What Does Differentiable Mean?


Posted on Thu, Jul 9, 2026
Tags math, calculus, cowork-with-llm
math, calculus, cowork-with-llm
📝 This article is a translation of the original Japanese post. View original

What Does Differentiable Mean?

When relearning analysis, the word “differentiable” shows up right at the beginning, yet it’s surprisingly easy to slide past it with the definition still vague. This article organizes systematically what differentiability is, the conditions for being differentiable, and how it differs from being “analytic.”

Conclusions

  • Differentiable means that the slope of the tangent line at a point (the derivative) is determined uniquely
  • Intuitively, it means that as you zoom into the graph it approaches a single straight line and the corner disappears
  • Being continuous with no corners is a necessary condition for differentiability, but not a sufficient one
  • Differentiability has a hierarchy of smoothness called class $C^k$ , meaning the function can be differentiated $k$ times and that derivative is continuous
  • “Infinitely differentiable (class $C^\infty$ )” and “analytic (class $C^\omega$ )” are different things; being $C^\infty$ does not imply being analytic
  • The difference between the two comes down to whether the remainder term of the Taylor expansion converges to $0$ so that the series agrees with the original function
  • In the world of complex functions, being differentiable once automatically makes a function analytic — a property not seen with real functions

Prerequisites

  • Intended reader: someone who knows the basic terminology of calculus but wants to systematically organize the definition of “differentiable” and the concepts around it

Definition of Differentiable

A function $f(x)$ is differentiable at a point $a$ when the following limit (the derivative) exists.

$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

For this limit to exist, the right-hand and left-hand limits must agree.

$$\lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$$

Intuition: Does It Look Like a Line When You Zoom In?

However far you zoom into the graph of a differentiable function, it eventually looks like a single straight line. Conversely, at a point where the function is not differentiable, a corner (a sharp point) remains no matter how far you zoom, and the slope isn’t determined uniquely.

A typical example that isn’t differentiable is $f(x) = |x|$ . At $x = 0$ we get the following.

  • Approaching from the right, the slope is $1$
  • Approaching from the left, the slope is $-1$

Because the right-hand and left-hand limits don’t agree, the function isn’t differentiable at $x = 0$ .

To be differentiable, at least the following two conditions are necessary.

  1. Being continuous (the graph is connected)
  2. Being smooth (no corners)

However, these two conditions are necessary, not sufficient. There exist functions that appear continuous with no corners yet are not differentiable (the Weierstrass function is a famous counterexample).

The Class $C^k$ Hierarchy of Smoothness

Differentiability has stages depending on “how many times it can be differentiated, and how far those derivatives stay continuous.” This is called class $C^k$ .

A function $f(x)$ is of class $C^k$ when it satisfies the following two conditions.

  1. It is $k$ times differentiable
  2. The $k$ -th derivative $f^{(k)}(x)$ is continuous

The $C$ is the initial of “continuous.” The key point is that it requires not just “can be differentiated” but also “the result of differentiating is continuous.”

NotationMeaning
Class $C^0$Continuous (corners are allowed)
Class $C^1$Differentiable once, and that derivative is continuous too
Class $C^2$Differentiable twice, and curvature is continuous. Important for things like acceleration in physics
Class $C^k$Differentiable $k$ times, and the $k$ -th derivative is continuous
Class $C^\infty$Differentiable any number of times

Concrete Example: The Smoothness Hierarchy Seen Through a Power Function

$$f(x) = |x|^3$$

Taking this as an example, let’s see how far smoothness is preserved as we differentiate repeatedly.

  • The original function is connected (class $C^0$ )
  • Differentiating once gives $f'(x) = 3x|x|$ , the corner disappears and it becomes smooth (class $C^1$ )
  • Differentiating twice gives $f''(x) = 6|x|$ , still connected (class $C^2$ )
  • Differentiating three times, a corner (a V shape) appears at $x = 0$ , and it can’t be differentiated further

So this function is “class $C^2$ but not class $C^3$ .”

The Difference Between Differentiable and Analytic

“Differentiable” and “analytic” are both words describing the smoothness of a function, but the level of smoothness they demand is completely different.

  • Differentiable: at that point, the slope of the tangent is determined uniquely (being differentiable once is enough)
  • Analytic (class $C^\omega$ ): not only differentiable any number of times around that point, but also the Taylor expansion agrees with the original function

By Taylor’s theorem, the remainder term of the Taylor expansion up to order $n$ can be written as follows.

$$R_n(x) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$$

Being class $C^\infty$ only guarantees that this $R_n(x)$ can be defined for any $n$ . Being analytic, on the other hand, additionally requires that $R_n(x) \to 0$ as $n \to \infty$ , so that the series agrees with the original function $f(x)$ .

AspectDifferentiableAnalytic
Number of differentiationsAt least onceInfinitely many
Series expansionNot necessarily possibleCan be expanded as a power series around the point
Binding powerProperties at the point don’t affect its neighborhoodInformation at the point alone determines values in the neighborhood

$e^x$ , $\sin x$ , $\cos x$ , and polynomials are all analytic.

An Example That Is Class $C^\infty$ but Not Analytic

This is the biggest pitfall: “infinitely differentiable” and “analytic” are not the same thing.

Consider the following function.

$$f(x) = e^{-1/x} \quad (x > 0)$$$$f(x) = 0 \quad (x \le 0)$$

However many times you differentiate this function at $x = 0$ , the derivative comes out $0$ . So it is infinitely differentiable (class $C^\infty$ ). But its Taylor expansion at $x = 0$ has every term equal to $0$ , yielding nothing but the expression “identically $0$ .” Meanwhile the actual function takes positive values for $x > 0$ , so the Taylor expansion and the original function don’t agree. Therefore this function is not analytic at $x = 0$ .

Seen through the remainder term: since every term of the Taylor expansion is $0$ , we have $R_n(x) = f(x)$ , and since $f(x) = e^{-1/x} > 0$ for $x > 0$ , $R_n(x)$ does not converge to $0$ even as $n \to \infty$ .

A Special Property of Complex Functions

In the world of real functions, there is this much of a gap between differentiable, class $C^\infty$ , and analytic. In the world of complex functions (complex analysis), however, things are very different.

If a complex function is differentiable once (holomorphic) on some region of the complex plane, it is known that it automatically becomes differentiable any number of times on that region, and moreover analytic. For real functions none of the arrows in “differentiable $\to$ class $C^\infty$ $\to$ analytic” hold, but for complex functions differentiability once guarantees analyticity in one leap.

Caveats

  • The discussion of class $C^k$ assumes single-variable real functions. For multivariable functions you need to distinguish partial differentiability from total differentiability
  • There are examples of functions that look “continuous with no corners” yet are not differentiable (such as the Weierstrass function), so it’s better not to judge differentiability by appearance alone
  • Confusing class $C^\infty$ with analytic (class $C^\omega$ ) can lead you to misjudge the range of convergence of a Taylor expansion

Summary

  • Differentiable means that at a point the slope of the tangent line (the derivative) is determined uniquely
  • Being continuous with no corners is a necessary condition for differentiability, but not a sufficient one
  • Class $C^k$ expresses the smoothness hierarchy of “differentiable $k$ times, and that derivative is continuous”
  • Class $C^\infty$ (infinitely differentiable) and analytic (class $C^\omega$ ) are different things; being $C^\infty$ does not imply being analytic
  • In the world of complex functions, being differentiable once automatically makes a function analytic

References

  • Teiji Takagi, Kaiseki Gairon (An Introduction to Analysis), Iwanami Shoten
  • Walter Rudin, “Principles of Mathematical Analysis”, McGraw-Hill
  • Lars V. Ahlfors, “Complex Analysis”, McGraw-Hill

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