AP Calculus AB and BC

Why Continuity Does Not Imply Differentiability

Q: What are the three ways differentiability fails?

A corner, where the one-sided slopes disagree, as with |x| at 0. A vertical tangent, where the difference quotient runs to infinity, as with x^{1/3} at 0. And a discontinuity, which rules out the derivative straight away by this theorem.

Q: Is every differentiable function continuously differentiable?

No, that is a further step. f(x) = x^2 \sin(1/x) with f(0) = 0 is differentiable everywhere, but f' is discontinuous at 0. Differentiable does not mean the derivative is well behaved.

Differentiability implies continuity, and the implication runs one way only. A function can be continuous everywhere and still have no derivative at a point: the absolute value function is continuous at 0 and has no derivative there, because the two one-sided slopes disagree.

Differentiability implies continuity

f \text{ differentiable at } a \implies f \text{ continuous at } a

If the derivative of f exists at a point, then f is continuous at that point. The converse fails.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
f is differentiable at the point a
$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \text{ exists}$
This is the only hypothesis, and the theorem is a one-way street. The useful question is what happens if you try to run it backwards and assume continuity buys you differentiability, which is the single most common error in Unit 2.
Drop it and the theorem fails
$f(x) = |x|$
f(x) = |x| on [-1, 1], continuous at 0 with no derivative there
Absolute value is continuous at every real number, including 0. If continuity implied differentiability, this function would be differentiable throughout $(-1,1)$ , and since $f(-1) = f(1)$ Rolle's Theorem would then hand us a point where the derivative is 0. The check confirms there is no such point: the slope is $-1$ on the left and $+1$ on the right, and undefined between. So continuity did not buy differentiability.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Write the difference $f(a+h) - f(a)$ as the product $\frac{f(a+h) - f(a)}{h} \cdot h$ , valid for $h \ne 0$ .
As $h \to 0$ the first factor tends to $f'(a)$ , which exists by hypothesis and is a finite number.
The second factor tends to 0, so the product tends to $f'(a) \cdot 0 = 0$ .
So $f(a+h) \to f(a)$ , which is exactly the statement that $f$ is continuous at $a$ . The finiteness of $f'(a)$ is what makes the product collapse, which is why an infinite derivative does not qualify.

What it does not say

Continuous means differentiable.

The implication only runs one way. $|x|$ at 0 is the standard corner, $x^{1/3}$ at 0 is the standard vertical tangent, and Weierstrass built a function that is continuous at every point of the line and differentiable at none.

A function with a vertical tangent is differentiable there.

It is not. $f(x) = x^{1/3}$ is continuous at 0 and its difference quotient runs off to infinity, so the derivative does not exist as a finite number. Continuity survives, differentiability does not.

If f is discontinuous at a, it might still be differentiable there.

It cannot be. This theorem, read as a contrapositive, says exactly that: no continuity, no derivative. That direction is the one you can rely on.

Frequently asked questions

What are the three ways differentiability fails?

A corner, where the one-sided slopes disagree, as with $|x|$ at 0. A vertical tangent, where the difference quotient runs to infinity, as with $x^{1/3}$ at 0. And a discontinuity, which rules out the derivative straight away by this theorem.

Is every differentiable function continuously differentiable?

No, that is a further step. $f(x) = x^2 \sin(1/x)$ with $f(0) = 0$ is differentiable everywhere, but $f'$ is discontinuous at 0. Differentiable does not mean the derivative is well behaved.