean theorem), but it relies on being able to cancel the $dx$s in $\frac {{dy}~/~{dx}}{{dx}~/~{dt}}$.

So, my question is, when, if ever, can you cancel differentials?

  • 2
    $\begingroup$ One does not simply cancel differentials. $\endgroup$
    – Pedro
    May 3, 2013 at 0:07
  • 1
    $\begingroup$ @PeterTamaroff: Exactly; it's not simplified until you cancel them :) $\endgroup$
    – wchargin
    May 3, 2013 at 0:10

1 Answer 1


You've clearly demonstrated one of the dangers of a naive notion of "cancelling" differentials: recall from elementary calculus that

$$\int_a^b{dx} = b-a$$

In particular, both equalities in $$\int_a^b{\frac {dx}{dt}~dt} = \int_a^b{dx} = x(b) - x(a)$$

are incorrect. The only reason you got the correct result at the end is because you remembered what you meant to do, and ignored what you actually did.

The "right" thing to do from an elementary calculus standpoint is the change of variable formula. In particular, you would have

$$ \int_a^b \frac{dx}{dt}~ dt = \int_{x(a)}^{x(b)}~dx$$

In a realistic problem, you might have actually believed yourself when you got to the expression $\int_a^b \, dx$ and gotten the wrong answer. A lot of beginning students get greatly confused over similar mistakes.

You've given yourself extra chances to get the wrong answer in your formula for arclength, since what you wrote isn't even a differential; e.g. you probably meant to multiply the integrand by $dt$ in the first formula for $\ell$.

When you're more comfortable with differenetial forms and a more differential-geometric-like notation (which I strongly suspect you implicitly are without knowing what that means), you will recognize that differential forms satisfy an identity

$$ \frac{dx}{dy} \, dy = dx $$

whenever $x$ is differentiable with respect to $y$ (and there is only one "degree of freedom"). And if $\gamma$ is some path through the "state space" that $x$ and $y$ measure, then

$$ \int_\gamma f \frac{dx}{dy} \, dy = \int_\gamma f \, dx $$

just as one might expect. $f$ is supposed to be any scalar field in the above. The important distinction between this and the calculus learned in elementary calculus is that we are thinking in terms of a path through "state space" rather than an integral between numeric limits, and that $x$ and $y$ implicitly have some meaning as (dependent) variables on "state space" rather than being indeterminate "dummy" variables.

  • $\begingroup$ Could you clarify "state space" and $\gamma$? I've only seen the term in relation to chaos theory. $\endgroup$
    – wchargin
    May 3, 2013 at 0:27
  • $\begingroup$ @WChargin: One common scenario is that some variable (e.g. $t$) is given special status, and the state space is implicitly defined to be the domain of values that $t$ ranges over. Then if we had a functional dependence like $v = v_0 - g t$, then $v$ would have value $v_0 - g$ at the point in the state space where $t$ has value $1$. But IMO it all works out best when the state space remains completely behind the scenes and you can work in terms of the variables. $\endgroup$
    – user14972
    May 3, 2013 at 0:34
  • $\begingroup$ In physics, one often explicitly talks about state spaces, which are often things that can be used for the purpose I describe. Things like "configuration space" can be used for this purpose too. $\endgroup$
    – user14972
    May 3, 2013 at 0:37

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