F'(t) = \frac(100)(t^2 + 4) - 100t(2t)(t^2 + 4)^2 = \frac100(t^2 + 4 - 2t^2)(t^2 + 4)^2 = \frac100(4 - t^2)(t^2 + 4)^2. - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$F'(t) = \frac{(100)(t^2 + 4) - 100t(2t)}{(t^2 + 4)^2} = \frac{100(t^2 + 4 - 2t^2)}{(t^2 + 4)^2} = \frac{100(4 - t^2)}{(t^2 + 4)^2}.$