Let the population of living fish at time t (years) be denoted by x(t) (tonnes). An initial quantity of x0 tonnes of fish will be put into the lake at time t = 0. The net rate of change of the fish population in tonnes of fish per year is dx - = 0.5x = x² - f, x(0) = = x0 dt where xo) and f are non-negative constants and f represents the fish-harvesting rate. (a) Taking f = 0, write down the resulting differential equation stating clearly the solution method used.

dy of y = (t + 6)(t +5)(t + 3) 1 Use logarithmic differentiation to find the derivative,

f(x) = 4" and g(x) = 3". What is f'(2) = What is g'(2) Which has a higher rate of change at z = 2? O f(x) O g(x) What is f'(– 2) = What is g'( – 2) = 2? Which has a higher rate of change at x = O f(x) O g(x) |