Propagation of Error question? If I have a function $$f(r,V,B)=\frac {2V}{r^2B^2} = 2Vr^{-2}B^{-2}$$ what is the propagation of error?
If I use the power rules and multiplication rules described here http://www.fas.harvard.edu/~scphys/nsta/error_propagation.pdf 
and
http://www.physics.pomona.edu/sixideas/labs/LRM/LR09.pdf
I get:
$$ \sigma_f = \mid f\mid*\sqrt{\frac{4}{r^2}(\sigma_r)^2 + \frac {(\sigma_V)^2}{V^2}+\frac{4}{B^2}(\sigma_B)^2} $$
Is this correct?
 A: Your answer is correct. The underlying theory of propogation is alluded to in the footnote of the first link you provide.
Generally, if you assume that the errors in your measurements (say of, r, V, B to use your example) are unbiased with variance $\sigma_r^2,\sigma_V^2,\sigma_B^2$, respectivley, then a given function of these variables, $f(r,V,B)$, will be affected by the uncertainty in the underlying variables through its first derivative (NOTE: this is only valid if the errors are assumed to be "small", so that the function is well approximated by a line over the expected range of error):
From calculus, the total derivative of $f$ is: $df = \frac{\partial f}{\partial r} dr + \frac{\partial f}{\partial V}dV +\frac{\partial f}{\partial B}dB$. 
Now, we re assuming that $dr, dV, dB$ are small, independent, random errors, with $E[dr]=E[dV]=E[dB]=0$ and variances $\sigma_r^2,\sigma_V^2,\sigma_B^2$.
From the linearity of expectation, we get the usual variance formula:
$\sigma_f^2 = \left(\frac{\partial f}{\partial r}\right)^2\sigma_r^2 + \left(\frac{\partial f}{\partial V}\right)^2 \sigma_V^2 +\left(\frac{\partial f}{\partial B}\right)^2 \sigma_B^2$
Therefore, the standard deviation of $f$ is just the square root of the above. If the errors are assumed to be normally distributed, then $f$ is normally distributed about the estiamted value with the above variance.
