I have trouble of understanding proof of work-energy formula. This formula is from Salas's and Hill's calculus textbook. There is one step that I don't understand. $\mathbf r''(t) \cdot \mathbf r'(t)=\frac{d}{dt}(\frac{1}{2} \mathbf r'(t)\cdot \mathbf r'(t))$ Why it is valid?
That is why aceleration vector and velocity vectors dot product is half of velocity vector squared?
Recall the product rule: $$\dfrac{\mathrm d}{\mathrm dt}(uv)=u\dfrac{\mathrm dv}{\mathrm dt}+v\dfrac{\mathrm du}{\mathrm dt}$$ Putting in $u=v=\mathbf{r}'$, we get $$\dfrac{\mathrm d}{\mathrm dt}\Big(\mathbf{r}'(t)\cdot\mathbf{r}'(t)\Big)=2\mathbf{r}'(t)\cdot\mathbf r''(t)$$ Hope this helps. Ask anything if not clear :)