First derivative via power rule

Using the power rule with $n = 1/3$: $f'(x) = \frac{1}{3}x^{1/3 - 1} = \frac{1}{3}x^{-2/3}$.

This can also be written as $f'(x) = \dfrac{1}{3x^{2/3}}$.

Differentiate $f'(x) = \frac{1}{3}x^{-2/3}$ term by term: multiply the coefficient by the exponent and reduce the exponent by 1, giving $f''(x) = \frac{1}{3}\cdot\left(-\frac{2}{3}\right)x^{-2/3-1}$.

Simplifying, $f''(x) = -\frac{2}{9}x^{-5/3}$, which can also be written as $-\dfrac{2}{9x^{5/3}}$.

Differentiate $f''(x) = -\frac{2}{9}x^{-5/3}$: multiply the coefficient by the exponent and subtract 1 from the exponent, giving $f'''(x) = -\frac{2}{9}\cdot\left(-\frac{5}{3}\right)x^{-5/3-1}$.

Simplifying, $f'''(x) = \frac{10}{27}x^{-8/3}$, which can also be written as $\dfrac{10}{27x^{8/3}}$.

Each derivative $f'(x) = \frac{1}{3}x^{-2/3}$, $f''(x) = -\frac{2}{9}x^{-5/3}$, and $f'''(x) = \frac{10}{27}x^{-8/3}$ involves $x$ raised to a negative power, meaning $x$ appears in the denominator.

As $x \to 0$, each denominator approaches 0, so the magnitude of each derivative grows without bound (the limits are infinite, and the one-sided limits may even differ in sign in some cases). Therefore none of $f'(0)$, $f''(0)$, or $f'''(0)$ exist — $f(x) = x^{1/3}$ has a vertical tangent line and increasingly singular behavior in its higher derivatives at $x=0$.

They alternate. The first derivative has coefficient one third, which is positive; the second derivative has coefficient negative two ninths; and the third derivative has coefficient ten twenty-sevenths, which is positive again. This alternating pattern arises because each differentiation step multiplies by the current exponent, and the exponents themselves alternate in sign as they decrease by one each time.