Xtykutl

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:

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

However, when I use derivative rules I get that:

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

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

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

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

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

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

You have $frac{sin(x)}{3sqrt[3]{x^2}}$, which is $frac 00$, or indeterminate. If you express sine in terms of its Taylor Series, and divide each term by ${3sqrt[3]{x^2}}$, 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 $x=0$.

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:
$$
f'(0) = limlimits_{x to 0} frac{sqrt[3]{x} sin(x)-0}{x-0}=0
$$

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

(2) does not contradict (1) since (2) is only valid for $xneq 0$.

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 $lim_{xto 0}f'(x)$ does not exists. In other words, $f'(x)$ is not continuous at $x=0$.
$$
lim_{xto 0} {sin(x) frac{1}{3sqrt[3]{x^2}}+cos(x)sqrt[3]{x}}
=lim_{xto 0}frac{sin x}{x}frac{x}{3sqrt[3]{x^2}}
+lim_{xto 0}cos(x)sqrt[3]{x}=1cdot 0+ 1cdot 0=0tag{3}
$$
which together with (1) implies that $f'$ is actually continuous at $x=0$.

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.