Is a function differentiable if it has a removable discontinuity

The map$$\begin{array}{rccc}f\colon&\mathbb R\setminus\{0\}&\longrightarrow&\mathbb R\\&x&\mapsto&\frac{x^2}x\end{array}$$is undefined at $0$, and therefore it is meaningless to ask whether or not it is differentiable there. It happens that we can extended it to one and only one continuous function $F\colon\mathbb R\longrightarrow\mathbb R$, which is defined by $F(x)=x$. And it happens that this function is differentiable at $0$.

However, if you take$$\begin{array}{rccc}g\colon&\mathbb R\setminus\{0\}&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x>0\\-x&\text{ if }x<0,\end{cases}\end{array}$$then you can extend $g$ to one and only one continuous map $G\colon\mathbb R\longrightarrow\mathbb R$, which is $G(x)=\lvert x\rvert$, but the function $G$ is not differentiable at $0$.