Variance of X/Y Suppose you have two normally distributed, independent random varianbles X and Y, where X has mean $\mu _x$ and variance $\sigma^2 _x$ and Y has mean $\mu _y$ and variance $\sigma^2 _y$.
For each, $\mu \gg \sigma^2$ and $\mu \gg 0$.
We know from sampling that the distribution of $X/Y$ is approximately normal with mean $\mu _x/\mu _ y$.
The question is: what is the approximate variance of this distribution?
Below is the sampling histogram:

 A: Technically, the variance is infinite because you are dividing by a variable with a positive density around $0$ while the numerator has a positive density away from $0$, and this forces the tails to be too large for the variance to exist. In fact, the expected value doesn't exist, either. In practice you may be able to ignore this because it is a very rare event for the denominator to be close to $0$.
Let $\Delta X = X-\mu_X$ and $\Delta Y = Y-\mu_Y.$
Let $r(X,Y) = \frac{X}{Y}$
$r(X,Y) = r(\mu_x + \Delta X, \mu_y + \Delta Y)$ 
$\approx r(\mu_X,\mu_y) + \frac{\partial r}{\partial X}(\mu_X,\mu_Y)\Delta X + \frac{\partial r}{\partial Y}(\mu_X,\mu_Y)\Delta Y + O((\Delta X)^2+(\Delta Y)^2)$.
If the variances of $X$ and $Y$ are small enough, then we can ignore the higher order terms and compute the variance of the linear approximation.
$\text{Var}\bigg(r(\mu_X,\mu_y) + \frac{\partial r}{\partial X}(\mu_X,\mu_Y)\Delta X + \frac{\partial r}{\partial Y}(\mu_X,\mu_Y)\Delta Y)\bigg)$
$=(\frac{\partial r}{\partial X}(\mu_X,\mu_Y))^2 \text{Var}(X) + (\frac{\partial r}{\partial Y}(\mu_X,\mu_Y))^2\text{Var}(Y)$
since this is just a constant plus a linear combination of $X$ and $Y$.
$\frac {\partial r}{\partial X}(x,y) = \frac 1 y$ so at $(\mu_x,\mu_y)$ it is $\frac{1}{\mu_y}$.
$\frac {\partial r}{\partial Y}(x,y) = \frac {-x} {y^2}$ so at $(\mu_x,\mu_y)$ it is $\frac{-\mu_x}{\mu_y^2}$.
Therefore, the variance of $\frac XY$ will appear to be about 
$$\frac{\sigma^2_X}{\mu_Y^2} + \frac{\mu_X^2 \sigma^2_Y}{\mu_Y^4}  = \frac{\mu_Y^2\sigma_X^2 + \mu_X^2 \sigma^2_Y}{\mu_Y^4}.$$
