Contradiction between first derivative formal definition and derivative rules?

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?

Solution 1:

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

Solution 2:

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

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 $x=0$.