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We have $W = 10 + 0.895m -12.33c^{0.18} + 12.33c^{0.18}$

we calculate the differential of m, which is $0.895$ and the differential of $c$, which is something like $-12.33/c^{0.82} + 12.33/c^{0.82}$ and what then?

what do we do with $c$. I am confused...

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    The way you've written it, the terms involving $c$ cancel out.2012-11-10

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I'm guessing that you have bounds on the errors in your measurements of $m$ and $c$, and you want to approximate the error in your estimate of $W$.

Use this:

\begin{equation} W(m + \Delta m, c + \Delta c) \approx W(m,c) + \frac{\partial W(m,c)}{\partial m} \Delta m + \frac{\partial W(m,c)}{\partial c} \Delta c. \end{equation}

This equation lets you see how large the quantity $|W(m + \Delta m, c + \Delta c) - W(m,c)|$ might be.

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    Bring $w(m, c) $ over to the left hand side, and you see the sum of two partials approximates the error in $w$.2012-11-11