Why dont you take derivative of force in definition of power ? P=F.v The derivative of work is $\bf F\cdot v .$
$$P(t)= \frac{\mathrm dW}{\mathrm dt}= \mathbf{F\cdot v}=-\frac{\mathrm dU}{\mathrm dt}\;.$$
But why not $$\frac{\mathrm{d}\mathbf{F}}{\mathrm{d}t}\cdot \mathbf x + \mathbf{F\cdot v}\;?$$
 A: This is because $P(t)$ as stated is the instantaneous power as a function of time and $W =\mathbf  F\Delta \mathbf x$ holds only for constant forces. More generally, recall that a definition of work is the integral:
$$W = \int_C\mathbf F(x)\mathrm d\mathbf x$$
Where $C = C(x,t)$ is some curve in space/time. Expressing in terms of time gives:
$$W = \int_t\mathbf F(t)\frac{\mathrm dx}{\mathrm dt}\mathrm dt = \int_t\mathbf F(t)\mathbf v(t)\mathrm dt$$
From the Fundamental Theorem of Calculus, we have:
$$P \equiv \frac{\mathrm dW}{\mathrm dt}= \mathbf F(t)\mathbf v(t)\;.$$
Apologies for sloppy notation, but that's the idea.
A: For a force to do work, it must act upon an object as that object moves some distance (with a component along or opposed to the direction of that force). Holding an object stationary, but carrying the strength of a given force, adds no more work done. Because of this, we say that 
$$W=\int_C\vec{F}\cdot d\vec{r}$$
where C is the curve in space along which the object moves. If power is the rate of change of work, then we would expect a product rule to be applied in the derivative, but no power is expended by a force that does no work, and without a displacement, there is no work and no power. 
