Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Many times I come across some new formula being used to work with and/or reduce partial differentials. As kleingordon said, these things are mysteriously not taught anywhere(atleast in physics courses). I can't find any list on the internet, either.

I'm talking about formulae like these:

$$\frac{\mathrm{d}}{\mathrm{d}\alpha}\int f(x,\alpha) \mathrm{d}x=\int\frac{\partial f(x,\alpha)}{\partial \alpha}\mathrm{d}x$$

$$\frac{\partial}{\partial x}\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\frac{\partial f}{\partial x}$$ (for continuous functions)

I've also seen that you can stuff a derivative inside a PD $$ \frac{\rm d}{\rm dt}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial \dot f}{\partial x}$$ (Note-$\dot f=\frac{\rm df}{\rm dt}$)

There's also a formula that allows one to split a function into a sum of partial derivatives. I think this is the multivariable chain rule.

I'd like a list of such formulae, or links to these lists. Books are also fine, though I'd prfer internet sources.

share|improve this question
The assertion that these things are mysteriously not taught anywhere is trivially wrong. It might be the case that some physics curricula skip over them too quickly but, for example, the two first identities you mention are a standard part of most (maths) differential calculus courses I am familiar with. The third identity is odd, and begs for some clarification (what is $\dot f$?). –  Did Mar 21 '12 at 6:58
@DidierPiau fixed. en.wikipedia.org/wiki/Newton%27s_notation . The dot signifies a time-derivative. –  Manishearth Mar 21 '12 at 7:37
I think that at least part of the reason why these rules are often found confusing is the persistent abuse of notation that goes on, for example when we write $x=x(t)$ (using the same symbol for a variable and a function). Clearly it makes sense to differentiate with respect to the variable $x$, but (in this context) it doesn't make sense to differentiate with respect to the function $x$. You might find it easier if you write down these formulas using $x=X(t)$ or something similar, to make it clear when you are dealing with a function and when you are dealing with a variable. –  Chris Taylor Mar 21 '12 at 8:38
add comment

2 Answers

$\def\p{\partial}$Here's a proof of your last statement. It uses the chain rule: for functions $x(t)$ and $g(x,t)$ you have

$$\frac{d}{dt} g(x,t) = \frac{\p g}{\p t} + \frac{\p g}{\p x} \frac{dx}{dt} \tag{1}$$

If you take $g=\p f/\p x$, then plugging into (1) gives

$$\frac{d}{dt} \frac{\p f}{\p x} = \frac{\p^2 f}{\p t \p x} + \frac{\p^2 f}{\p x^2} \frac{dx}{dt}$$

On the other hand, if first take $g=f$ and then take the partial derivative with respect to $x$:

$$\frac{\p}{\p x} \frac{df}{dt} = \frac{\p}{\p x} \left( \frac{\p f}{\p t} + \frac{\p f}{\p x} \frac{dx}{dt} \right) = \frac{\p^2 f}{\p x\p t} + \frac{\p^2f}{\p x^2} \frac{dx}{dt}$$

You can compare the right-hand sides of these expressions and see that they are equal (since partial derivatives commute). Therefore

$$\frac{d}{dt} \frac{\p f}{\p x} = \frac{\p}{\p x} \frac{df}{dt}$$

so the partial derivative wrt $x$ commutes with the total derivative wrt $t$.

share|improve this answer
Yeah, I know that PDs commute, I was surprised that they seem to commute with total derivatives. So you're saying that bringing the dot inside is just a notation and cannot be written as a total derivative inside? –  Manishearth Mar 21 '12 at 8:15
Oh, hold on - I've just realized what the notation means. I'll rewrite my answer. –  Chris Taylor Mar 21 '12 at 8:20
Guess it was physics prejudice on my part--I thought dots (being widely used by Newton) would be understood easily. –  Manishearth Mar 21 '12 at 8:28
Well, I've seen the dot notation before - I was just having a slow moment! It can be a useful notation, but it's prone to abuse. for example, in Lagrangian mechanics its common to see both $q$ and $\dot{q}$ treated as independent variables in the same equation. While you can make the math work out, I think that this practice tends to obscure what's going on. –  Chris Taylor Mar 21 '12 at 8:30
Yeah, when I checked out Lagrangian mechanics a while ago, that confused me as well. But this answer only proves the third statement. What I would like is a list of as many such formulae as possible. Or a link to one :) –  Manishearth Mar 21 '12 at 8:45
add comment

An identity related to the first formula that might come in handy is when you have to differentiate under the integral sign, but the limits of integration are functions of the variable you're differentiating with respect to. Then:

$F = F(\alpha) = \int_{x_1(\alpha)}^{x_2(\alpha)}f(x,\alpha)dx$

$\frac{dF}{d\alpha} = F'(\alpha) = f(x_2,\alpha)\frac{dx_2}{d \alpha} - f(x_1,\alpha)\frac{dx_1}{d \alpha} + \int_{x_1(\alpha)}^{x_2(\alpha)}\frac{\partial f(x,\alpha)}{\partial \alpha}dx.$

share|improve this answer
Used that one as well, but thanks for posting! –  Manishearth Mar 21 '12 at 9:54
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.