Differentiate the function. $$ F(t)=(\ln (t))^{2} \cos (t) $$ $$ F^{\prime}(t)=\square $$

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We have the function
$$
F(t) = (\ln t)^2 \cos t.
$$

**Step 1: Differentiate $$ (\ln t)^2 $$.**

Using the chain rule:
$$
\frac{d}{dt} (\ln t)^2 = 2 (\ln t) \cdot \frac{d}{dt} (\ln t) = 2 (\ln t) \cdot \frac{1}{t} = \frac{2 \ln t}{t}.
$$

**Step 2: Differentiate $$\cos t$$.**

$$
\frac{d}{dt} (\cos t) = -\sin t.
$$

**Step 3: Apply the product rule to differentiate $$F(t) = u \cdot v$$ where $$u(t) = (\ln t)^2$$ and $$v(t)= \cos t$$.**

The product rule states:
$$
F'(t) = u'(t)v(t) + u(t)v'(t).
$$

Substitute the derivatives:
$$
F'(t) = \left(\frac{2 \ln t}{t}\right) \cos t + (\ln t)^2 (-\sin t).
$$

**Step 4: Simplify the expression.**

$$
F'(t) = \frac{2 \ln t}{t} \cos t - (\ln t)^2 \sin t.
$$

Thus, the derivative is
$$
\boxed{F'(t)=\frac{2 \ln t}{t} \cos t - (\ln t)^2 \sin t.}
$$
$$ F^{\prime}(t) = \frac{2 \ln t}{t} \cos t - (\ln t)^2 \sin t $$

Differentiate the function.

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