Differentiability Implies Continuity

Statement	Truth Value
Differentiable at $x_0$ $\Rightarrow$ Continuous at $x_0$	True
Continuous at $x_0$ $\Rightarrow$ Differentiable at $x_0$	False in general
Discontinuous at $x_0$ $\Rightarrow$ Not differentiable at $x_0$	True (contrapositive)

The Theorem

The following theorem tells us that only continuous functions are differentiable. Notice, however, that not all continuous functions are differentiable.

Proof

To show $f$ is continuous at $x=x_0$, we need to show that $f(x_0)$ exists and $$\underset{h\to 0}{lim}f(x_0+h)=f(x_0).$$ We use the identity (which holds for $\mathrm{\Delta }x\ne 0$): $$f(x_0+\mathrm{\Delta }x)=f(x_0)+\mathrm{\Delta }x\cdot \frac{f(x_0+\mathrm{\Delta }x)-f(x_0)}{\mathrm{\Delta }x}.$$ Taking the limit of both sides as $\mathrm{\Delta }x\to 0$: \begin{aligned} \lim_{\Delta x \to 0} f(x_0 + \Delta x) &= \lim_{\Delta x \to 0} \left[f(x_0) + \Delta x \cdot \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}\right] \\ &= \lim_{\Delta x \to 0} f(x_0) + \left(\lim_{\Delta x \to 0} \Delta x\right)\left(\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}\right) \\ &= f(x_0) + 0 \cdot f'(x_0). \end{aligned} By hypothesis, f'(x_0) exists and is finite. Therefore 0 \cdot f'(x_0) = 0, the right-hand side equals $f(x_0)$, and $$\underset{\mathrm{\Delta }x\to 0}{lim}f(x_0+\mathrm{\Delta }x)=f(x_0).$$ This completes the proof.

Implications

The above theorem tells us:

$$\boxed{\text{Differentiable}\Rightarrow \text{Continuous}}$$

It follows immediately that if a function $f$ is discontinuous at a point $x_0$, then f'(x_0) does not exist:

$$\boxed{\text{Not continuous}\Rightarrow \text{Not differentiable}}$$

The Converse Is False

It is important to remember the theorem: a differentiable function is continuous. However, it is just as important to remember that the converse is not true: a continuous function need not be differentiable.

For example, the absolute value function $f(x)=|x|$ is continuous everywhere, but its graph has a sharp corner at $x=0$ and thus is not differentiable there. Keeping this example in mind will help you avoid confusing which statement is true.

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Constructing continuous functions that are not differentiable at multiple points, or even at an infinite number of points, is straightforward. Historically, mathematicians believed that continuous functions were differentiable except possibly at isolated points. This notion was upended in the late 19th century when Karl Weierstrass introduced a class of functions that are continuous everywhere but nowhere differentiable.

The graph of this function displays an infinite number of corners, at which tangent lines cannot be constructed. Consequently, the function is not differentiable at these infinitely many points. For further details, refer to the article on the Weierstrass function.

Frequently Asked Questions

Why does a corner prevent differentiability?

At a corner, the curve approaches the point from two different directions, giving two different one-sided tangent slopes. Since the left-hand and right-hand limits of the difference quotient are unequal, the two-sided limit does not exist, and therefore the derivative does not exist.

If I know a function is not continuous at a point, can I still compute its derivative there?

No. The theorem guarantees that differentiability implies continuity, so its contrapositive says: if $f$ is not continuous at $x_0$, then $f$ is not differentiable at $x_0$. There is no need to set up the limit; discontinuity alone rules out differentiability.

Are there functions that are continuous everywhere but differentiable nowhere?

Yes. The Weierstrass function is the classic example. It is continuous at every real number but has no tangent line at any point. Such functions surprised 19th-century mathematicians, who had assumed continuity was "almost always" sufficient for differentiability.