Poisson Distribution - Expressing Variance that is free of Mu.

Let X1...Xn be a random sample from a Poisson distribution with mean $\displaystyle \mu$. Let $\displaystyle Y = \Sigma X_i$ has a Poisson distribution with mean $\displaystyle n \mu$. $\displaystyle \overline{X} = Y/n$ is approximately $\displaystyle N(\mu, \mu/n$ for large n. Show that u(Y/n) = $\displaystyle \sqrt{Y/n}$ is a function of Y/n whose variance is free of $\displaystyle \mu$

$\displaystyle u(Y/n) = u(\overline{X}) = u(\mu) + u(\mu)(\overline{X}) $

$\displaystyle Var[u(\overline{X})] = [u'(\mu]^2(\mu/n) = c$

$\displaystyle u'(\mu) = c_1/\sqrt{\mu}$ (c1 = cn)

$\displaystyle u(\mu) = c2 \sqrt{\mu}$ (c2 = 2c1)

If c2 = 1, then $\displaystyle u(\overline{X}) = \sqrt{X} = \sqrt{Y/n}$

Is this correct, or have I made a mistake somewhere?