real analysis - why here $x^2$ is used ? why not $x ?$

Does there exists a function $f : \mathbb{R }\rightarrow \mathbb{R}$ which is differentiable only at the point $0.$?

My attempt : I found the answer here Is there a function $f: \mathbb R \to \mathbb R$ that has only one point differentiable?

But i didn't understands the answer , my doubts given below

Answer

Because while $x p(x)$ is continuous at $0$, it is not differentiable.

In particular, the fraction
$$
\frac{(0+h)p(0+h)-0p(0)}{h}
$$

has value $0$ or $1$ depending on whether $h$ is rational or not. So it has no limit as $h\to 0$, which by definition of derivative means that $xp(x)$ had no derivative at $0$.

On the other hand, the fraction
$$
\frac{(0+h)^2p(0+h)-0^2p(0)}{h}
$$
has value $h$ or $0$ depending on whether $h$ is rational or irrational. Thus it does have a limit as $h\to0$, which is to say that $x^2p(x)$ has derivative $0$ at $x=0$.