1. The problem is to find the derivative of the function $F(t) = (\ln(t))^2 \cos(t)$.\n\n2. We apply the product rule for differentiation: if $F(t) = u(t) v(t)$, then $F'(t) = u'(t) v(t) + u(t) v'(t)$. Here, $u(t) = (\ln(t))^2$ and $v(t) = \cos(t)$.\n\n3. Calculate $u'(t)$: Using the chain rule, $u'(t) = 2 \ln(t) \cdot \frac{1}{t} = \frac{2 \ln(t)}{t}$.\n\n4. Calculate $v'(t)$: The derivative of $\cos(t)$ is $-\sin(t)$.\n\n5. Substitute back into the product rule:\n$$F'(t) = u'(t) v(t) + u(t) v'(t) = \frac{2 \ln(t)}{t} \cos(t) + (\ln(t))^2 (-\sin(t))$$\n\n6. Simplify the expression:\n$$F'(t) = \frac{2 \ln(t)}{t} \cos(t) - (\ln(t))^2 \sin(t)$$\n\nThis matches the corrected form of the derivative.