How to differentiate $\left(\frac{d}{dt}\left(x\right)\right)^2$?

Question

How does

$\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right)$ = 0

equals to

m$\frac{d^2}{dt^2}\left(x\right)$ + kx = 0

Answer 1

\begin{align}\frac{d}{dt}\left(\frac{1}{2}m\left(\frac{d}{dt}\left(x\right)\right)^2+\frac{1}{2}kx^2\right) &= 0 \\ \Longrightarrow \frac{1}{2} m \frac{d}{dt}\big(\frac{d}{dt}x\big)^2+\frac{1}{2}k \frac{d}{dt}x^2 &= 0 \\ \Longrightarrow m \frac{d}{dt} x\frac{d}{dt}\frac{d}{dt}x+kx \frac{d}{dt}x &= 0 ~\text{ (chain rule)}\\ \Longrightarrow m \frac{d}{dt}\frac{d}{dt}x+k x &= 0 ~\text{ (assuming that $\dot x \neq 0$)} \\ \Longrightarrow m\frac{d^2}{dt^2}x + kx &= 0\end{align}

Answer 2

Hint: $$1/2m(2\dot{x})\ddot{x}+1/2k(2x)\dot{x}=0$$