Assuming we are talking about non-relativistic speeds, then defining $\vec{F}=m\vec{a}$, your proposal (of sorts) is that: $$P=\frac{d\vec{F}}{dt} \cdot \vec{r}+\vec{F} \cdot \frac{d\vec{r}}{dt}=m\left(\vec{r} \cdot \frac{d\vec{a}}{dt}+\vec{a} \cdot \vec{v}\right)$$ with $\vec{v}:=d\vec{r}/dt$.
$d\vec{a}/dt$ is referred to as the "jerk" (or "jolt"), with $\vec{a}$ the acceleration vector. It must occur whenever a force is ramped from $\vec{0}$ to $\vec{F}$. You can learn more about the jerk on its Wikipedia article.
Whether or not the jerk appears in the formula for power depends upon how you define the work: if you use $dW=\vec{F} \cdot d\vec{r}$, then the jerk does not appear. However, if you define $W=\int \vec{F} \cdot d\vec{r}$, then, by applying differentiation under the integral you will have the equation you derived except that $d\vec{F}/dt$ will be replaced by $\partial\vec{F}/\partial t$: According to this differentiation under the integral rule, $$\frac{dW}{dt}=\frac{d}{dt} \int {\vec{r}(t_0)}^{\vec{r}(t)}\vec{F}(\vec{r},t') \cdot d\vec{r}$$ $$=\vec{F}(\vec{r}(t),t) \cdot \frac{d(\vec{r}(t))}{dt}-\vec{F}(\vec{r}(t_0),t_0) \cdot \frac{d(\vec{r}(t_0))}{dt}+\int {\vec{r}(t_0)}^{\vec{r}(t)}\frac{\partial\vec{F}}{\partial t} \cdot d\vec{r}$$ $$=:\vec{F} \cdot \vec{v}-\int _{\vec{r}_0}^{\vec{r}}\frac{\partial\vec{F}}{\partial t} \cdot d\vec{r}.$$ If one then assumes that $\frac{\partial\vec{F}}{\partial t}$ is independent of distance (which is not generally the case), then $$\frac{dW}{dt}=:\vec{F} \cdot \vec{v}-\frac{\partial \vec{F}}{\partial t} \cdot \Delta \vec{r}.$$ However, if one considers what would be required for the jolt to be independent of distance (think of the simple example of ramping a force such that the source of the force is always in contact with the object experiencing the force), one can quickly convince oneself that this is highly unusual. 766