Law of Conservation of Energy

Theorem. Let $\vec{c} (t)$ as a position of the particle of mass $m$ at time $t$. Let $V : \mathbb {R}^3 \rightarrow \mathbb{R}$ as a potential function, having $- \nabla V$ as a force field. Then

$$\frac 1 2 m \| \vec{c}'(t) \|^2 + V( \vec{c}(t))$$  

is constant.

proof. Let above equation as $E(t)$, and set our aim as showing $E'(t) = 0$. Thereby approaching $V(\vec{c}(t) )$ first.

By chain rule, we get following equation. Note that $\cdot$ is an inner product.

$$\frac {d}{dt} V( \vec{c} (t) ) ) = \nabla V (\vec{c} (t)) \cdot \vec{c} ' (t)$$  

Since force field is $-\nabla V$, we can use Newton's law.

$$- \nabla V ( \vec{c}(t)) = m \vec{c}'' (t)$$

Using this fact,

$$\frac {d}{dt} V(\vec{c} (t) ) =\nabla V (\vec{c} (t)) \vec{c} ' (t) = - m \vec{c} '(t) \cdot \vec{c}''(t)$$

Now consider the other part.

$$\frac {1}{2} m \left( \frac {d}{dt} \| \vec{c}'(t) \|^2 \right) = \frac 1 2 m \frac {d}{dt} ( \vec{c}'(t)  \vec{c}'(t) ) = \frac 1 2 m ( 2 \vec{c}'(t) \cdot \vec{c}''(t) )= m \vec{c}' (t) \cdot \vec{c}'' (t)$$

Therefore $E'(t)=0$ and $E(t)$ is constant among $t$.