Finding the stochastic differential equation satisfied by process Y Consider the stochastic process $X$ that satisfies the SDE 
\begin{equation}
X_t=1+\mu\int_0^t X_s \, ds+\sigma\int_0^t X_s \, dB_s,
\end{equation}
with $\mu$ and $\sigma$ constants. Consider another process $Y$ given by
\begin{equation}
Y_t=X_t^\beta,
\end{equation}
with $\beta\ge 2$. Find the stochastic differential equation for process $Y$. 
I am looking at the following solution:
\begin{equation}
dY_t=\beta X_t^{\beta-1}dX_t+\beta(\beta-1)X_t^{\beta-2}(dX_t)^2\\[10pt]
dY_t=\beta X_t^{\beta-1}(\mu X_t \, dt+\sigma X_t \, dB_t)+\beta(\beta-1) X_t^{\beta-2} \sigma^2 X_t^2 \, dt\\[10pt]
dY_t=\beta\mu X_t^\beta \, dt+\sigma\beta X_t^\beta \, dB_t + \beta(\beta-1) \sigma^2 X_t^\beta \, dt\\[10pt]
dY_t=(\beta\mu+\beta(\beta-1)\sigma^2)Y_t \, dt+\sigma\beta Y_t \, dB_t
\end{equation}
Assuming this solution is right, my $\textbf{question}$, or the problem I'm having difficulty fully understanding is, why is the second spatial derivative not multiplied by $\frac{1}{2}$? Unless I'm totaly spacing, should that term not be
\begin{equation}
\left[\frac{\beta(\beta-1)}{2}\right]X_t^{\beta-2}(dX_t)^2
\end{equation}
from an application of the Ito formula? Thank you.

You have a typo in your first equation, it should be 
$$dY_t=\beta X_t^{\beta-1}dX_t+\beta(\beta-1)X_t^{\beta-2}(dX_t)^2$$
But it looks to me that you are correct, and the above equation is wrong.
If $Y_t =f(X_t)$, with $f(x)=x^{\beta}$, I do expect the $\frac{1}{2}$:
$$dY_t=df(X_t)=\frac{\partial f(X_t)}{\partial x}dX_t + \color{red} {\frac{1}{2}} \cdot\frac{\partial^2 f(X_t)}{\partial x^2}(dX_t)^2$$
