Contradiction between first derivative formal definition and derivative rules?

Question

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?

Answer 1

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

Answer 2

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 $x=0$.

Answer 3

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

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 $x=0$.

Answer 4

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.