Why $\frac{\partial f}{\partial v}=\sum_{i=1}^n\alpha_i\frac{\partial f}{\partial x_i}$

Question

Let $U\subset \mathbb R^n$ be an open set and $f:U\to \mathbb R^m$ and suppose $\alpha=(\alpha_1,\ldots,\alpha_n)\in \mathbb R^n$

I'm trying to prove that $\frac{\partial f}{\partial v}(\alpha)=\sum_{i=1}^n\alpha_i\frac{\partial f}{\partial x_i}(\alpha)$.

I know that the directional derivative is $$\frac{\partial f}{\partial v}(\alpha)=\lim_{t\to 0}\frac{f(\alpha+tv)-f(\alpha)}{t}$$ and the partial derivatives are $$\frac{\partial f}{\partial x_j}(\alpha)=\lim_{t\to 0}\frac{f(\alpha+te_j)-f(\alpha)}{t}$$

I've just written down these formulas without any success.

Answer 1

Let's call the point $a$ and the vector $v = \sum_i v_i e_i$ to make the notation a bit better. Then the directional definition of the derivative says $$ f(a+tv) = f(a) + t\frac{\partial f}{\partial v}(a) + o(t). $$ But $ tv = \sum_{i=1}^n (tv_i) e_i $, and we can expand $f(a+tv)$ one partial derivative at a time: $$ \begin{align} f(a+tv) &= f\left(a + \sum_{i=1}^n (tv_i) e_i \right) \\ &= f\left(a + \sum_{i=2}^n (tv_i) e_i + tv_1 e_1 \right) \\ &= f\left(a + \sum_{i=2}^n (tv_i) e_i \right) + tv_1 \frac{\partial f}{\partial x_1} \left(a + \sum_{i=2}^n (tv_i) e_i \right) + o(t) \end{align}$$ Assuming the partial derivative is continuous in a neighbourhood of $a$, we have $$ \frac{\partial f}{\partial x_1} \left(a + \sum_{i=2}^n (tv_i) e_i \right)= \frac{\partial f}{\partial x_1} (a) + O\left( \sum_{i=2}^n (tv_i) e_i \right) = \frac{\partial f}{\partial x_1} (a) + o(1), $$ so we have $$ f(a+tv) = f\left(a + \sum_{i=2}^n (tv_i) e_i \right) + tv_1 \frac{\partial f}{\partial x_1} (a) + o(t) $$

Repeating this procedure, assuming the other partial derivatives are continuous, we obtain $$ f(a+tv) = f(a) + t\sum_{i=1}^n v_i \frac{\partial f}{\partial x_i} (a) + o(t) $$ Now, subtracting $f(a)$, dividing by $t$, and taking $t \to 0$, we find that $$ \frac{\partial f}{\partial v}(a) = \sum_{i=1}^n v_i \frac{\partial f}{\partial x_i} (a) $$ as desired.