How is delta method used here in approximating the square root of a normal random variable?

I am reading this example where the distribution is given by $$Y=\frac{\sigma^2\chi^2_{n-1}}{n-1}.$$ By CLT, $$Y\sim\mathcal{N}(\sigma^2,\frac{2\sigma^4}{n-1}).$$ Up to here it was all clear to me.

Then my textbook said the sitribution of $$2\sqrt{Y}$$, using Delta Method, is approximately $$\mathcal{N}(2\sigma,\frac{2\sigma^2}{n-1}).$$

However I cannot seem to use Delta Method to get the aforementioned variance. I thought the statement maybe missing out a $$n$$ term? I thought the variance should be $$\frac{2\sigma^2}{(n-1)n}$$ because we have to divide by $$n$$ at the end when applying Delta Method?

The statement from the textbook is correct. If we define $$f(y)=2\sqrt y$$, we have $$f'(y)=1/\sqrt y$$. By the delta method, we approximate:
\begin{align} \text{Var}(2\sqrt Y)&=f'(\mathbb E[Y])^2\cdot\text{Var(Y)}\\ &=\frac{1}{\sigma^2}\cdot\frac{2\sigma^4}{n-1}\\ &=\frac{2\sigma^2}{n-1} \end{align}