# Simple exercise with differentials

Given the diferentials $$$$d'Q=K(x-x')d'x$$$$ $$$$d'Q'=K(x'-x)d'x'$$$$

where $K$ is a constant, I need to show that

$$$$\frac{d'Q}{dt}+\frac{d'Q'}{dt}=\frac{d}{dt}\left[\frac{K}{2}\left(x-x'\right)^{2}\right]$$$$

How can I do so?

Attempt:

$$\begin{array}{ll} \frac{d}{dt}\left(d'Q+d'Q'\right) & =K\frac{d}{dt}\left[(x-x')d'x+(x'-x)d'x'\right]\\ & =K\left[(x-x')\frac{d}{dt}\left(d'x\right)+d'x\frac{d}{dt}(x-x')\right.\\ & \qquad\left.(x'-x)\frac{d}{dt}\left(d'x'\right)+d'x'\frac{d}{dt}(x'-x)\right]\\ & =K\left[(x-x')\frac{d}{dt}\left(d'x-d'x'\right)+\left(d'x-d'x'\right)\frac{d}{dt}(x-x')\right]\\ & =? \end{array}$$

• What is $d'$?.. – amsmath Aug 29 '18 at 16:21
• I am sorry, but as long as you cannot explain the symbols you use, it is not possible to help you. – amsmath Aug 29 '18 at 18:20
• @amsmath Ok, it's meant to represent an infinitesimal. From mathworld: "An infinitesimal is some quantity that is explicitly nonzero and yet smaller in absolute value than any real quantity". – Igaturtle Aug 29 '18 at 18:29
• So, $d'Q$ is the same as $dQ$? – amsmath Aug 29 '18 at 18:30
• @amsmath $d'Q$ is in the context of thermodynamics usually called a differential; here in particular it means the average heat transferred over some stochastic process between $t$ and $t+dt$. – Igaturtle Aug 29 '18 at 18:37

## 1 Answer

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*}

• I tried that before but I was unsure if it was on good grounds to work the infinitesimals with such freedom... Since you have more of a formal knowledge maybe you can assure me it is? – Igaturtle Aug 29 '18 at 19:15
• Since it's physics, it is. ;-) – amsmath Aug 29 '18 at 19:18
• -_- ... Thanks! – Igaturtle Aug 29 '18 at 19:20