Assuming you mean f(t) = g(t) × h(t), notice that
f(t) = g(t) × h(t) = cos(t) sin(t) = 1/2 sin(2t)
Then the difference quotient of f is

Recall the angle sum identity for sine:
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
Then we can write the difference quotient as

or

(As a bonus, notice that as h approaches 0, we have (cos(2h) - 1)/(2h) → 0 and sin(2h)/(2h) → 1, so we recover the derivative of f(t) as cos(2t).)