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$\alpha_L = \frac{1}{a}\frac{da}{dT}$

I know error in $a$, i.e., $da$

I need to find out $d\alpha$ from data of $da$.

$d\alpha_L = -\frac{1}{a^2}\frac{da}{dT}da$

Is this correct?


$\alpha_L$ = Linear thermal expansivity

$a$ = Lattice constant

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closed as off-topic by Kyle Kanos, Brandon Enright, ja72, John Rennie, Emilio Pisanty Jan 13 '14 at 17:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, Brandon Enright, ja72, John Rennie, Emilio Pisanty
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Possible duplicate of – Emilio Pisanty Jan 13 '14 at 18:13

You are confusing yourself with two different uses of $da$, it would be better to make the error $\delta a$ Then if you consider $\frac {da}{dT}$ to be a constant you will be correct.

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