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
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 $x=0$.
Note that $f(0)=0$ and $\frac{f(x)}{x}=xp(x)$
And the limit of $xp(x)$ at zero is zero.
If $f=xp(x)$ then $\frac{f(x)}{x}=p(x)$
and the limit of $p(x)$ at zero does not exist.
With $f(x)=xp(x),$ the derivative at $0$ would be $\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0}\dfrac{xp(x)}x=\lim\limits_{x\to0}p(x),$
which does not exist, since $p(x)$ takes values of $0$ and $1$ for arbitrarily small $x$.
