Showing $\ddot{x} = \frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{2} \dot{x}^2)$ In Nonlinear Ordinary Differential Equations: An Introduction for Scientists and Engineers one of the very first stated equations are, as in the title of the question,
$$
  \ddot{x} = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2} \dot{x}^2 \right).
$$
However, I'm having trouble seeing why this should be true. Could anyone clarify this? Thank you for your time in advance.
 A: On segments where the function $x(t)$ is monotonous, you can invert the direction of dependence and find $t(x)$. Then also the derivative can be parametrized by $x$, $\dot x=u(x)$. To this equation one can apply the chain rule for the time derivative
$$
\ddot x = u'(x(t))\dot x = u'(x(t))u(x(t))=\frac12\left.\frac{d(u(x)^2)}{dx}\right|_{x=x(t)}.
$$
By abuse of notation one can now replace the function $u(x)$ with $\dot x(t)$ omitting the arguments and leaving it to the reader to insert the correct independent arguments on-the-fly and write this equation in the claimed form.
A: $\dfrac{d}{dx} \left (\dfrac{1}{2} \dot x^2 \right ) = \dfrac{d \dot x}{dx}\dot x  = \dfrac{d\dot x}{dx} \dfrac{dx}{dt}  = \dfrac{d \dot x(x(t))}{dt} = \dfrac{d \dot x}{dt} = \dfrac{d^2 x}{dt^2} = \ddot x \tag 1$
by the chain rule.
The key here is the observation that, for one-dimensional motion (as (1) appears to describe), the variables $x$ and $t$ may both be taken to be parameters along the curve $x(t)$.
A: Try putting $$ \frac{dx}{dt} = y(x,t).$$
Notice that $y$ is a function of $x$ and $t$. You can then use the chain rule for $$\frac{dy(x,t)}{dt}.$$
From this you should be able to find the expression you want.
A: By the chain rule, $$\frac{d}{dx}\left(\frac12\dot{x}^2\right)=\frac{1}{\dot{x}}\frac{d}{dt}\left(\frac12\dot{x}^2\right)=\frac{1}{\dot{x}}\times\dot{x}\ddot{x}=\ddot{x}.$$
