Time derivative of a function of $x(t)$

Question

I have that $$\dot{x}^2=g\frac{(2bx-x^2)}{(b-x)}$$

where $x$ is a function of time ; $x(t)$. I am trying to compute $\ddot{x}$. I apply the chain rule, as in $\frac{d}{dx}\sqrt{f(x)}=\frac{1}{2\sqrt{f(x)}}\cdot f'(x)$. I have $$\ddot{x}=\frac{1}{2\sqrt{g\frac{(2bx-x^2)}{(b-x)}}}\cdot\frac{d}{dt}g\frac{(2bx-x^2)}{(b-x)}$$

Is this the correct way to proceed? This expression should clear up to $$\ddot{x} = g \enspace+\enspace g\frac{(2bx-x^2)}{2(b-x)}$$

Answer 1

You're on the right track.

$$\begin{align}\frac{d}{dt}\left[g\frac{2bx-x^2}{b-x}\right] &= \frac{d}{dx}\left[g\frac{2bx-x^2}{b-x}\right]\cdot\frac{dx}{dt}\\\\ &=g\frac{(2b-2x)(b-x)-(2bx-x^2)(-1)}{(b-x)^2}\cdot\frac{dx}{dt}\\\\ &=g\left[2 + \frac{2bx-x^2}{(b-x)^2}\right]\sqrt{f(x)}\end{align}$$

Now,

$$\frac{1}{2\sqrt{f(x)}}\frac{d}{dt}\left[g\frac{2bx-x^2}{b-x}\right] =\frac{g}{2}\left[2 + \frac{2bx-x^2}{(b-x)^2}\right] = g\left[1 + \frac{2bx-x^2}{2(b-x)^2}\right] $$

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Alternately,

$\begin{align}&(\dot x)^2 =g\frac{2bx-x^2}{b-x} \\ \Rightarrow& 2\dot x\ddot x= \frac{d}{dx}\left[g\frac{2bx-x^2}{b-x}\right]\cdot\frac{dx}{dt} =g\left[2 + \frac{2bx-x^2}{(b-x)^2}\right]\dot x \\ \Rightarrow& \ddot{x} =g\left[1 + \frac{2bx-x^2}{2(b-x)^2}\right] \end{align}$

Answer 2

I would do the derivative without changing the $x^\prime$ by the real value. $$\dfrac{d}{dt}[(x^\prime)^2] = \dfrac{d}{dx}\left[g\left(\dfrac{2bx - x^2}{(b-x)}\right)\right]$$ $$2x^\prime x^{\prime\prime} = g\left(\dfrac{(2bx^\prime-2xx^\prime)(b-x)+(2bx-x^2)(x^\prime)}{(b-x)^2}\right)$$ $$x^{\prime\prime} =\dfrac{g}{2x^\prime}\left(\dfrac{(2bx^\prime-2xx^\prime)(b-x)}{(b-x)^2}\right) +\dfrac{g}{2x^\prime}\left(\dfrac{(2bx-x^2)(x^\prime)}{(b-x)^2}\right)$$ $$x^{\prime\prime} = g + \dfrac{g(2bx-x^2)}{2(b-x)^2}$$