$$\frac{d^2}{dx^2}xy$$
I know it equals zero but I don't know the in-between steps.
I'm using it to prove that Newton's Laws work in any frame of reference. So, say, two guys start from the same point and one starts moving with velocity $v$. Newton's second law should be the same for both guys.
\begin{align} F &= F_1, \\ a &= a_1, \\ \frac{d^2}{dt^2}x &= \frac{d^2}{dt^2}(x+vt), \\ a &= a + \frac{d^2}{dt^2}(vt),\\ \end{align}
so that last term should equal zero.
$\endgroup$1 Answer
$\begingroup$In the equation $\frac{d^2}{dt^2}x = \frac{d^2}{dt^2}(x+vt)$, you have to know that $v$ is constant (because that's how Galilean transformations work).
Then it's trivial $$\frac{d^2}{dt^2}(x+vt) = \ddot x+v\frac{d^2}{dt^2}t = \ddot x$$
Note that the second term is a special case of $\frac{d^2}{dx^2}(xy)$, where $y$ is a function of $x$:
$$\begin{align}\frac{d^2}{dx^2}(xy) &= \frac{d}{dx}\left[\frac{d}{dx}(xy)\right] \\ &= \frac{d}{dx}\left[y+x\frac{dy}{dx}\right] \\ &= \frac{dy}{dx}+\frac{dy}{dx}+x\frac{d^2y}{dx^2} \\ &= 2\frac{dy}{dx}+x\frac{d^2y}{dx^2}\end{align}$$
If you take $y(x) = \text{constant}$, then this just $=0$ as in your problem.
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