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It has been along time since I did real analysis. Someone asked me this question and I could not respond. I am not sure if this is the right place to ask.

He said:

The function $f: [0, \infty)\mapsto [0, \infty)$ given by $f(x)=\sqrt{x}$ is not differentiable at $0$ because its derivative $f'(x)=1/(2\sqrt{x})$ is not continuous at $0$.

I said (what I could remember):

It is not differentiable at $0$ because $\lim_{x\to 0}(f(x)-f(0))/(x-0)=\infty$.

He said that I know that but what I told is it true?

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    $\begingroup$ Hopefully your natural reaction would be that this is not true, since in general one can't show that a function doesn't exist by showing "if it did exist, then it would not be continuous". $\endgroup$ – Dave L. Renfro Aug 3 '16 at 19:13
  • $\begingroup$ @TonyS.F. Actually, no; they are not saying the same thing. See my posted answer for a classical example of a function that is everywhere differentiable, but not continuous everywhere. $\endgroup$ – Mark Viola Aug 3 '16 at 19:25
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    $\begingroup$ A meta/rule-of-thumb argument: if that were true, why on Earth would we have two separate concepts for "differentiable" and "continuously differentiable"? $\endgroup$ – Clement C. Aug 3 '16 at 19:42
  • $\begingroup$ Possible duplicate of Discontinuous derivative. $\endgroup$ – Mehrdad Aug 3 '16 at 20:14
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    $\begingroup$ Your friend is wrong: as $f'$ isn't defined for $x=0$ it is neither continuos nor discontinuous, it simply doesn't exist. $\endgroup$ – Michael Hoppe Aug 4 '16 at 10:36
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The derivative $f'(x)$ of a differentiable function, $f(x)$, need not be continuous itself. A classical example is the function

$$f(x)=\begin{cases}x^2\sin(1/x)&,x\ne0\\\\0&,x=0\end{cases}$$

Then, we have

$$f'(x)=\begin{cases}2x\sin(1/x)-\cos(1/x)&,x\ne0\\\\0&,x=0\end{cases}$$

Clearly, $\lim_{x\to 0}f'(x)$ does not even exist while $f'(0)=0$ as shown by

$$f'(0)=\lim_{h\to 0}\frac{h^2\sin(1/h)-0}{h}=0$$

Hence, we have an example of a function that is differentiable at $x=0$ although it derivative is discontinuous at $x=0$.

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Another counterexample, which has the additional property $\limsup\limits_{x\to 0^+} f'(x)=\infty$ and $\liminf\limits_{x\to 0^+} f'(x)=-\infty$ is $$f(x):=x^2\left( \sin\frac1x\right)\ln\lvert x\rvert$$ whose derivative (basically the same reason as DR.MV's answer) is $$f'(x)=\begin{cases}2x\left( \sin\frac1x\right)\ln\lvert x\rvert-\color{blue}{\left(\cos\frac1x\right)\ln\lvert x\rvert}+x\sin\frac1x&\text{if }x\ne0\\ 0&\text{if }x=0\end{cases}$$

However, there are restrictions on how a derivative can be discontinuous at a given point. For instance, an interesting question for the sake of your problem is:

Let $f$ be a differentiable function $[0,\varepsilon)\to \Bbb R$. Can $\lim\limits_{x\to0^+} f'(x)=\infty$ hold?

The answer is no: in fact, assume as a contradiction that this were the case. Then, by extending $f$ to $$\overline f(x):=\begin{cases} f(x)&\text{if }x\in[0,\varepsilon)\\ f'(0)x+f(0)&\text{if }x<0\end{cases}$$

we can assume that $f$ is a differentiable function on $(-\infty,\varepsilon)\ni 0$ with constant derivative for $x\le0$.

Now, due to Darboux's theorem, the image under $f'$ of any interval $(-\delta,\delta)$ must be an interval $I$ containing $f'(0)$. But this cannot be the case, because, since $\lim\limits_{x\to 0^+}f'(x)=\infty$, $\{f'(0)\}\subsetneqq f'(-\delta,\delta)\subseteqq \{f'(0)\}\cup [f'(0)+1,\infty)$ for $\delta$ sufficiently small. Absurd.

With the same idea, you can also prove that the derivative of a function, though it can be discontinuous, it cannot have jump discontinuities.

Added: To spill the beans, at each point $x$ it must hold $$\liminf_{t\to x^+} f'(t)\le f'(x) \le \limsup_{t\to x^+}f'(t)\\ \liminf_{t\to x^-} f'(t)\le f'(x) \le \limsup_{t\to x^-}f'(t)$$

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    $\begingroup$ For reference: Darboux's theorem states that if $f$ is a real-valued, differentiable function, then $f'$ has the intermediate value property. $\endgroup$ – Clement C. Aug 3 '16 at 19:51
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    $\begingroup$ I was going to point out both of your points. An example of a differentiable function with unbounded derivative that seems simpler to me is given by $x^2\sin(1/x^2)$. $\endgroup$ – David C. Ullrich Aug 3 '16 at 19:53
  • $\begingroup$ @DavidC.Ullrich In fact, while I was juggling with logarithms, I started wondering if I was overcomplicating. How could I not think of it? :D $\endgroup$ – user228113 Aug 3 '16 at 19:55
  • $\begingroup$ What does the subset symbol with the equal/different sign below it mean? "$\subsetneqq$" $\endgroup$ – someonewithpc Aug 3 '16 at 23:09
  • $\begingroup$ @someonewithpc "$A\subsetneqq B$" (or "$A\subsetneq B$") means "$A\subseteq B$ but $A\neq B$". $\endgroup$ – user228113 Aug 3 '16 at 23:15
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Just to add a thought to what others already said.

As Dr MV pointed out, a function is differentiable at a point if the limit of the difference quotient exists and is finite. That's it. Continuity of the derivative is something more. Therefore your answer was correct.

However, continuity is required when we say that a function $f$ is in $\mathcal{C}^1(I)$ (here $I$ is an interval): $f$ is in $\mathcal{C}^1(I)$ if it is differentiable in $I$, and the derivative is also continuous in $I$.

In this context, the example given by Dr MV is an example of a function which is differentiable everywhere, but does not belong to $\mathcal{C}^1(\mathbb{R})$.

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