Derivable product of derivable and non-derivable function, not satisfying product rule

Question

Are there functions $f, g : \mathbb{R} \to \mathbb{R}$ satisfying the following:

• $f(0) = g(0) = 0$,
• $f'(0)$ and $(fg)'(0)$ are defined, but $g'(0)$ is not defined,
• $(fg)' \neq 0$

If both functions were required to be derivable, there is no such example (due to the product rule). If neither is required to be derivable, we can take $|x|$ and $sgn(x)$. But what if one is derivable and the other is not?

Answer 1

Let $f:\mathbb{R} \to \mathbb{R}$, $f(x)=x^2$ and $g: \mathbb{R} \to \mathbb{R}$, $g(x)=\frac{1}{x} \forall x \neq 0 \land g(0)=0$.
We have that $f(0)=g(0)=0$, and $f'(0)=0$, but $\nexists g'(0)$.
But we have that $(fg)(x)=x^2\frac{1}{x}=x \forall x \neq 0$, and $(fg)(0)=0$.
So $(fg)(x)=x$, hence $(fg)'(0)=1.$