MHB Differentiating $\mathcal{E}$: How to Reach $\dot{x}(m\ddot{x} + kx)$?

[Dustinsfl]

$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
The derivative is
$$
\frac{d\mathcal{E}}{dt} = \frac{1}{2}m\ddot{x} + kx\dot{x}
$$
but the solution is suppose to be
$$
\frac{d\mathcal{E}}{dt} = \dot{x}(m\ddot{x} + kx).
$$
How?

 

[I like Serena]

$\mathcal{E} = \frac{1}{2}m\dot{x} + \frac{1}{2}kx^2$
The derivative is
$$
\frac{d\mathcal{E}}{dt} = \frac{1}{2}m\ddot{x} + kx\dot{x}
$$
but the solution is suppose to be
$$
\frac{d\mathcal{E}}{dt} = \dot{x}(m\ddot{x} + kx).
$$
How?

Hi dwsmith! :)

The first part of your energy is the kinetic energy.
However, kinetic energy contains speed squared instead of just speed.
So your energy formula should be:

$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$

 

[Dustinsfl]

Hi dwsmith! :)

The first part of your energy is the kinetic energy.
However, kinetic energy contains speed squared instead of just speed.
So your energy formula should be:

$\qquad \mathcal{E} = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$

The book has a typo then.