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

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 $$x\to1^+$$.

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

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 $]0,+\infty[$.

• That makes sense, thank you very much. 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 $x_0$.

• 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. Nov 27, 2016 at 11:50
• 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. Nov 27, 2016 at 12:04
• I understand, thank you. Nov 27, 2016 at 12:05