When I try to find the derivative of $f(x) = \sqrt[3]{x} \sin(x)$ at $x=0$, using the formal definition of first derivative, I get this:

$$ f'(0) = \lim\limits_{x \to 0} \frac{\sqrt[3]{x} \sin(x)-0}{x-0},$$ which gives zero.

However, when I use derivative rules I get that:

$$ f'(x) = {\sin(x) \frac{1}{3\sqrt[3]{x^2}}+\cos(x)\sqrt[3]{x}} $$

and thus $f'(0)$ doesn't exist.

Why does this happen? what's the reason behind it?

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    I'm not sure I follow. Note that $\sin(x)\sim x$ around $x=0$, and so $\sin(x)/x^{2/3}\sim x^{1/3}$, which goes to $0$ as $x\to0$. So $f'(x)\to0$, as expected. – AccidentalFourierTransform Nov 8 at 16:56
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    @AccidentalFourierTransform your comment actually constitutes an answer and shouldn't be just a comment. – Ister Nov 9 at 9:17
  • +1 for a good question. – Randall Nov 9 at 13:25
up vote 43 down vote accepted

The rule $(fg)'=f'g+fg'$ works where $f$ and $g$ are differentiable. And $\sqrt[3]x$ is not differentiable at $x=0$.

  • Oh, alright I see. But, why is the first derivative able to find it? I mean, I have never really understood the differences between direct derivative rules and formal definition. – Just_Cause Nov 8 at 15:16
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    It is possible that $g$ is not differentiable and $fg$ is. – ajotatxe Nov 8 at 15:18
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    It makes sense now. – Just_Cause Nov 8 at 15:19

$$ f'(x) = {\sin(x) \frac{1}{3\sqrt[3]{x^2}}+\cos(x)\sqrt[3]{x}} $$

and thus $f'(0)$ doesn't exist.

You have $\frac{\sin(x)}{3\sqrt[3]{x^2}}$, which is $\frac 00$, or indeterminate. If you express sine in terms of its Taylor Series, and divide each term by ${3\sqrt[3]{x^2}}$, you will see that you get a series that evaluates to zero at $x=0$. The lesson here is just because you can put something in an indeterminate form, doesn't mean that it doesn't exist. For instance, one can rewrite $x^2$ as $\frac {x^3} x$; that doesn't mean that $x^2$ doesn't exist when $x=0$.

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    Your argument is appealing but incorrect. One can write $x^2$ as $x^3/x$ only when $x\neq 0$. What you are trying to do here is to find limit $\lim_{x\to 0}f'(x)$ and show that it equals $f'(0) $. This works here only because $f'$ is continuous in this case. Consider another example where $g(x) =x^2\sin(1/x),g(0)=0$ then $g'(0)=0$ but $\lim_{x\to 0}g'(x)$ does not exist. – Paramanand Singh Nov 8 at 18:53
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    @ParamanandSingh And the OP's application of the product rule is valid for $x \neq 0$. So my example is analogous. – Acccumulation Nov 8 at 18:56
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    I think the intent of your answer is to explain how we can make sense of the formula obtained by product rule even when $x=0$. This is something which I believe is not the right way to think about the situation. – Paramanand Singh Nov 8 at 18:58
  • Wait @Paramanand Singh, how's $f'(x)$ continous here? And in your example why is $g(0) = 0$ ? I'm a bit confused – Stephen Alexander Nov 9 at 5:55
  • @StephenAlexander: note that $f'(0)=0$ and $\lim_{x\to 0}f'(x)=0 $ (check this!) so that makes $f'$ continuous. For my example I have chosen $g(0)=0$ this ensures that $g$ is continuous at $0$. – Paramanand Singh Nov 9 at 5:59

When I try to find the derivative of $f(x) = \sqrt[3]{x} \sin(x)$ at $x=0$, using the formal definition of first derivative, I get this: $$ f'(0) = \lim\limits_{x \to 0} \frac{\sqrt[3]{x} \sin(x)-0}{x-0}=0 $$

This is correct since: $$ \lim_{x \to 0} \frac{\sqrt[3]{x} \sin(x)-0}{x-0} =\lim_{x \to 0} \sqrt[3]{x}\cdot \frac{ \sin(x)}{x}=0.\tag{1} $$

However, when I use derivative rules I get that: $$ f'(x) = {\sin(x) \frac{1}{3\sqrt[3]{x^2}}+\cos(x)\sqrt[3]{x}}\tag{2} $$

(2) does not contradict (1) since (2) is only valid for $x\neq 0$.

and thus $f'(0)$ doesn't exist.

This implication is false:

[Edited later (thanks to comments by MMASRP63 and Paramanand Singh)] (2) only implies that the limit $\lim_{x\to 0}f'(x)$ does not exists. In other words, $f'(x)$ is not continuous at $x=0$. $$ \lim_{x\to 0} {\sin(x) \frac{1}{3\sqrt[3]{x^2}}+\cos(x)\sqrt[3]{x}} =\lim_{x\to 0}\frac{\sin x}{x}\frac{x}{3\sqrt[3]{x^2}} +\lim_{x\to 0}\cos(x)\sqrt[3]{x}=1\cdot 0+ 1\cdot 0=0\tag{3} $$ which together with (1) implies that $f'$ is actually continuous at $x=0$.

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    I believe that $f'$ is continuous at $x=0$ in this instance? – MMASRP63 Nov 8 at 18:13
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    Your last two sentences are incorrect and one should observe that using continuity of $f'$ at $0$ to evaluate $f'(0)$ is a very silly/roundabout method. – Paramanand Singh Nov 8 at 18:56
  • @MMASRP63: I should really not do the calculation in my head. What a silly mistake. Thanks for pointing it out! – user587192 Nov 8 at 19:49
  • @ParamanandSingh: thank you for pointing it out! – user587192 Nov 8 at 19:49

I want to add something to user587192's answer.

Indeed, there is no contradiction, and

$$\lim_{x \to 0} f'(x) = 0$$

as shown before. However, if you don't want to evaluate the derivative using the formal definition, you can use a theorem:

Suppose $f'(x)$ exists in a deleted neighbourhood of $a$ and $\lim_{x \to a} f'(x) $ exists and equals $L$. Then $f'(a)$ exists and equals $\lim_{x \to a} f'(x)$.

This is an immediate consequence of L'Hospital's Rule. Notice that both $f(x)-a$ and $x-a$ are differentiable on a deleted neighbourhood of $a$, and $\lim_{x \to a} \frac{f'(x)}{1}$ exists and equals $L$, we conclude that $\lim_{x \to a} \frac{f(x)-a}{x-a}$ exists and equals $L$. Hence $f'(a)=L$.

Indeed, in your example, it would be much faster to calculate $f'(0)$ using the formal definition. However, in some cases the above theorem does help.

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