Question:
Let \(f(x) = \frac{\sqrt[3]{x^4}}{\sqrt[3]{x^2}},\ x \neq 0\). Then the value of \(f'(27)\) is equal to
Let \(f(x) = \frac{\sqrt[3]{x^4}}{\sqrt[3]{x^2}},\ x \neq 0\). Then the value of \(f'(27)\) is equal to
Show Hint
\(\sqrt[3]{x^m} = x^{m/3}\). Simplify before differentiating.
Updated On: Apr 27, 2026
- \(\frac{1}{9}\)
- \(\frac{2}{9}\)
- \(\frac{1}{3}\)
- \(\frac{4}{9}\)
- \(\frac{5}{9}\)
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The Correct Option is B
Solution and Explanation
Step 1: Concept:
• Simplify \(f(x)\) using laws of exponents.
Step 2: Detailed Explanation:
• Simplify function: \[ f(x) = \frac{x^{4/3}}{x^{2/3}} = x^{(4/3 - 2/3)} = x^{2/3} \]
• Differentiate: \[ f'(x) = \frac{2}{3} x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \]
• Substitute \(x = 27\): \[ f'(27) = \frac{2}{3\sqrt[3]{27}} = \frac{2}{3 \cdot 3} = \frac{2}{9} \]
Step 3: Final Answer:
• \[ f'(27) = \frac{2}{9} \]
• Simplify \(f(x)\) using laws of exponents.
Step 2: Detailed Explanation:
• Simplify function: \[ f(x) = \frac{x^{4/3}}{x^{2/3}} = x^{(4/3 - 2/3)} = x^{2/3} \]
• Differentiate: \[ f'(x) = \frac{2}{3} x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \]
• Substitute \(x = 27\): \[ f'(27) = \frac{2}{3\sqrt[3]{27}} = \frac{2}{3 \cdot 3} = \frac{2}{9} \]
Step 3: Final Answer:
• \[ f'(27) = \frac{2}{9} \]
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