Isn't the equation for power more complicated? I’ve been told that work is equal to $Fs$ and power is work done  per second $dW/dt$ so looking at the equation, isn't power more like this? $\frac{dF}{dt}s + F\frac{ds}{dt}$
Edit: Now that I'm not in high school, I've learned that work is not $Fs$, it's $\int F \,\mathrm{d}s$, a line integral, apparently. Making it's derivative somewhat more nuanced, I'll add a bit more after I know a bit more vector calculus
 A: Your definition for Power is correct, however, it is more intuitive to think of it as amount of energy transferred per unit time. Now, since the energy transferred per unit time is just the work done, i.e. $$ dW = \mathbf{F} \cdot d\mathbf{s}$$ Here, you have to understand a very essential point, when we wrote $\mathbf {F} \cdot d\mathbf{s}$ we considered that $\mathbf{F}$ was constant through the displacement $d\mathbf{s}$. Therefore, energy transferred per unit time in moving a object through distance of $ds$ with a force $F$ is $$ P = \frac{dW}{dt}$$
Considering, the force and displacement to be in same direction  we have
$$ P = \frac{F~ds}{dt}$$
$$P = F \frac{ds}{dt} \implies P = F~v$$
Now, let's understand this formula a little, we said that our force caused a displacement $ds$, but we should say that force at $t$ caused a displacement $ds$ and if write the force at $t$ as $F(t)$ and surely $ds/dt$ can be written as $v(t)$ (this only means that what was the velocity when the force acted at time $t$) and therefore we can write Power at time $t$ (energy transferred in unit time at time $t$) as $$ P(t) = F(t) ~v(t)$$ The intuition that Power depends on velocity is that whenever we apply a force on an object it's velocity will change, more the velocity change (more the change in Kinetic Energy, $ W = \Delta KE = 1/2 m (v_2^2-v_1^2)$ ) means more energy has been transferred and hence, more the Power.
A: The expression
$$ P = F v \;, $$
is usually used in the context of questions like 

"How much power does the engine need to be able to provide to maintain the vehicle at a constant speed $v$ if the resistive forces have total magnitude $F$?". 

In that case you can say, that the engine must supply a force of magnitude $F$ to establish equilibrium and the power of the engine is:
$$ P = \frac{W}{\Delta t} = \frac{F \, \Delta x}{\Delta t} = F \frac{\Delta x}{\Delta t} = F v \;.$$
And in that context your objection (which is of course generally correct) is a non-issue because $\frac{\mathrm{d}F}{\mathrm{d}t} = 0$.
There are two things that are going on here.


*

*We have chosen a particular force/energy source to evaluate. Notice that the work-energy theorem tells us that the net work done is zero (because we have assumed constant speed), so we must be looking at a selected subset.1

*Because we are assuming a steady state situation we can interchange average and instantaneous quantities with impunity. 
So, if it applies to such a tightly defined condition what use is this expression and why bother to tabulate it in a chapter that mostly contains very broadly applicable principles? Good question. If you know (are given/can guess/have measured) how the resistive forces depend on velocity then you can find


*

*The maximum sustainable speed given maximum power (or a power-curve in the case of, for example, geared internal combustion vehicles)

*"Gas mileage" or other measures of efficiency.


both of which are interesting from a day-to-day perspective as well as from an engineering perspective.

1 Neglecting to state what work is being considered is one of the most common oversights I see in talking about energy/work/power. By students, sure, but also by textbook authors and other who really should know better.
