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Sunday, 01 March 2026
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If \(y = 6x^3 + 2x^{-2} - x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\).

If \(y = 6x^3 + 2x^{-2} - x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
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  • A \(\frac{\mathrm d y}{\mathrm d x} = 15x^2 - 4x^{-2} - 3x^{-2}\)
  • B \(\frac{\mathrm d y}{\mathrm d x} = 6x + 4x^{-1} - 3x^{-4}\)
  • C \(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)
  • D \(\frac{\mathrm d y}{\mathrm d x} = 12x^2 + 4x^{-1} - 3x^{-2}\)
Correct Answer: Option C
Explanation:
\(y = 6x^3 + 2x^{-2} - x^{-3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)

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