Q:

Differential of Gauss map

Let $S$ we a regular surface, differential of Gauss map is $\mathrm{d} \mathrm{N}_{\mathrm{p}}: \mathrm{T}_{\mathrm{p}}(\mathrm{S}) \rightarrow \mathrm{T}_{\mathrm{p}}(\mathrm{S})$.
To evaluate the differential at some point, we can choose a curve $\alpha(t) = x(u(t),v(t))$ on $S$ as a image of plane curve $(u(t),v(t))$ under the chart map $\textbf{x}$.
So we have :
$$\begin{aligned}
d N_{p}\left(\alpha^{\prime}(0)\right) &=d N_{p}\left(\mathbf{x}_{u} u^{\prime}(0)+\mathbf{x}_{v} v^{\prime}(0)\right) \\
&=\left.\frac{d}{d t} N(u(t), v(t))\right|_{t=0} \\
&=N_{u} u^{\prime}(0)+N_{v} v^{\prime}(0)
\end{aligned}$$

The first equality and the third one is easy, what I don't understand is the second one, we know in general $df_p(v) = \frac{d}{dt}_{t=0}(f\alpha(t))$ so for here shouldn't it be $\left.\frac{d}{d t} N(x(u(t), v(t)))\right|_{t=0}$.

This is the result from Do Carmo's differential geometry textbook page 142-143.

The second question is  the book says in particular $d
   N_{p}\left(\mathbf{x}_{u}\right)=N_{u} \text { and } d
   N_{p}\left(\mathbf{x}_{v}\right)=N_{v}$. I don't know how to get it. My idea is to choose plane curve $(u(t),v(t))$ vary only along the u-axis so $N_u =dN_p(x_u)$ in this case.Is my interpretation right?

A:

You are correct in your first bullet point. For instance, for the second equality he could have wrote $=\frac{d}{d t} N\left(t)\right|_{t=0}$  which would be true in the sense that $\textbf{N}$ depends on $t$ (in the case of our curve), but this way of writing it wouldn't have been useful to us. Equally he could have wrote $\frac{d}{d t} \textbf{N}\left(\textbf{x}(u(t), v(t)))\right|_{t=0}$ as you say, (which is the most explicit form as it shows all dependencies of variables).
But  Do Carmo chooses to write it the way he does because we care about how $\textbf{N}$ changes as we change the parameters $u$ and $v$ (i.e. $N_u$ and $N_v$).
Then finally in your second bullet point, the answer follows from the fact that $d N_{p}$ is a linear operator in conjunction with the fact that $d N_{p}\left(\mathbf{x}_{u} u^{\prime}(0)+\mathbf{x}_{v} v^{\prime}(0)\right) = N_{u} u^{\prime}(0)+N_v v^{\prime}(0)$.

