Normal vector of $\Gamma \times \mathbb{R}^+$ where $\Gamma$ is compact hypersurface Let $\Gamma$ be a smooth boundaryless hypersurface of dimension $n-1$ in $\mathbb{R}^n$. Define $Q=\Gamma \times \mathbb{R}^+$.
What does a normal vector of $Q$ look like? Because I want to compute the normal derivative of a function $f:Q \to \mathbb{R}$.
Do we need a normal vector of $\Gamma$ find this?

$Q$ is a hypersurface in $\mathbb R^{n+1}$ with boundary (a cylinder to be more specific). On the interior of $Q$ the normal vector is nothing else than the normal vector to $\Gamma$ with a zero $n+1$ component. The boundary of $Q$ has codimension $2$ hence you have two independent normal vectors. One is again the vector defined before, and the other one is $e_{n+1}$ the unit versor of the last coordinate in $\mathbb R^{n+1}$.
addendum after more details given in the comments. It seems that you are interested in the vector which is tangent to the the hypersurface $Q$ and normal to the boundary $\partial Q$. Such a vector is $(0,-1)$ with $0\in \mathbb R^n$. If the variables of $\mathbb R^n \times \mathbb R^+$ are $(y,x)$ then the normal derivative of a function $u$ would be $-\partial u / \partial x$.
