standard error of transformed regression coefficient I have the regression $y= \beta_0 + \beta_1 \,x + e$, along with the standard error of $\beta_1$
I would like to find the standard error of the elasticity at $\bar{x},\bar{y}$, which is given by $\beta_1 \, \bar{x}/\bar{y}$
Is that simply $\text{SE}(\beta_1)\cdot (\bar{x}/\bar{y})$
 A: Here, you can apply the Delta Method. Denote $\omega^2$ as the asymptotic variance of $\hat{\beta}$. Then, for the regression coefficients holds $\sqrt{n}(\hat{\beta} - \beta) \xrightarrow{d} N(0, \omega^2)$. The statement of the Delta Method is that if you transform an estimator by a function $g$, the following property holds: 
$\sqrt{n}(g(\hat{\beta}) - g(\beta)) \xrightarrow{d} N(0, \omega^2g'(\beta)^2)$. Where $g'$ denotes the first derivative of $g$.
This implies $V[g(\hat{\beta})] = V[\hat{\beta}] \cdot g'(\hat{\beta})^2$.
In your case, $g(\hat{\beta_1}) = \hat{\beta_1} \cdot (\bar{x}/\bar{y})$ and $g'(\hat{\beta_1}) =(\bar{x}/\bar{y})$. Hence, your standard error is $SE[\hat{\beta_1} \cdot (\bar{x}/\bar{y})] = \sqrt{(\bar{x}/\bar{y})^2 \cdot V[\hat{\beta_1}]} =  \sqrt{(\bar{x}/\bar{y})^2} \cdot SE[\hat{\beta_1}] $. 
Note: What I defined as asymptotic variance for introducing the Delta Method is not equal to the variance in $\hat{\beta} \sim N(\beta, \sigma^2 \cdot (X'X)^{-1}$) which is the correct distribution.
A: Your answer is correct: $\sigma_{\beta_1}\frac{\bar x}{\bar y}$.
Here your $\bar x$ and $\bar y$ are given (not random), unless you used bar accent to denote the sample means of $x,y$ (which is common in physics, but not economics).
To those who are wondering what's elasticity: it's the sensitivity of a percentage change over a percentage change. For instance, if $y$ is demand, and $x$ is income, then income elasticity of demand: 
$$\frac{\frac{\Delta y}{y}}{\frac{\Delta x}{x}}\approx\frac{x}{y}\frac{dy}{dx}$$
When you plug the the linear regression and the derivatives you get $$\beta_1\frac{x}{y}$$
