I think a picture can help explain what's going on here. Below is a graph of a function (red) and its derivative (blue). The original function is differentiable, which we can see by its lack of "sudden turns". However, the derivative is not, as evidenced by the "pointy turn" at $(0,0)$.

\begin{align} \text{(red)} \qquad f(x) &= x \cdot |x| \qquad\qquad \qquad \\ \text{(blue)} \quad\,\,\, f'(x) &= 2 \cdot |x| \qquad\qquad \qquad \end{align}

Part of the reason why this idea can be confusing is that the original (red) function looks smooth, but it's not smooth by the math definition since its first derivative isn't differentiable. While $f(x)$ has no pointy bits, it has the spirt of one inside, just waiting to be released.

(By the way, if you're wondering how we derive $f'(x)$ you can see that here.)

I think a picture can help explain what's going on here. Below is a graph of a function (red) and its derivative (blue). The original function is differentiable, which we can see by its lack of "sudden turns". However, the derivative is not, as evidenced by the "pointy turn" at $(0,0)$.

\begin{align} \text{(red)} \qquad f(x) &= x \cdot |x| \qquad\qquad \qquad \\ \text{(blue)} \quad\,\,\, f'(x) &= 2 \cdot |x| \qquad\qquad \qquad \end{align}

Part of the reason why this idea can be confusing is that the original (red) function looks smooth, but it's not smooth by the math definition since its first derivative isn't differentiable. While $f(x)$ has no pointy bits, it has the spirit of one inside, just waiting to be released.

(By the way, if you're wondering how we derive $f'(x)$ you can see that here.)