Calculating total derivative out of partial derivative

Question

I am currently learning about differentiation with multiple variables and i am wondering about the proof that the total derivative consists of a matrix of the partial derivatives. Let $U\subseteq R^n$ be open. $f: U\to R^m$, $A \in R^{ m\times n}$ is the total derivative and $\phi(h)$ is the error in $$f(x+h) = f(x) + A \cdot h + \phi(h)$$ Now every row of $f$ can be written individually as $$f_{i}(x+h)=f_{i}(x)+\sum_{j=1}^{n}a_{ij}h_{j} + \phi_{i}(h) $$ for $i =1,...,m$ and $a_{ij}\in A$. With $h = t\cdot e_{j}$ where $e_{j}$ is the $j$-th unit vector it is: $$f_{i}(x+t\cdot e_{j})=f_{i}(x)+a_{ij}t + \phi_{i}(t\cdot e_{j})$$ Thats the beginning of the proof. Later it shows that $a_{ij}=\frac{\partial f_{i}}{\partial x_{j}}(x)$ but i can not comprehend what happened to the sum from the second to the third equation.

Answer 1

In the second equation $h$ is a vector and $h_j$ is the jth component of $h$ so they have a sum over all components. In the third equation they have replaced $h$ by $t\cdot e_j$, t times the "j-th unit vector" which is the vector that has 1 as the jth component and 0 as all other components so the sum collapses to the single j-th term.