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

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 is C

\(y = 6x^3 + 2x^{-2} - x^{-3}\)

\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)