I have $F = F(x_1(t),x_2(t),\dotsc,t)$, where $x_1,x_2,...$ are (unknown) functions of $t$. Everything is continuous, differentiable, etc.

Is it possibly, necessarily, or never true that $$\dfrac{d}{dt}\left(\dfrac{\partial F}{\partial x_i}(x_1, x_2, \dotsc, t)\right)=\dfrac{\partial}{\partial x_i}\left(\dfrac{d F}{d t}\right)?$$

Edit: expanding the total time derivative: $$\dfrac{d F}{d t} = \sum_j\dfrac{\partial F}{\partial x_j}\dfrac{dx_j}{dt} + \dfrac{\partial F}{\partial t}$$

I don't see an easy way to reduce this, though.

  • $\begingroup$ Try writing out $\dfrac{d \Phi}{d t}$ using the chain rule and see what happens $\endgroup$ Sep 2, 2015 at 22:28
  • $\begingroup$ @DeltaG: After I rewrote the question, I notice that $\frac{\partial}{\partial x} \frac{d\Phi(t)}{dt}$ is constant $0$. If that is not in your intention, I am sorry. I would revert the question to the original state. $\endgroup$
    – user251257
    Sep 2, 2015 at 22:32
  • $\begingroup$ Not sure that I follow why you changed some of the F's to Phis, but not others. I don't see why the RHS is necessarily zero, either, if that's what you're saying. $\endgroup$
    – DeltaG
    Sep 2, 2015 at 22:35

1 Answer 1


For notational convenience, I will write $x$ instead of $x_1,x_2,\dotsc$ As it is not important how many $x_i$'s we have. Also, I write $\partial_x$ instead $\frac\partial{\partial x}$ and so on.

There are two ways to interpret your question.

  1. $x$ is unknown but fixed and you're interested in the derivatives of the function $G(t) = F(x(t), t)$.
  2. $x$ is a variable and you're interested in the derivatives of the operator $H(x, t) = F(x(t), t)$.

In case 1: $\partial_{x} G'(t) = 0$ as $G'$ is not a function of $x$.

In case 2: $\partial_{x} \partial_t H$ is a bilinear operator, namely $$\partial_{x} \partial_t H(x,t)[\Delta x, \Delta t] = \partial^2_{x} F(x(t), t) x'(t) \Delta x(t) \Delta t + \partial_x F(x(t), t) \Delta x'(t) \Delta t.$$

In both cases it is in general not the same as $$ \frac d{dt} \partial_x F(x(t), t) = \partial^2_x F(x(t), t) x'(t) + \partial_t \partial_x F(x(t), t). $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .