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[tex] \large \bold{SOLUTION:} [/tex]
[tex] \small \begin{array}{l} \bold{Given:}\: f(t) = (t^3−2t+1)(2t^2+3t) \\ \\ f'(t) = \dfrac{d}{dt}\Big[(t^3−2t+1)(2t^2+3t)\Big] \\ \\ \textsf{By Product Rule,} \\ \\ f'(t) = (3t^2 - 2)(2t^2 + 3t) + (t^3 - 2t + 1)(4t + 3) \\ \\ \textsf{At }t = -3, \\ \\ f'(-3) = (3(-3)^2 - 2)(2(-3)^2 + 3(-3)) + ((-3)^3 -2(-3) + 1)(4(-3) + 3) \\ \\ f'(-3) = \boxed{405} \\ \\ \: \end{array} [/tex]
[tex] \small \begin{array}{l} \textsf{Or we can expand the RHS of given equation then} \\ \textsf{take the derivative with respect to }x. \\ \\ f(t) = (t^3−2t+1)(2t^2+3t) \\ \\ f(t) = 2t^5 + 3t^4 - 4t^3 - 6t^2 + 2t^2 + 3t \\ \\ f(t) = 2t^5 + 3t^4 - 4t^3 - 4t^2 + 3t \\ \\ f'(t) = \dfrac{d}{dt}(2t^5 + 3t^4 - 4t^3 - 4t^2 + 3t) \\ \\ \textsf{By Chain Rule,} \\ \\ f'(t) = 10t^4 + 12t^3 - 12t^2 - 8t + 3 \\ \\ \textsf{At }t=-3, \\ \\ f'(-3) = 10(-3)^4 + 12(-3)^3 - 12(-3)^2 - 8(-3) + 3 \\ \\ f'(-3) = \boxed{405} \end{array} [/tex]