= Stationary points with real cube-root factors
{title2=$[(x-a)\sqrt[3]{q(x)}]'=[3q+(x-a)q']/[3(\sqrt[3]q)^2]$}
Where $q(x)\ne0$, the <product rule> and <chain rule> give the displayed real derivative. Its stationary points therefore satisfy $3q+(x-a)q'=0$. The denominator uses the square of the real <cube root>, which stays positive for negative $q$. Points with $q=0$ must be checked in the original function: a simple zero of $q$ with $x\ne a$ creates a vertical tangent, not a stationary point. If the prefactor also vanishes, a cancellation can change differentiability and must be analyzed separately.
Back to article page
