A function has a derivative at $a$ only if the limit in The Derivative as a Function exists there. This lesson shows the ways it can fail, and connects differentiability with Continuity.
Key idea
If $f$ is differentiable at $a$, then $f$ is continuous at $a$. The converse is false: a continuous function can fail to be differentiable at a corner, a cusp, or a vertical tangent. And a discontinuity always prevents differentiability.
Why it works
Here is the proof that differentiable implies continuous. For $x \ne a$, write $f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a)$. As $x \to a$, the first factor tends to $f'(a)$, a finite number, and the second tends to 0. So $f(x) - f(a) \to f'(a) \cdot 0 = 0$, which says $\lim_{x \to a} f(x) = f(a)$. The argument needs $f'(a)$ to be a number, which is why it does not run backward.
| Failure | What the graph does | Example at 0 |
|---|---|---|
| corner | the left and right slopes are different numbers | $\lvert x \rvert$ |
| cusp | the slopes blow up to $\infty$ on one side and $-\infty$ on the other | $x^{2/3}$ |
| vertical tangent | the slope blows up to $\infty$ (or $-\infty$) on both sides | $x^{1/3}$ |
| discontinuity | the graph breaks | a jump |
Worked example: A corner
Is $f(x) = \lvert x - 3 \rvert$ differentiable at $x = 3$?
- From the right, $f(x) = x - 3$, so the difference quotients are all $1$.
- From the left, $f(x) = -(x - 3)$, so they are all $-1$.
- The one-sided limits of the quotient disagree, so $f'(3)$ does not exist. The function is still continuous at 3.
Answer: No: there is a corner at $x = 3$.
Worked example: A vertical tangent
Is $f(x) = x^{1/3}$ differentiable at 0?
- $\frac{f(h) - f(0)}{h} = \frac{h^{1/3}}{h} = \frac{1}{h^{2/3}}$.
- As $h \to 0$ from either side, $h^{2/3}$ is a tiny positive number, so the quotient tends to $\infty$.
- The tangent line is vertical; its slope is not a real number.
Answer: No: a vertical tangent at $x = 0$.
Worked example: Smooth gluing
Is $$f(x) = \begin{cases} x^2 & x \le 1 \\ 2x - 1 & x > 1 \end{cases}$$ differentiable at 1?
- Values: both pieces give 1 at $x = 1$, so $f$ is continuous.
- Slopes: the left piece has slope $2x = 2$ at 1; the right piece has slope 2.
- The slopes match, so $f'(1) = 2$. (The line is the parabola's tangent line at 1.)
Answer: Yes, $f'(1) = 2$.
Common mistake
"It's continuous, so it's differentiable." The absolute value function is continuous everywhere but has no derivative at 0. Check by computing the slope from each side: both one-sided limits must exist and agree.