If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ has coordinates $f^1 \ldots f^m$, and each $f^i$ is differentiable at $0$... ...then does it follow that $f$ is differentiable at 0?
My motivation for asking this is as follows: in Spivak's Calculus on Manifolds, in theorem 2.9, he uses this with the additional condition that each $f^i$ is continuously differentiable in a nbh of 0, to conclude that $f$ is differentiable and I don't think is necessary.
Namely, if each $f^i$ has derivative $Df^i$, I claim the matrix with $i^{th}$ row $Df^i$ will serve as $Df$. Indeed, let $v_j$ be a sequence tending to 0 in $\mathbb{R}^n$,  we have (by the triangle inequality, if you wish)$$\frac{| f(v_j) - f(0) - \sum_i Df^i(v) |}{|v_j|} \leqslant \frac{\sum_i |f^i(v_j) - f^i(0) - Df^i(v_j)|}{|v_j|}$$Taking the limit as $j \rightarrow \infty$, each summand goes to 0 by the differentiability of $Df^i$ (and there's only $m$ of them), hence the limit is 0.
Is this wrong? Thanks in advance!
EDIT: btw, conditional on the above proof being right and/or the claim being right, does anyone know maybe what Spivak was going for? 

The "continuously differentiable" condition for the partial derivatives is essential to showing that $f$ is differentiable $0$. This is the canonical counterexample $$f(x,y) = \left\{
     \begin{array}{lr}
       \frac{x^2y}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\
       0 & \text{if } x = (0,0)
     \end{array}
   \right.$$
