Given $f(x)=\frac{x^2 - 1}{x-1}$
Using $a^2 - b^2 = (a+b)(a-b)$, we get $f(x) = \frac{(x+1)(x-1)}{(x-1)}$
$\rightarrow f(x)=x+1$.
Since we divided out numerator and denominator, we have a hole at $x = 1$, so the function is discontinuous at $x=1$
(A function which is not continuous isn't differentiable either).
Option D.
(You can refer to the video to know why this is discontinuous):