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.

• What is $\dot x$ in this case? Is it the derivative to another variable, say $t$? – WarreG Jan 7 at 18:54
• My guess, you want to know, why $$\frac{d\dot{x}}{dx}\frac{dx}{dt}=\frac{d}{dx}\Big(\frac{1}{2}\dot{x}^2\Big)$$ Right? – Fakemistake Jan 7 at 19:06
• Yes! Is it a simple consequence of the chain rule? Christmas has made me rusty. – Erik André Jan 8 at 14:38

$$\dfrac{d}{dx} \left (\dfrac{1}{2} \dot x^2 \right ) = \dfrac{d \dot x}{dx}\dot x = \dfrac{d\dot x}{dx} \dfrac{dx}{dt} = \dfrac{d \dot x(x(t))}{dt} = \dfrac{d \dot x}{dt} = \dfrac{d^2 x}{dt^2} = \ddot x \tag 1$$
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 $$x(t)$$.
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.
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.
By the chain rule, $$\frac{d}{dx}\left(\frac12\dot{x}^2\right)=\frac{1}{\dot{x}}\frac{d}{dt}\left(\frac12\dot{x}^2\right)=\frac{1}{\dot{x}}\times\dot{x}\ddot{x}=\ddot{x}.$$