When we say $f \in C^1$, we mean that $f$ is continuously differentiable. Isn't the continuity a redundant word? I mean, we have a theorem that says if $f$ is differentiable then it is continuous. So why in most of the textbooks they always mention them two?

So these are all equivalent:

  • $f \in C^1$
  • $f$ is continuously differentiable
  • $f'$ exists
  • 40
    $\begingroup$ A function $f$ is continuously differentiable if $f$ is differentiable and its derivative is continuous. No, those are not all equivalent. The first two are; the third is weaker. The first two require that $f'$ not merely exist, but also be continuous. $\endgroup$ Jan 24, 2015 at 5:57
  • 7
    $\begingroup$ Note the ending "-ly", which makes it an adverb, not an adjective. So "continuously differentiable" means "differentiable in a continuous way". $\endgroup$ Jan 24, 2015 at 9:22

1 Answer 1


No, they are not equivalent.

A function is said to be differentiable at a point if the limit which defines the derivate exists at that point. However, the function you get as an expression for the derivative itself may not be continuous at that point. A good example of such a function is $$f(x) = \begin{cases}x^2(\sin(\frac{1}{x^2})) &\quad x \neq 0 \\ 0 &\quad x = 0 \end{cases}$$ which has a finite derivative at $x=0,$ but the derivative is essentially discontinuous at $x=0.$

A continuously differentiable function $f(x)$ is a function whose derivative function $f'(x)$ is also continuous at the point in question.

In common language, you move the secant to form a tangent and it may give you a real tangent at that point, but if you see the tangents around it, they will not seem to be approaching this tangent in any sense. Might sound counter intuitive, but it is possible. Such a function is not a continuously differentiable.

  • 3
    $\begingroup$ $f(x)=x^2 \sin\frac{1}{x^2}$ (you didn't specify $x$) does not have a finite derivative at $x=0$ because it is not even continuous there. If you change it to: $f(x)=x^2\sin\frac{1}{x^2}$ for $x\ne 0$ and $f(x)=0$ for $x=0$, then $f$ is continuous at $x=0$ and it also has a finite derivative there. $\endgroup$
    – Redbox
    Jun 20, 2021 at 12:13

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