Discussion
Let \(f(x) = \frac{\sqrt[3]{x^{4}}}{\sqrt[3]{x^{2}}}, x \ne 0\). Then the value of \(f'(27)\) is equal to
Answer: B
\(f'(27) =\) \(\frac{2}{9}\).
Simplify first: \(f(x) = \frac{x^{4/3}}{x^{2/3}} = x^{2/3}\). Then
\[ \begin{aligned} f'(x) &= \frac{2}{3}x^{-1/3} \\ f'(27) &= \frac{2}{3}\cdot\frac{1}{3} = \frac{2}{9} \end{aligned} \]
since \(27^{1/3} = 3\). \(\frac{1}{9}\) drops the factor 2, \(\frac{4}{9}\) doubles it, and \(\frac{1}{3}\) and \(\frac{5}{9}\) come from wrong exponents.
The trap is to differentiate numerator and denominator separately; simplifying the radicals to a single power first avoids the quotient rule entirely.
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