Setup: Let $M^n$ be an oriented smooth manifold with boundary. We want to show that this induces an orientation on $\partial M$.
Proof:
Let $\mathcal{A} := \{(U_i,\varphi_i):i \in I\}$ be an atlas of $M$ and $$\mathcal{A}_{\partial M} := \{(V_i,\psi_i):i \in I\} := \{(U_i \cap \partial M,\varphi_i|_{U_i \cap \partial M}): i \in I\}$$ be the induced atlas on $\partial M$.
For $$x = (0,x') \in \varphi_i(U_i \cap U_j \cap \partial M) = \varphi_i(U_i \cap U_j) \cap \partial \mathbb{R}^n\_ = \psi_i(U_i \cap U_j)$$ we have $$det\Big(D(\varphi_j \circ \varphi_i^{-1})\Big)\Big|_x > 0$$ by another theorem, since $M$ is orientable is equivalent to this condition.
Now, we have that $$(\varphi_j \circ \varphi_i^{-1})^1(x) = (\varphi_j \circ \varphi_i^{-1})^1 (0,x') = 0 \in \mathbb{R}.$$
Here is where I get lost. In the notes I have, it says that $$\frac{\partial (\varphi_j \circ \varphi_i)^1}{\partial x^k}(x) = 0$$ for $k \geq 2$, which seems right, since the value is constantly $0$ on the domain. But, then it says that we have $$(\varphi_j \circ \varphi_i^{-1})^1(x+te_1) < 0, \ t < 0$$ and that this implies that $$\frac{\partial (\varphi_j \circ \varphi_i^{-1})^1}{\partial x^1}> 0.$$ Why do we have this implication?