The differential of the Gaussian map AKA the Weingarten map (with a minus sign) in three dimensions is defined as :
\begin{align} W: T_{p} M \subset \mathbb{R}^3 \rightarrow T_{p} M \subset \mathbb{R}^3 \end{align} A linear map from a tangent plane to itself.
I was reading a proof in a book on curves and surfaces and I saw that \begin{align*} W(\sigma_{u}) = -N_{u} \ \text{and} \ W(\sigma_{v}) = - N_{v} \end{align*} Where $\sigma_{u}$ is a surface patch differentiated w.r.t. the variable u and $N$ is the standard unit normal.
I have been looking at the above equality, I assume that it is quite obvious, but I don't really get why that is the case?