Derivation of Conservation of Energy from Newton's Second Law Given Newtons's Second Law: $$ \frac {d}{dt} (m \boldsymbol{\dot r}) = \mathbf F $$
How is it possible to derive the conservation of energy equation with a constant mass?
That is how can you derive $ \mathbf F = - \nabla V(\mathbf r) $ where $V(\mathbf r)$ is shown to be the potential energy when the force is conservative?
Attempted Proof:
Let $KE = T = \frac {1}{2} m\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} $
or
$$\frac{dT}{dt} = \frac{1}{2}m[\boldsymbol{\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt} + \boldsymbol{\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt}]  = m \boldsymbol {\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt}$$
and $\nabla V = \frac {\partial V} {\partial{\mathbf r}} $
Also, a conservative force says $\frac{dE}{dt} = 0$
Newton's Second Law could also be written as: $$m\boldsymbol{\dot r} \cdot \frac{d\boldsymbol{\dot r}}{dt} = \mathbf F \cdot \boldsymbol{\dot r}$$ 
My question is how is $ \mathbf F = - \nabla V(\mathbf r) $ introduced to Newton's second law properly and then integrated (? maybe) to obtain the energy? 
Because I can easily prove that $ \mathbf F = - \nabla V(\mathbf r) $ if $E = T + V(\mathbf r)$ but I am trying to conclude that $V(\mathbf r)$ is the potential energy, not assume it 
 A: Defining $\vec{v}=\dfrac{d\vec{r}}{dt}$ and $\vec{a}=\dfrac{d\vec{v}}{dt},$ we have:
$$\dfrac{dE}{dt}=\dfrac{d}{dt} \left(\frac{1}{2}m\vert \vec{v} \vert^2+V  \right)=m\vec{v} \cdot \vec{a}+\dfrac{dV}{dt}.$$
Next notice that because of the chain rule we have:
$$\dfrac{dV}{dt}=\nabla V \cdot \vec{v},$$ so that we have:
$$\dfrac{dE}{dt}= m\vec{v} \cdot \vec{a}+\nabla V \cdot \vec{v}$$
Next we use Newton's second Law, which states $m\vec{a}=\vec{F}.$ If we assume that the force field is conservative, which means that the force is the gradient of a scalar field (the potential energy), we further have $F=-\nabla V,$ which finally yields:
$$\dfrac{dE}{dt}=-\vec{v}\cdot \nabla V+\nabla V \cdot \vec{v} = 0$$
A: Here is another approach to the question.
We know that work done by the force, $dW = \vec{F}\cdot d\vec{r}$. 
Now, if the force is conservative, by definition it means that work done is independent of the path taken and only depends on the end state. Hence $dW = \vec{F}\cdot d\vec{r}$  should be exact, i.e. writable as $V(\vec r_f) - V(\vec r_o) $.
This translates to $\vec{F}\cdot d\vec{r} $ being equal to total derivative of some function $V(\vec r)$ that depends on position only.
Hence, $$\vec{F}\cdot d\vec{r} = dV(\vec r)$$
$$\vec{F} = \frac{dV(r)}{d\vec{r}} = \nabla{V(\vec r)} $$
Now, we see that work done by the conservative force $F$ is expressed by $V(r)$, which we define as the potential energy.
A: As stated, the answer is: you cannot. There are force fields which are not conservatives, i.e. they cannot be written as the gradient of a scalar function. In such situations energy is not conserved. Such examples are quite common, consider for example friction.
Note that, by definition, a force is said to be conservative if it can be written as the gradient of a scalar function.
