Let $(M,g)$ be an oriented Riemannian manifold with boundary $\partial M$. Let $j\colon \partial M\to M$ denotes the natural embedding and $\mu_M$ the canonical volume form.
If $\psi\in C^\infty(M)$ then $j^\ast(\star \mathrm d\psi)$ is a $(n-1)$-form of $\partial M$, so there exists $\varphi\in C^\infty(\partial M)$ such that $$ j^\ast(\star \mathrm d\psi)=\varphi\,\mathrm{d}\mu_{\partial M}\,, $$ where $\mu_{\partial M}$ is the volume form of $\partial M$ induced by $\mu$.
Here my question:
Is $\varphi$ the outward normal derivative of $\psi$ along the boundary?
Here my attempt. I choose a local boundary chart $(U,(x^1,\dots,x^n))$ such that the outward unit normal field $\nu$ along $\partial M$ has the following expression $$\nu=-\frac{\partial}{\partial x^n}\,.$$ The normal derivative of $\psi$ is defined as $\dot \psi=j^\ast\mathrm{d}\psi(\nu)$, thus $$ \dot \psi=-\frac{\partial \psi}{\partial x^n}\,.$$ By definition of hodge star operator, I have $$ \star \mathrm d\psi=\sum_{i=1}^n (-1)^{i-1} \frac{\partial \psi}{\partial x^i}\sqrt{\det g_{ij}}\,\mathrm d x^1\wedge \dots \wedge \widehat{\mathrm{d}x^i}\wedge \dots \wedge \mathrm d x^n\,. $$ Because of $j^\ast \mathrm dx^i=\mathrm d(x^i\circ j)$, I obtain $j^\ast \mathrm dx^n=0$. Therefore $$j^\ast\star \mathrm d\psi=(-1)^{n-1}\frac{\partial \psi}{\partial x^n}\sqrt{\det g_{ij}}\, \mathrm d(x^1\circ j)\wedge \cdots \wedge \mathrm d(x^{n-1}\circ j)$$ In conclusion, $(-1)^n\sqrt{\det g_{ij}}\, \mathrm d(x^1\circ j)\wedge \cdots \wedge \mathrm d(x^{n-1}\circ j)$ should be the local expression of $\mu_{\partial M}$ (up to a positive factor?)