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JAMB 1999 · UME · Q39

What is the derivative of t2sin⁡(3t−5)t^2\sin(3t - 5) with respect to tt?

Worked solution (try it first)
  1. Product rule with u=t2u = t^2 and v=sin⁡(3t−5)v = \sin(3t - 5).
  2. Then u′=2tu' = 2t and, by the chain rule, v′=3cos⁡(3t−5)v' = 3\cos(3t - 5).
  3. So the derivative is 2tsin⁡(3t−5)+3t2cos⁡(3t−5)2t\sin(3t - 5) + 3t^2\cos(3t - 5), option C.

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