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$$j^{q}=\frac{1}{2} n v[\varepsilon(T[x-v \tau])-\varepsilon(T[x+v \tau])]$$

To this:

$$j^{q}=n v^{2} \tau \frac{d \varepsilon}{d T}\left(-\frac{d T}{d x}\right)$$

At first I was thinking of using the fundamental theorem of calculus but I can't seem to do it. Any words of advice would be appreciated.

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closed as off-topic by Chair, Aaron Stevens, M. Enns, Buzz, ZeroTheHero Feb 9 at 5:25

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Use:

$$\frac{d\epsilon}{dT}=\frac{\epsilon(T+dT)-\epsilon(T-dT)}{2dT}$$

and:

$$dT=\frac{dT}{dx}dx=\frac{dT}{dx}v\tau$$

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  • $\begingroup$ I was blind but now I can see, thanks jonny $\endgroup$ – JohnBenz Feb 9 at 3:23
  • $\begingroup$ you are welcomed, lol! $\endgroup$ – Wolphram jonny Feb 9 at 3:24
  • $\begingroup$ But where does the - come from ? edit: nope I got it. $\endgroup$ – JohnBenz Feb 9 at 3:37

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