Let y = 4t/(t^2 + 5). Find dy/dt, writing your answer in it's simplest form, and find all values of t for which dy/dt = 0

We use the quotient rule to find dy/dt. Let u = 4t and v = (t^2 + 5). Then, u' = 4 and v' = 2t. Hence,

dy/dt = u'v - v'u / v² = 4(t^2 + 5) - 4t x 2t / (t^2 + 5)² = 20 - 4t² / (t^2 + 5)².  Now, we need to find all t such that dy/dt < 0 i.e.

20 - 4t² / (t^2 + 5)² < 0 which rearranges to give t² > 5, so, t > 5^1/2 and t < -5^1/2.