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If \(y = sinx\ sin2x\), and \(t = cosx,\) then \(\frac{dy}{dt}\) is

  • A.\(2(3t^{2}-1)\)
  • B.\(1-3t^{2}\)
  • C.\(\frac{1}{2}(1-3t^{2})\)
  • D.\((3t^{2}-1)\)
  • E.\(2(1-3t^{2})\)

Answer: E

\(\frac{dy}{dt}\) is \(2(1 - 3t^{2})\).

Write y in terms of t. Since \(\sin 2x = 2\sin x\cos x\),

\[ \begin{aligned} y &= 2\sin^{2}x\cos x = 2(1 - t^{2})t \\ &= 2t - 2t^{3} \\ \frac{dy}{dt} &= 2 - 6t^{2} = 2(1 - 3t^{2}) \end{aligned} \]

\(1 - 3t^{2}\) and \(\frac{1}{2}(1 - 3t^{2})\) lose the factor 2 from the double angle formula, and the options with \(3t^{2} - 1\) have the wrong sign.

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