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?
On
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.
On
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 $$ \sqrt[3]{28}=f(27+1)\approx f(27)+1\cdot f'(27)=3+\frac{1}{3\sqrt[3]{27^2}}=3+\frac{1}{27}. $$
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.