Apply Leibniz integral rule to derive the formula for power $P=\mathbf{F}\cdot\mathbf{v}$ I have posed this problem for myself. I am asking it here because I need to know about the calculus specifically, not the physics.

Suppose that $\mathbf{r}$ varies with $t$ and that $\mathbf{F}$ varies with $\mathbf{r}$ and $t$. Using the definitions of power and work, derive $P=\mathbf{F}\cdot\mathbf{v}$.

When integrating across constant limits $a$ and $b$, Leibniz integral rule reduces to $$\frac{d}{dx}\int_a^b f(x,t)\,dt=\int_a^b\frac{\partial}{\partial x}f(x,t)\,dt$$
Using this rule and the definitions of power and work—
$$\begin{align}
P &= \frac{dW}{dt} \\[2ex]
W &= \int\mathbf{F}\cdot d\mathbf{r} \\
\end{align}$$
—I am trying to proceed with the derivation thusly:
$$\begin{align}
P &= \frac{dW}{dt} \tag1\\[2ex]
&= \frac{d}{dt}\int\mathbf{F}\cdot d\mathbf{r} \tag2\\
\end{align}$$
Here is where I get stuck: is the next step
$$P=\int\frac{\partial\mathbf{F}}{\partial t}\cdot d\mathbf{r} \tag{3a}$$
or is it
$$P=\int\frac{\partial}{\partial t}\left(\mathbf{F}\cdot d\mathbf{r}\right) \tag{3b}$$
These are the vibes I am getting for how to execute the derivation:


*

*$\partial/\partial t$ and $d\mathbf{r}$ somehow come together to make $\mathbf{v}$

*$\partial/\partial t$ needs to end up with $\mathbf{F}$ as well somehow

*the integral will somehow “undo” the time-derivative of $\mathbf{F}$


I am also not sure what to do with the extra differential of $\mathbf{r}$: will it become $d(\partial\mathbf{r})$ or $\partial(d\mathbf{r})$, and how does this interact with the integral?
 A: Raskolnikov's comment above sums it up pretty nicely, but I think it is worth looking into why what you wanted to accomplish wasn't getting anywhere. The main problems with your approach, as I see it, is that


*

*The definition of work that you are trying to start from is inadequate; 


but in it is possibly hidden a second problem, namely


*You might be mixing two different variables in one symbol ($t$).


Let's start with the first. Work cannot be defined as a simple indefinite integral. In fact, work (in general) is not a function of finitely many variables. The best you can do is define work done in a particular process, but with explicit limits - from some initial state to some final state. In the case of point mass acted on by a resultant force ${\bf F}(t,{\bf r})$,
$$
W(t) = \int_{{\bf r}(t_0)}^{{\bf r}(t)}\mathbf{F}\cdot d\mathbf{r}
$$
is the total work done on that point mass from time $t_0$ to time $t$. But the dependence of the integrand on time is not an explicit dependence on $t$; the latter only appears in the upper limit of the integral. Finding the time derivative of the integral ultimately has to do exactly with that upper limit, but it is not as easy to recognize that in this form, which is still not a simple one-variable integral, but a line integral; it depends on the particular trajectory which takes us from ${\bf r}(t_0)$ to ${\bf r}(t)$. For non-conservative forces, this is crucial. To make the time-dependence a bit clearer, it is best to follow Raskolnikov's suggestion and rewrite as
$$
W(t) = \int_{t_0}^t \mathbf{F}\cdot \frac{d\mathbf{r}}{dt} dt $$
Now the trajectory information is contained entirely inside the integrand through ${\bf r}(t)$, and the integral is only over time. But the way I've written it is very lazy, and it exposes me to the danger of making a very serious error: I've called both the upper limit and the integration variable $t$. While physicists sometimes write it like that, trusting that they can mentally keep track of which $t$ is which, mathematically it is absurd. So, more properly,
$$
W(t) = \int_{t_0}^t \mathbf{F}(s, {\bf r}(s))\cdot \frac{d\mathbf{r}(s)}{ds} ds $$
where I renamed the dummy variable to $s$ and wrote out the arguments of $\bf F$ and $\bf r$ explicitly.
We now see quite clearly that the only way in which $W(t)$ depends on $t$ is through the upper limit. For this, the rule is
$$
\frac{d}{dt} \int_{t_0}^t \mathbf{F}(s, {\bf r}(s))\cdot \frac{d\mathbf{r}(s)}{ds} ds = \left[\mathbf{F}(s, {\bf r}(s))\cdot \frac{d\mathbf{r}(s)}{ds} \right]_{s = t} = {\bf F}(t, {\bf r}(t)) \cdot {\bf v}(t)$$
so that finally $P = \bf F \cdot v$, where all quantities are evaluated at the current moment $t$.
