Derivative of $x\dot x$

Question

I found that the derivative of $\mathbf {x \dot x}$ is $\mathbf {x\ddot x + \dot x^2}$. I understand the first part of the result is due to the product rule, but what about the square power in the other part? It should be the product rule AND the chain rule, but how are they applied in this case exactly?

Answer 1

The product rule is

$$(uv)^{\prime} = u^{\prime}v + v^{\prime}u$$

We have

$$u = x, \; v = \dot x \implies u^{\prime} = \dot x, \; v^{\prime} = \ddot x$$

Therefore,

$$(x\dot x)^{\prime} = \dot x\dot x + \ddot xx = \boxed{x\ddot x + {\dot x} ^2}$$

Answer 2

By the product rule: $$\frac{\mathrm d }{\mathrm d t}(uv) = \dot u v + u\dot v.$$

In this case: $u= x$, $v = \dot x$ and $$\frac{\mathrm d }{\mathrm d t}(x \dot x) = \dot x \dot x + x\ddot x = x\ddot x + \dot x ^2.$$

Answer 3

The product rule is the following: $(fg)' = f'g + fg'$. So $(xx')'= x'x' + xx'' = (x')^2 + xx''$

You don't need the chain rule, I guess.