I am using an equation of the following form:
$$\ln{cY} = -\frac{X}{a} + b$$
I am interested in $a$, which is related to the slope of the plot $\ln{cY}$ versus $X$. If I can generate this plot and I know $\sigma_{Y}$, the error in $Y$, how can I find the resulting error in $a$?
When you have a straight line, the error in the slope can be calculated from (link):
$$SE = \frac{\sqrt{\sum(y_i-\hat{y_i})^2/(n-2)}}{\sqrt{\sum(x_i-\bar{x})^2}}$$
This assumes that the errors in all the data points is the same, and that the distribution of the errors is normal. When you take the logarithm of a normally distributed number, the result is no longer quite normal.
When $Y$ is normally distributed with a standard deviation $\sigma_Y$, then the error in $\log(Y)$ can be approximated by noting that
$$\frac{d(\log(Y)}{dY} = \frac{1}{Y}$$
So a variation of $\sigma_Y$ around $Y$ gives a variation of $\frac{\sigma_Y}{Y}$ about $\log(Y)$ (assuming that $\sigma_y\ll Y$). The constant $c$ in your equation can be taken outside of the logarithm and will just appear as an offset on $b$, so we can ignore that for the purpose of determining the slope.
Now given that the standard deviation of an individual measurement is $\sigma_Y$, you can probably assume that the expected value of an individual $(y_i-\hat{y_i})^2$ is equal to the variance, $\sigma_y^2$. In that case we can rewrite the above equation for the standard error of the slope:
$$SE = \frac{ \sqrt{ \sum{\frac{\sigma_Y^2}{y_i(n-2)}}{}}}{{\sqrt{\sum(x_i-\bar{x})^2}}}$$
Finally, since your expression has $\frac{1}{a}$ for the slope, you would have to invert this expression if you wanted the error in $a$.
Check my math - it looks right but it's possible I made a mistake.
If you are OK with a number, and don't need a formula, there are a variety of methods. One such is the bootstrap method. I have applied it with success in a variety of situations.
It's very simple with partial derivatives. For any well behaved function of $n$ independent variables $f_(x_1, \ldots, x_n)$, then the uncertainty in $f$ is given by the total derivative added in quadrature weighted by uncertainties. That is,
$$\Delta f= \sqrt{\left(\frac{\partial f}{\partial x_1}\right)^2 \Delta x_1^2 + \cdots +\left(\frac{\partial f}{\partial x_n}\right)^2 \Delta x_n^2 }$$
where $\Delta x_i$ is the uncertainty in the variable $x_i$. So my suggestion to you is that you simply solve for $a(X,Y,b,c)$ and apply this formula to the quantity $\Delta a$ (instead of $\Delta f$). If you don't know the uncertainties in any one of $X,Y,b,c$ then I'm unsure of another way to find this uncertainty.
