getting elasticity after sqrt transformation My model is $y^.5= \beta_0 + \beta_1 * x^.5 + e$
After taking partial derivatives, I find that $\beta_1$ represents $(dy/y^.5)/(dx/x^.5)$
Now my question is how I can transform $\beta_1$ to be an elasticity. Should I multiply by $x^.5/y^.5$ to get the desired form? If so, what does that mean intuitively and how do I multiply by that factor with real data? 
 A: When $x$ has a causal effect on $y$, the elasticity is

the ratio of the percentage change in one variable to the percentage change in another variable

(Wikipedia).
When $x$ is changed to $x + dx$, causing $y$ to change to $y+dy$, the percentage changes are $100 dx / x$ and $100 dy / y$, respectively.  Their ratio is
$$\frac{100\, dy/y}{100\, dx/x} = \frac{dy}{dx} / \frac{y}{x}.$$
Letting $y$ stand for the fitted value at $x$, we may compute
$$y = (\beta_0 + \beta_1\sqrt{x} )^2$$
and
$$\frac{dy}{dx} = \frac{2 \sqrt{y} \beta_1 }{2\sqrt{x}} = \beta_1 \frac{\sqrt{y}}{\sqrt{x}}.$$
Consequently the elasticity is
$$\beta_1 \frac{\sqrt{y}}{\sqrt{x}} / \frac{y}{x} = \beta_1 \frac{\sqrt{x}}{\sqrt{y}} =  \frac{\beta_1\sqrt{x}}{\beta_0 + \beta_1 \sqrt{x} }.$$
When analyzing data, you can only estimate the elasticity using estimates $b_0$ and $b_1$ instead of the true (but unknown) coefficients $\beta_0$ and $\beta_1$.  Therefore the estimated elasticity at any point $x_0$ will be
$$\frac{b_1\sqrt{x_0}}{b_0 + b_1 \sqrt{x_0} }.$$

There are many ways to compute confidence intervals for those estimates. In the absence of the data--knowing only the variance-covariance matrix of the estimates--use the Delta method.  When fitting using another procedure, such as Maximum Likelihood, you might be able to obtain confidence limits directly (such as by profiling the likelihood for a fixed $x_0$).
Note that the estimated elasticity is not constant unless $b_0=0$, or at least (approximately) when $|b_0| \ll |b_1 x_0|$.

As a double-check, consider the limiting case $\beta_0=0$.  The original model for the fit
$$\sqrt{y} = \beta_1 \sqrt{x}, $$
can be rewritten
$$\log(y) = \log(\beta_1^2) + \log(x).$$
In a log-log model, the elasticity is the coefficient of $\log(x)$, here shown to be $1$.  Indeed,
$$\frac{\beta_1 \sqrt{x}}{\beta_0 + \beta_1 \sqrt{x}} = \frac{\beta_1\sqrt{x}}{\beta_1 \sqrt{x}} = 1$$
agrees with that.
