Is it possible that $f_1,f_2,\dots,f_n$ are not all differentiable, but $\alpha:I\to \Bbb{R^n}$ is differentiable?

Question

Consider the parametrized curve $\alpha:I\to \Bbb{R^n}$. These notes say that $f_1,f_2,\dots f_n$ being differentiable $\implies$ $\alpha$ is differentiable.

I wonder why the converse is not true. Is it possible that not all of $f_1,f_2,\dots f_n$ are differentiable, but $\alpha$ is differentiable? I know that this is not possible if we have a function of multiple variables. Here we only have a parametrized curve.

Thanks.

Answer 1

If a function $f: \mathbb R^n \rightarrow \mathbb R^m$ is differentiable, then the partials exist, and the derivative/differential is given by the Jacobian, the matrix whose entries are the partial derivatives. Same applies to your case, where the derivative is given by the vector ( a $1 \times n $ -matrix ) whose entries are $f'_i$.

Answer 2

I'm assuming that $\alpha(t)=(f_1(t),\ldots,f_n(t))$. Another way to think about it (but in this case it is equivalent to the other answer) is:

If you know that the composition of differentiable functions is differentiable, and that the canonical projections $\pi_i:\mathbb{R}^n\longrightarrow \mathbb{R}$ are differentiable then, since $f_i=\pi_i \circ\alpha$ and $\alpha$ is differentiable, we have that $f_i$ is differentiable for all $i$.