I am trying to prove a statement which says if you have a function $f : \mathbb{R} \to \mathbb{R^m}$ and $\partial f_i/\partial x$ exists for $i = 1, \dots, m$ then $Df$ exists.
I was wondering if I could get some comments on my proposed proof, I can pretty much only use the definition of the derivative and partial derivative?
We can write $f(x) = (f_1(x), f_2(x), \dots , f_m(x))$ where each $f_i : \mathbb{R} \to \mathbb{R}$. Then since $\partial f_i/ \partial x$ exists we know
$$\frac{\partial f_i}{\partial x}(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} = Df_i(x)$$
since $f_i$ goes from $\mathbb{R} \to \mathbb{R}$, $Df_i = \partial f_i / \partial x$, which means
$$\lim_{x\to x_0} \frac{|f_i(x) - f_i(x_0) - Df_i(x_0)(x-x_0)|}{|x-x_0|} = 0$$
by definition of derivative. Let
$$Df = \begin{pmatrix} \frac{\partial f_1}{\partial x} \\ \vdots \\ \frac{\partial f_m}{\partial x} \end{pmatrix}$$
Then we have
$$\left|\left|\frac{f(x) - f(x_0) -Df(x_0)(x-x_0)}{x-x_0}\right|\right| \leq \sum_{i=1}^m \left|\frac{f_i(x) - f_i(x_0)-Df_i(x_0)(x-x_0)}{x-x_0}\right| \to 0, x \to x_0$$
Hence,
$$\lim_{x\to x_0}\left|\left|\frac{f(x) - f(x_0) -Df(x_0)(x-x_0)}{x-x_0}\right| \right| = 0$$
and $Df$ satisfies the definition of the derivative.