This question was already asked in the following link:
Proving Differentiability of a Function (Spivak)
But I tried to do it by myself and I had some doubts that I would like to be clarified.
The question says
Let $g$ be a continuous, real-valued function on the unit circle, such that $g(-x)=-g(x)$ and $g(0,1)=g(1,0)=0$. Now define $f:\mathbb{R}^2\rightarrow\mathbb{R}$ by $$ f(x) =\begin{cases} \|x\|\cdot g\Big(\frac{x}{\|x\|}\Big) & x\neq 0 \\ 0 & x=0 \end{cases} $$ fix $x\in\mathbb{R}^2$ and define $h:\mathbb{R}\rightarrow\mathbb{R}$ by $h(t)=f(tx)$. Show that $h$ is differentiable.
It is not difficult to see (after considering the cases) that $h(t)$ has the following form: $$ h(t) = t\cdot\|x\|\cdot g\Big(\frac{x}{\|x\|}\Big) $$ And from there it's obvious that $h'(t)=f(x)$. But before I was thinking in solve the problem without considered the cases where $t$ it's positive or negative (no directly) as do the guy in the question. instead I defined the function $I:\mathbb{R}\rightarrow\mathbb{R}$ by $$ I(t) =\begin{cases} 1 & t > 0 \\ -1 & t<0 \end{cases} $$ And write $h$ as $$ h(t) =\begin{cases} |t|\cdot\|x\|\cdot g\Big(I(t)\cdot\frac{x}{\|x\|}\Big) & t\neq 0 \\ 0 & t=0 \end{cases} $$ If we differentiated $h$ we respect to $t$ have that the derivative has the form $$ h'(t)= \dfrac{t}{|t|}\cdot\|x\|\cdot g\Big(I(t)\cdot\frac{x}{\|x\|}\Big)+|t|\cdot\|x\|\cdot g_t\Big(I(t)\cdot\frac{x}{\|x\|}\Big)\cdot I'(t)\\=\dfrac{t}{|t|}\cdot\|x\|\cdot g\Big(I(t)\cdot\frac{x}{\|x\|}\Big)\\=I(t)\cdot\|x\|\cdot g\Big(I(t)\cdot\frac{x}{\|x\|}\Big)\\= \|x\|\cdot g\Big(\frac{x}{\|x\|}\Big)=f(x) $$ If $t\neq0$ and for the case where $t=0$ we can calculate the derivative of $h$ by definition and get that $h'(t)=f(x)$.
I know that is easier if I just notice that the fuction $h$ has the form $h(t)=t\cdot f(x)$. But my first idea was do it like this and now I have the doubt of how can I be sure that $\dfrac{d}{dt}g\Big(I(t)\cdot\frac{x}{\|x\|}\Big)$ exist.
So this is my question.
When can we know that a continuous function $g:S^1\to\mathbb{R}$ have a derivative with respect to $t\in\mathbb{R}$? i.e, When the derivative of $g\left(\dfrac{t}{|t|}\cdot\dfrac{x}{\|x\|}\right)$ with respect to $t$ exist?
I know that is basically the same as the exercise but i would like to know is there exist a general way to do this and if it isn't exist in general, i would like to know if there is a way to be sure that in the case of this $g$ the derivative exist.