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I get how to use differentials to compute error, but why is it a "good" method? For example, a standard problem is something like:

If the radius of a circle is $3 \pm 0.1$ cm, find the area with error.

What's wrong/undesirable with just finding the area for $r=2.9$ and $r=3.1$ to determine the error?

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    $\begingroup$ In some cases, the maximum error might not occur at the maximum deviation from the given value. To give a pathological example, find the error of the function $$ f(x)= \frac{1}{x-2} $$ on the interval $2.1\pm 0.2$. The biggest error (so to speak, I guess) occurs at $x-2$, which is not an endpoint of the error interval. $\endgroup$ – Ian Coley Mar 2 '14 at 21:24
  • $\begingroup$ @IanColey I disagree: simply calculate the increment $\Delta f(x) = f(2.1) - f(2.1 \pm 0.2)$ and see which of the two errors is the largest. If the input is 2.3, then the error is $10-3.33=6.67$, and if the input is 1.9, the error is $10- (-10)=20$. Clearly, the latter yields the largest error. This has the additional benefit of avoiding figuring out what the derivative of $f(x)$ is. $\endgroup$ – Ius Klesar Jun 13 at 11:26
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The nice thing about using differentials is that as you adjust the error, you don't need to compute the highest and lowest again; you just need to multiply some coefficient.

So, for your example, we might have a situation where we ask "if the radius $3 \pm 0.1$ cm, what's the error?" But the more useful question is "if we want the area to be within $.01$ of $3 \text{cm}^2$, how small does the error in the radius need to be?" In this case, we could solve for $\Delta r$, saying that if our radius is $3 \pm \Delta r$ cm, then $$ \Delta A \approx 2 \pi (3 \text{cm})\Delta r \leq .01 \text{cm} $$ and solve for $\Delta r$.

On the other hand, doing things the other way, we would have to solve something like $$ \Delta A = \pi (3\text{cm} + \Delta r)^2 - \pi (3\text{cm})^2 \leq .01 \text{cm} $$ In this situation, we simply have a quadratic equation, but in others we could wind up with something much more complicated.

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If I understand your question correctly, there are a couple of reasons I can see. First, you can use a general formula instead of computing with specific numbers. Second, you can usually see how the error in measuring (say) $r$ affects the error in the computed value of $A(r)$.

For $A(r)=\pi r^2$, if the measured radius is $r\pm\epsilon$, then the computed area is $A(r\pm\epsilon)\approx A(r) \pm A'(r)\epsilon= \pi r^2\pm 2\pi r\epsilon$. This shows that the error in area is about $2\pi r$ times the error in measuring the radius.

I hope that's the sort of thing you were looking for.

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Sometimes you cannot compute the error exactly. For instance, how do you compute the cubic root of 28? If you have a calculator, it is easy. Or you can do it by hand noting that $\sqrt[3]{27}=3$. Let $f(x)=\sqrt[3]{x}$. Then $$ f'(x)=\frac{1}{3\sqrt[3]{x^2}} $$ and $$ \sqrt[3]{28}=f(27+1)\approx f(27)+1\cdot f'(27)=3+\frac{1}{3\sqrt[3]{27^2}}=3+\frac{1}{27}. $$

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