Energy transformation for pendulum

Question

The equation of motion of a mass on a pendulum is

$$ \ddot x + \omega^2\sin x = 0 $$

where $t$ = time and $x$ = inclination from the vertical.

According to my book, it is possible to re-write $ \ddot x $ in terms of $ \dot x $ and $ x $ by using:

$$ \begin{align} \ddot x & = \frac{d \dot x}{dt} \\ & = \frac{d \dot x}{dx} \frac{dx}{dt} \\ & = \frac{d}{dx}\left(\frac{1}{2}\dot x^2\right) \end{align}$$

I don't understand where the $ \frac{1}{2} $ comes from. It seems to me that since $\dot x = \frac{dx}{dt}$ we should end up with

$$\frac{d}{dx} \left( \dot x \frac{dx}{dt} \right) = \frac{d}{dx} (\dot x \dot x) = \frac{d}{dx} (\dot x^2) $$

Can anyone tell me where the $1/2$ comes from? Thank you in advance.

Answer 1

let assume that $$\dot x(t)=f(x(t)) $$

You agree that $$\ddot x(t)=\frac{df(x(t))}{dt}=f'(x(t))\dot x(t) $$

In the other hand ,

$y(t)=\frac{1}{2}f(x(t))^2$ has for derivative with respect to x

$$\frac{dy(t)}{dx}=\frac{1}{2}*2*f(x(t))*f'(x(t))$$

Answer 2

Up to the lack of rigor in your book, note that $$ \frac{d \dot x}{dx} \frac{dx}{dt} = \frac{d \dot x}{dx} \dot x=\frac{d}{dx}\left(\frac{1}{2}\dot x^2\right), $$ thinking of $\dot x$ as $\dot x=\dot x(x)$ (the new variable being $x$).

Answer 3

You have to consider the product rule when looking at this equation:

$$ \frac{d}{dx}(u*v) = v*\frac{du}{dx} + u*\frac{dv}{dx}\\ $$ Since, in this case, $u = v$ you will obtain the following equation: $$ \frac{d}{dx}(\frac{1}{2}\dot{x}^2) =\frac{1}{2}*2\dot{x}*\frac{d\dot{x}}{dx} =\frac{d\dot{x}}{dx}\frac{dx}{dt} $$