The derivative at a point is a limit, so a derivative fails to exist when the corresponding limit of difference quotients fails to exist as a finite two-sided limit. That gives several familiar cases. A function may have a corner, a cusp, a vertical tangent, or a discontinuity, and the graphs look different, but the underlying question is the same. What do the one-sided difference quotients do?

Four ways differentiability can fail

Each function in the visualization has a problem at \(x=0\).

from the left
from the right

Both secants share the anchor, so the two numbers are the one-sided difference quotients at the same h. Shrink h and watch them. A corner sends them to two different finite numbers. A cusp sends them to infinities of opposite sign. A vertical tangent sends them to the same infinity, which is still not a number. A discontinuity is the odd one out: the function itself has a gap, one quotient stays put while the other runs away, and no repair to the slope is possible because the failure is not about slope at all.

The two reported values are one-sided difference quotients at the same distance from the point. Shrink \(h\) and compare them.

Corner

For \(f(x)=\vert x\vert\), the left-hand difference quotient is \(-1\) and the right-hand difference quotient is \(1\). Both are finite, and they simply disagree, so the two-sided derivative does not exist. Graphically, this produces a sharp corner.

Cusp

For \(f(x)=x^{2/3}\), the one-sided difference quotients grow without bound with opposite signs. One side tends toward \(-\infty\) and the other toward \(+\infty\). The curve narrows into a cusp, and there is no finite derivative.

Vertical tangent

For \(f(x)=x^{1/3}\), the one-sided difference quotients both tend toward \(+\infty\). The two sides agree in direction, but they do not approach a finite number. The graph has a vertical tangent line, and a vertical line has undefined slope, so the ordinary derivative does not exist there.

Discontinuity

The last case fails even earlier. If the function is not continuous at the point, it cannot be differentiable there, and there is no need to continue to the difference quotient once continuity has already failed.

Differentiability implies continuity

If a function is differentiable at \(x=c\), then it is continuous at \(x=c\). The contrapositive is often the useful version: if a function is not continuous at \(c\), then it is not differentiable at \(c\).

The converse is false. A function may be continuous but not differentiable, and the corner, cusp, and vertical-tangent examples are all continuous at the point where the derivative fails. So the relationship is

\[\text{differentiable at }c \Longrightarrow \text{continuous at }c\]

but in general \(\text{continuous at }c \not\Longrightarrow \text{differentiable at }c\). It is also useful to remember that if \(c\) is not in the domain of \(f\), then \(c\) cannot be in the domain of \(f'\).

Testing a piecewise join

For a piecewise function, check continuity before checking derivatives. Consider

\[f(x) = \begin{cases} x^2, & x\le1,\\ 2x-5, & x>1 \end{cases}\]

The derivative of the first expression at \(x=1\) is 2, and the derivative of the second expression is also 2, so the one-sided slopes match. But the function values do not. From the left, \(f(1)=1\), and from the right, the expression approaches \(2(1)-5=-3\). The function is not continuous at \(x=1\), so it is not differentiable there, regardless of the matching derivative formulas.

The correct order is:

  1. Check continuity at the join.
  2. If continuity holds, compare the one-sided derivatives.

Matching slopes across a jump does not produce differentiability.

A useful self-test is to compare the cusp and vertical tangent in the visualization. Both involve unbounded one-sided quotients, and the difference is the sign pattern. Then ask what happens for \(-x^{2/3}\) and for \(\sqrt{\vert x\vert}\). The classification comes from the behavior of the two one-sided difference quotients.