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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?

Note:

$\alpha_L$ = 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 at 17:48

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Possible duplicate of physics.stackexchange.com/q/93514 –  Emilio Pisanty Jan 13 at 18:13

1 Answer 1

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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