Find all $x$ for which the derivative of $f(x)=\sqrt[3]{x}|\sin{x}|$ doesn't exist

Question

The task is to find all $x$ for which the derivative of $f(x)=\sqrt[3]{x}|\sin{x}|$ doesn't exist. $\sqrt[3]{x}$ doesn't cause any problems. The derivative can be problematic because of $|\sin{x}|$. I think that the values for which the derivative doesn't exist are $x=k\pi, k=\pm1, 2,...$. However, I came up with my solution by intuition and I haven't proven that no other $x$, which fit the condition, exist. If I try to write my thoughts, they would be something like this: $f'(x)=\frac{|\sin{x}|}{3\sqrt[3]{x^2}}+\sqrt[3]{x}\cos{x}$ and $\lim\limits_{x\rightarrow \pi+}{f'(x)}\neq \lim\limits_{x\rightarrow \pi-}{f'(x)}$. But then comes the problem about the zero. I think the derivative at that points exists because $f'(0)=\frac{|\sin{x}|x}{3x\sqrt[3]{x^2}}+\sqrt[3]{x}\cos{x}=\frac{|\sin{x}|\sqrt[3]{x}}{3x}+\sqrt[3]{x}\cos{x}=0$.

But I think that what I've written is very vague and not enough. What is the formal way of solving the problem? How sould I write the solution on a test?

Answer 1

You approached the problem correctly.

Hint

$$f(x)= \sqrt[3]{x}|sin{x}|$$ $$\implies f'(x)=\frac{|\sin{x}|\sqrt[3]{x}}{3x} + \frac{|\sin{x}|*\cos{x}}{\sin{x}}*\sqrt[3]{x}$$

Wondering from where the $ \frac{|\sin{x}|*\cos{x}}{\sin{x}}$ came?

Well you made a mistake while differentiating $f(x)$. The correct differentiation of $|x|$ is $\frac{|x|}{x}$.

So apply chain rule to get $f'(x)$.

The rest approach would remain same. Just see when the function shows different values when approaching from $x^+$ and $x^-$.