To differentiate the function $F(t) = \frac{2 - 3 \cos t}{t^2}$, you can use the quotient rule. The quotient rule is:

$\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}$

For the function $F(t) = \frac{2 - 3 \cos t}{t^2}$, let:

• $u = 2 - 3 \cos t$ so $u' = 3 \sin t$
• $v = t^2$ so $v' = 2t$

Applying the quotient rule:

$F'(t) = \frac{t^2 \cdot (3 \sin t) - (2 - 3 \cos t) \cdot (2t)}{t^4}$

Simplify the expression:

$F'(t) = \frac{3t^2 \sin t - 4t + 6t \cos t}{t^4}$

Finally, divide each term by $t^4$:

$F'(t) = \frac{3t \sin t - 4 + 6 \cos t}{t^3}$

This gives the derivative of the function.