Answer-
The interval where the function is diffrentiable is
![[-\infty,1)\ \bigcup\ (1,\infty]](https://img.qammunity.org/2019/formulas/mathematics/high-school/yabdg3ekbswtnf4uchktwzew3v5reqs5bu.png)
Solution-
The given expression is,

The function will be differentiable where it is continuous and it will not be differentiable, where the function is not continuous.
The function continuous everywhere except at x = 1, because

at x = 1, its limit does not exist.
Therefore, apart from x=1, this function is differentiable everywhere. The interval will be
![[-\infty,1)\ \bigcup\ (1,\infty]](https://img.qammunity.org/2019/formulas/mathematics/high-school/yabdg3ekbswtnf4uchktwzew3v5reqs5bu.png)