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I think this is a simple question. If I have that $E(L)=\tau L$ and we are told that $\tau=BTL$ would this mean that $E=BTL^2$ implies $dE=(2BTL)dL$ or should I sub $\tau$ straight into the second law giving $dE=\tau dL=(BTL)dL$?

$\tau=tension \ L=length \ T=temperature \ B= constant$

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    $\begingroup$ Please explain your notations as well. What all $B,L...$ represent. $\endgroup$ – ABC May 30 '13 at 8:06
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Both ways give you the same results:

(1) $E(L)=\tau L$ , $dE(L)=d\tau L + \tau dL = BTLdL + BTLdL = 2BTLdL$

(2) $E(L)=BTL^2$ and then $dE=2BTLdL$

Hope that helps,

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