Given the diferentials $$ \begin{equation} d'Q=K(x-x')d'x \end{equation} $$ $$ \begin{equation} d'Q'=K(x'-x)d'x' \end{equation} $$
where $K$ is a constant, I need to show that
$$ \begin{equation} \frac{d'Q}{dt}+\frac{d'Q'}{dt}=\frac{d}{dt}\left[\frac{K}{2}\left(x-x'\right)^{2}\right] \end{equation} $$
How can I do so?
Attempt:
$$ \begin{array}{ll} \frac{d}{dt}\left(d'Q+d'Q'\right) & =K\frac{d}{dt}\left[(x-x')d'x+(x'-x)d'x'\right]\\ & =K\left[(x-x')\frac{d}{dt}\left(d'x\right)+d'x\frac{d}{dt}(x-x')\right.\\ & \qquad\left.(x'-x)\frac{d}{dt}\left(d'x'\right)+d'x'\frac{d}{dt}(x'-x)\right]\\ & =K\left[(x-x')\frac{d}{dt}\left(d'x-d'x'\right)+\left(d'x-d'x'\right)\frac{d}{dt}(x-x')\right]\\ & =? \end{array} $$
In formal terms (I set $K=1$ and $d' = d$), \begin{align*} \frac 1 2\frac d{dt}(x-x')^2 &= (x-x')\left(\frac{dx}{dt}-\frac{dx'}{dt}\right)\\ &= \frac{1}{dt}\big((x-x')dx + (x'-x)dx'\big)\\ &= \frac{1}{dt}(dQ+dQ'). \end{align*}