How can we find these expressions: $\nabla_X \psi^T = X + \rho AX$ and $ \nabla \rho = -A(\psi^T), X \in \mathfrak{X}(M)$?

Question

Let $M$ be an immersed orientable hypersurface of the Euclidean space $\mathbb{R}^{n+1}$ with unit normal field $N$, shape operator $A$ and $\psi:M\rightarrow\mathbb{R}^{n+1}$ the immersion. Then we have $$\psi = \psi^T + \rho N, (1) $$ where $\rho = \langle \psi , N\rangle$ is the support function os the hypersurface $M$ and $\psi^T \in \mathfrak{X}(M).$ I tried to do this: taking the covariante derivate of Euclidean space $\overline{\nabla}$ in $(1)$, I get: \begin{eqnarray*} \overline{\nabla}_X\psi &= &\overline{\nabla}_X\psi ^T + \overline{\nabla}_X(\rho N)\\ \Rightarrow \overline{\nabla}_X\psi & = &B(X, \psi^T) + \nabla _X \psi^T + (X\rho )N + \rho\ \overline{\nabla}_XN \\ \Rightarrow \nabla_X \psi^T &=& \overline{\nabla}_X \psi - B(X, \psi^T) - (X\rho )N - \rho\ \overline{\nabla}_XN, \end{eqnarray*} where $\nabla$ is the the covariant derivative in $M$ and $B$ is the second fundamental form $B(X,Y) = \overline{\nabla}_XY - \nabla_XY$. Taking product with a arbitrary $Y \in \mathfrak{X}(M)$, we get: \begin{eqnarray*} \langle \nabla_X \psi^T,Y \rangle &=& \langle \overline{\nabla}_X \psi - B(X, \psi^T) - (X\rho )N - \rho\ \overline{\nabla}_XN,Y \rangle\\ &=& \langle \overline{\nabla}_X \psi - \rho(\overline{\nabla}_XN)^T,Y \rangle \\ &=& \langle \overline{\nabla}_X \psi + \rho AX,Y \rangle,\text{where I used that } (\overline{\nabla}_XN)^T = -AX ( \text{Weingarten Formula}). \end{eqnarray*} Then $$\nabla_X \psi^T = \overline{\nabla}_X \psi + \rho AX.$$ I don't know how to continue and even if what I did is right. I couldn't do anything in the other equation.