When a Function Is Not Differentiable at a Point

Case	Description	Example
Discontinuity	$f$ has a jump or infinite blow-up	Heaviside step function, $1/x^2$
Corner / Cusp	One-sided derivatives exist but differ	$|x|$ at $x=0$
Vertical tangent	Difference quotient $\to \pm \mathrm{\infty }$	$x^{1/3}$ at $x=0$
Oscillation	Difference quotient has no limit	$x\sin (1/x)$ at $x=0$

The Four Cases of Non-Differentiability

Here are the cases in which $f$ is not differentiable at a point $x_0$.

Case 1: $f$ is discontinuous at $x=x_0$.

If there is a jump in the graph of $f$ at $x_0$, or $\underset{x\to x_0}{lim}f(x)=+\mathrm{\infty }$ or $-\mathrm{\infty }$, the function is not differentiable at the point of discontinuity.

For example, the Heaviside step function

H(x)=\begin{cases}1 & \text{if }0\leq x\\0 & \text{if }x<0\end{cases}

is not continuous at $x=0$; therefore H'(0) does not exist.

Another example is $F(x)=1/x^2$. Because $\underset{x\to 0}{lim}F(x)=+\mathrm{\infty }$, $F$ is not differentiable at $x=0$.

Case 2: f'_+(x_0) and f'_-(x_0) both exist but f'_+(x_0)\neq f'_-(x_0).

In this case, there is a corner in the graph of $f$.

Case 3: The tangent line is vertical.

In this case,

$$\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x_0+\mathrm{\Delta }x)-f(x_0)}{\mathrm{\Delta }x}=+\mathrm{\infty }\text{or}-\mathrm{\infty }$$

or, more generally, \lim_{\Delta x\to0}|f'(x)|=+\infty.

For example, consider $f(x)=x^{1/3}$. The tangent to its graph at $(0,0)$ is vertical. This function is not differentiable at $x=0$, for

\begin{aligned} \left.\frac{d(\sqrt[3]{x})}{dx}\right|_{x=0} &=\lim_{\Delta x\to0}\frac{\sqrt[3]{0+\Delta x}-\sqrt[3]{0}}{\Delta x}\\ &=\lim_{\Delta x\to0}\frac{\sqrt[3]{\Delta x}}{\Delta x}\\ &=\lim_{\Delta x\to0}\frac{1}{\Delta x^{(1-1/3)}}\\ &=\lim_{\Delta x\to0}\frac{1}{\sqrt[3]{(\Delta x)^{2}}}\\ &=+\infty. \end{aligned}

Case 4: The difference quotient approaches neither a number nor $\pm \mathrm{\infty }$.

Consider

f(x)=\begin{cases}x\sin\dfrac{1}{x} & \text{if }x\neq0\\0 & \text{if }x=0\end{cases}

This function is not differentiable (although it is continuous) at $x=0$, because

\begin{aligned} f'(0)&=\lim_{\Delta x\to0}\frac{f(0+\Delta x)-f(0)}{\Delta x}\\ &=\lim_{\Delta x\to0}\frac{\Delta x\sin\frac{1}{\Delta x}-0}{\Delta x}\\ &=\lim_{\Delta x\to0}\sin\frac{1}{\Delta x} \end{aligned}

does not exist. The graph of $f(x)$ is shown below.

Frequently Asked Questions

Can a function be continuous but not differentiable?

Yes. The function $f(x)=|x|$ is continuous everywhere but not differentiable at $x=0$ because of the corner there. Similarly, $f(x)=x\sin (1/x)$ (extended by $f(0)=0$) is continuous but not differentiable at $x=0$ because the difference quotient oscillates.

Does differentiability imply continuity?

Yes, always. If $f$ is differentiable at $x_0$, then $f$ is also continuous at $x_0$. The converse is false, as the examples above show.

What does a vertical tangent look like on the graph?

A vertical tangent appears as a point where the curve goes straight up or down with infinite steepness. The function $y=x^{1/3}$ has a vertical tangent at the origin: the curve passes through $(0,0)$ but its slope is infinite there.