Differentiability of x·∛x

Problem

Show that $f(x) = x\,\sqrt[3]{x}$ is differentiable on the whole of $\mathbb{R}$, whereas $g(x) = \sqrt[3]{x}\,/\,x$ has a pathological behaviour at $0$ (specify which).

Solution

Function $f$: $f(x) = x\cdot x^{1/3} = x^{4/3}$. Then $f'(x) = \frac{4}{3}x^{1/3} = \frac{4}{3}\sqrt[3]{x}$, which is defined and continuous for every $x\in\mathbb{R}$ (in particular $f'(0)=0$). Hence $f$ is differentiable on the whole of $\mathbb{R}$.

Function $g$: $g(x) = \dfrac{x^{1/3}}{x} = x^{1/3-1} = x^{-2/3} = \dfrac{1}{\sqrt[3]{x^2}}$. As $x\to 0$ we have $g(x)\to +\infty$: the function is neither defined nor continuous at $0$ (vertical asymptote). The question of differentiability therefore does not even arise: the pathology is a discontinuity (divergence) at $0$.

Links

Topics: Derivatives
Concepts: Continuity · Non-differentiable point
Skills: Differentiate · Reason by cases
Exercise type: Derivative computation