**TWO**

y = (1 - bt)^a

Take logs to base e of both sides =>

ln(y) = a*ln(1 - bt)

Differentiate both side wrt t =>

(1/y)(dy/dt) = a*(-b/(1 - bt)) =>

dy/dt = a*(-b/(1 - bt))*(1 - bt)^a = -ab(1-bt)^(a - 1)

**ONE**

y = e^(ut + s^2t^2/2)

This is a function of a function - has form y = e^(f(t)) so differential is dy/dx = f'(t)*e^(f(t)) - so we have:

dy/dt = (u + s^2t)e^(ut + s^2t^2/2)

**THREE**

y = e^(u(e^t - 1)) - same thing again -

dy/dt = (u*e^t - 1)e^(u(e^t - 1))