Continuously Differentiable Function

Continuously Differentiable Function

A function which has a derivative that is itself a continuous function.

Key Formula

f \in C^1 \iff f'(x) \text{ exists for all } x \text{ in the domain and } f' \text{ is continuous}

Where:

$f$ = The original function
$f'$ = The first derivative of f
$C^1$ = The class of all continuously differentiable functions; the superscript 1 indicates one continuous derivative

Worked Example

Problem: Determine whether f(x) = x³ − 6x is continuously differentiable on all of ℝ.

Step 1: Compute the derivative of f.

f'(x) = 3x^2 - 6

Step 2: Check whether f'(x) exists for every real number x. Since 3x² − 6 is a polynomial, it is defined for all x ∈ ℝ.

Step 3: Check whether f'(x) is continuous. Every polynomial is continuous on all of ℝ, so f'(x) = 3x² − 6 is continuous everywhere.

Answer: Yes, f(x) = x³ − 6x is continuously differentiable (C¹) on all of ℝ because its derivative 3x² − 6 exists everywhere and is continuous.

Another Example

Problem: Show that the function g(x) = x² sin(1/x) for x ≠ 0 and g(0) = 0 is differentiable everywhere but NOT continuously differentiable at x = 0.

Step 1: For x ≠ 0, compute g'(x) using the product rule.

g'(x) = 2x\sin\!\left(\frac{1}{x}\right) - \cos\!\left(\frac{1}{x}\right)

Step 2: At x = 0, use the limit definition of the derivative.

g'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h\sin\!\left(\frac{1}{h}\right) = 0

Step 3: Check continuity of g' at x = 0. As x → 0, the term −cos(1/x) oscillates between −1 and 1 and does not approach g'(0) = 0.

\lim_{x \to 0} g'(x) \text{ does not exist}

Answer: g is differentiable at every point, including x = 0, but g' is not continuous at x = 0. Therefore g is NOT continuously differentiable — it is differentiable but not C¹.

Frequently Asked Questions

Is every differentiable function also continuously differentiable?

No. A function can be differentiable everywhere yet have a derivative that is not continuous. The classic counterexample is g(x) = x² sin(1/x) (with g(0) = 0), which is differentiable at every point, but its derivative oscillates wildly near x = 0 and fails to be continuous there. Continuously differentiable (C¹) is a strictly stronger condition than just differentiable.

What does C¹ mean, and how does it relate to C⁰, C², etc.?

C⁰ denotes the class of continuous functions. C¹ means the first derivative exists and is continuous. C² means the first and second derivatives both exist and are continuous. In general, Cⁿ means all derivatives up to and including the nth are continuous. C∞ describes functions that are infinitely differentiable with every derivative continuous — polynomials, eˣ, and sin x are all C∞.

Differentiable function vs. Continuously differentiable (C¹) function

A differentiable function has a derivative at every point in its domain, but that derivative might not be continuous. A continuously differentiable function satisfies the additional requirement that the derivative itself is continuous. Every C¹ function is differentiable, but not every differentiable function is C¹. In most standard calculus courses, nearly every function you encounter (polynomials, trig functions, exponentials) is actually C∞, so the distinction rarely arises in practice — but it matters in analysis and advanced applications.

Why It Matters

Continuous differentiability is a key hypothesis in many important theorems. For instance, the Inverse Function Theorem and the Implicit Function Theorem require C¹ conditions, not mere differentiability. In physics and engineering, modeling with C¹ functions ensures that quantities like velocity (the derivative of position) change smoothly without sudden jumps, which reflects physically realistic behavior.

Common Mistakes

Mistake: Assuming that if a function is differentiable, its derivative must be continuous.

Correction: Differentiability does NOT guarantee continuity of the derivative. A derivative satisfies the Intermediate Value Theorem (Darboux's theorem), but it can still be discontinuous. You must separately verify that f' is continuous to claim f is C¹.

Mistake: Confusing 'continuous' with 'continuously differentiable.'

Correction: A continuous function (C⁰) need not even be differentiable — for example, f(x) = |x| is continuous everywhere but not differentiable at x = 0. Continuously differentiable (C¹) is two levels stronger: the function must be differentiable, and its derivative must be continuous.

Related Terms

Derivative — The core operation whose continuity defines C¹
Continuous Function — C¹ requires the derivative to be continuous
Function — The general object being classified
Differentiable — Weaker condition — derivative exists but may not be continuous
Smooth Function — C∞ functions — infinitely differentiable
Mean Value Theorem — Foundational theorem requiring differentiability
Polynomial — Polynomials are always C∞, hence C¹