If $\frac{df}{dx}$ is continuous on $[a,b]$ is $f$ continuous on $[a,b]$ as well?

Let's consider a special case. If $\frac{df}{dx}=\sqrt{1-x^2}$ then $f(x)=\frac{1}{2}\left(x\sqrt{1-x^2}+\arcsin(x)\right)$ should be continuous on $[a,b]$ since differentiability implies continuity. However, I believe that $f(x)$ is not defined for $|x|> 1$ since it depends on $sin^{-1}(x)$ and in addition to that it becomes complex at that point and I don't know complex analysis. So I suppose my initial statement sort of holds true, but how can $f(x)$ be differentiable at $x=\pm 1$ if $[-1,1]$ is its domain of definition? The high school tangent interpretation of derivatives does not apply anymore when $x\to-1^-$ and $x\to1^+$.

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    $\begingroup$ A differentiable function is always continuous, is it not ? $\endgroup$
    – Astyx
    Nov 27, 2016 at 11:34
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    $\begingroup$ As for differentiability at the endpoints, there is a notion of one-sided derivative which makes perfect sense here. $\endgroup$
    – Wojowu
    Nov 27, 2016 at 11:36

2 Answers 2


On open sets it is true that if $f$ is differentiable then it is continuous. Hence if we replace $[a,b]$ with $]a,b[$ in your original question everything works.

However, for closed intervals of the form $[a,b]$, it does not make sense to ask if $f$ is differentiable (in the standard sense) at the endpoints, but only if it is differentiable from the left/right, i.e. if it admits a one-sided derivative. And if a one-sided derivative exists at one endpoint then the original function $f$ is continuous from the left/right there.

Notice that if $f \colon A \to \mathbb{R}$ is a differentiable function, then the derivative is defined as a function $f' \colon A \to \mathbb{R}$ with the same origin. Strictly speaking it does not make sense to ask if $f'$ is continuous outside $A$, and on $A$ $f$ is clearly continuous.

Let me make an example: $f = \log$. Then $f \colon ]0,+\infty[ \to \mathbb{R}$ and its derivative is $f'(x) = 1/x \colon ]0,+\infty[ \to \mathbb{R}$. You can see that the function $1/x$ is defined on all $\mathbb{R}\setminus \{0\}$, so you can extend the derivative by the same formula there. But formally $f'$ is defined only on $]0,+\infty[$.

  • $\begingroup$ That makes sense, thank you very much. $\endgroup$
    – David
    Nov 27, 2016 at 11:59

For $x$ outside of $[-1, 1]$ your derivative is already undefined, so you can't argue in this way.

If a function is differentiable at a point, then it is also continuous at this point - regardless whether $f'$ is continuous of not. This can easily be seen in the definition of the derivative, as the limit $f'(x_0) = \lim \limits_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$ can only exist if $f(x_0 + h) - f(x_0)$ goes to zero for $h \to 0$, i.e. $f$ is continuous in $x_0$.

  • $\begingroup$ Yes but if $x_0 = -1$ and $h \to 0^-$ then $f(x_0+h)$ is no longer defined and since $h\to 0$ implies that one might as well take the limit as $h \to 0^-$ then it cannot be differentiable att $x = \pm 1$ no? Yet that appears to be the case. $\endgroup$
    – David
    Nov 27, 2016 at 11:50
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    $\begingroup$ No, the definition of a limit takes the domain of the function into account. Otherwise, you would already have problems to define what continuity on the border of the domain even means. $\endgroup$
    – Dominik
    Nov 27, 2016 at 12:04
  • $\begingroup$ I understand, thank you. $\endgroup$
    – David
    Nov 27, 2016 at 12:05

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