Q.If $f(x) = \begin{cases} ax^2-b, & -1<x<1 \\ \dfrac{1}{|x|}, & \text{elsewhere} \end{cases}$ is differentiable at $x=1$, then

A function can be continuous at a point yet still fail to have a derivative there — differentiability is a strictly stronger requirement than continuity, never the other way around.

Three ways a derivative can fail to exist at $x_0$:

1. A corner or cusp (the graph comes to a sharp point $\vee$ or $\wedge$). Example: $f(x)=|x-2|$ at $x=2$ — continuous there, but $f'(2^-)=-1\ne1=f'(2^+)$, so $f'(2)$ does not exist.
2. A vertical tangent. Example: $f(x)=x^{1/3}$ at $x=0$ — continuous, but $f'(0)=\lim_{x\to0}x^{-2/3}=+\infty$, not a finite number.
3. A discontinuity. Example: $f(x)=\lfloor x\rfloor$ at any integer $n$ — not even continuous there, so certainly not differentiable; or a jump function like $f(x)=x$ ($x\le0$), $f(x)=x+1$ ($x>0$), where $f'(0^+)$ fails to exist because the right-hand difference quotient blows up as the jump is approached.

These three cases are exhaustive: a function fails to be differentiable at a point of its domain precisely when one of them holds. In short, discontinuity always forces non-differentiability — but continuity alone never guarantees differentiability, as cases (1) and (2) show.

The one implication that always holds:

Theorem 10.1. If $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$. …