Conservative vector fields and Energy

Question

If one knows that $\vec{F}=m\vec{a}$ is a conservative force, which means that $\vec{F}=\nabla\phi$, and that energy $E(t) = \frac{1}{2}mv(t)^2 + \phi(\vec{x}(t))$, where $v(t) = \|\vec{v}(t)\|$ then it can be deduced that $E'(t) = m\frac{d}{dt}(v\cdot v) + \frac{d\phi}{dt} = m(\vec{v}\cdot \vec{a} - \vec{a}\cdot \vec{v}) = 0$.

(Corrections were made as the discussion went).

Answer 1

Where did you get $\vec x$ on RHS of your derivative of energy? I get:

$E'(t) = m \vec a \cdot \vec v+ \nabla \phi \cdot \vec v = \vec v \cdot (m \vec a - \vec F)=0$

Answer 2

I think by definition $\vec F = -\nabla \phi$

I am not sure how you $E'(t)$, but to me it looks like this - $$ E'(t) = m\frac{d}{dt}(\vec v.\vec v) + \frac{d\phi}{dt} = m(\vec v.\vec a + \frac{\nabla\phi.\vec v}{m} + \frac{\partial \phi}{\partial t}) = \vec v.(m\vec a + \frac{\nabla\phi}{m}) = 0 $$