C'(t) = \frac(3)(t^2 + 4) - 3t(2t)(t^2 + 4)^2 = \frac3t^2 + 12 - 6t^2(t^2 + 4)^2 = \frac-3t^2 + 12(t^2 + 4)^2. - AMAZONAWS

📅 March 11, 2026 👤 scraface

Mar 11, 2026

$C'(t) = \frac{(3)(t^2 + 4) - 3t(2t)}{(t^2 + 4)^2} = \frac{3t^2 + 12 - 6t^2}{(t^2 + 4)^2} = \frac{-3t^2 + 12}{(t^2 + 4)^2}.$