Propagating uncertainty in logarithmic calculation I have a calculation using the following pseudo-formula:
$ y = -ln(X/X_o)$
where both $X$ and $X_o$ have an associated error with them. I have propagated the error out simply using:
$ \delta y = y\sqrt{\frac{\delta X}{X} + \frac{\delta X_o}{X_o} } $
However, I'm not sure how to account for the use of the natural log - I feel like this is adequate since my general understanding is that the deviation $S_p$ in a measurement $p$ is found with:
$S_X = \frac{S_p}{p}$
My other thought is to calculate the error as:
$ \delta y = y\sqrt{\ln({\frac{\delta X}{X})} + \ln{(\frac{\delta X_o}{X_o} })} $
 A: The error from a logarithmic function can be estimated by a series of its derivatives. In a truely mathematical form, this should be
$ln(x+\Delta x) = ln(x)+\sum_{n=1}^{\infty} \frac{(-1)^n\Delta x}{x^n}$
but really we just use
$ln(x+\Delta x) = ln(x)+\frac{\Delta x}{x} $
A: The "Guide to the expression of uncertainty in measurement" [ISO/IEC Guide 98-3] is very clear on this: If $Y=f(X_1, X_2, \ldots, X_k)$ then the variance is approximated by a Taylor series
    \begin{align}
  \sigma^2_y &\approx 
\sum_{i=1}^{k}  \sum_{j=1}^k 
               \frac{\partial f}{\partial x_i}
               \frac{\partial f}{\partial x_j}
               Cov[x_i, x_j]
\\
%%%
&=\sum_{i=1}^k 
               \left(%
               \frac{\partial f}{\partial x_i}
               \right)^2  
               \sigma^2_{x_i}  
               +  2 \sum_{i=1}^{k-1}  \sum_{j=i+1}^k 
               \frac{\partial f}{\partial x_i}
               \frac{\partial f}{\partial x_j}
               Cov[x_i, x_j] %+ \mathcal{O}(\sigma^4)    
\end{align} 
where 
$Cov[x_i, x_j] = \rho[x_i, x_j] \, \sigma_{x_i} \sigma_{x_j}$ is the covariance of the two variables $x_i, x_j$, and $\rho[x_i, x_j]$ is their correlation. So, if the two random variables are independent, then use
\begin{align}
y &= -\ln{(x/x_0)} = \ln{(x_0)}-\ln{(x)}\\
\Rightarrow \sigma_y^2 &\approx
(1/x_0)^2 \sigma_{x_0}^2 - 
(1/x)^2 \sigma_{x}^2
= \left(\frac{\sigma_{x_0}}{x_0}\right)^2
+ \left(\frac{\sigma_{x}}{x}\right)^2
\end{align}
