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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
[tex]\dfrac{\frac12 \sin(2(t+h)) - \frac12 \sin(2t)}h = \dfrac{\sin(2t+2h) - \sin(2t)}{2h}[/tex]
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
[tex]\dfrac{\sin(2t)\cos(2h) + \cos(2t)\sin(2h) - \sin(2t)}{2h}[/tex]
or
[tex]\boxed{\sin(2t)\dfrac{\cos(2h)-1}{2h} + \cos(2t)\dfrac{\sin(2h)}{2h}}[/tex]
(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).)