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We are given $y = \cos^3(\sec^2 2t)$. We need to find $\frac{dy}{dt}$. We will use the chain rule repeatedly.
First, differentiate $\cos^3 u$ with respect to $u$, where $u = \sec^2 2t$. $$ \frac{d}{du} (\cos^3 u) = 3 \cos^2 u \cdot (-\sin u) = -3 \cos^2 u \sin u $$
Next, differentiate $\sec^2 v$ with respect to $v$, where $v = 2t$. $$ \frac{d}{dv} (\sec^2 v) = 2 \sec v \cdot (\sec v \tan v) = 2 \sec^2 v \tan v $$
Finally, differentiate $2t$ with respect to $t$. $$ \frac{d}{dt} (2t) = 2 $$
Now, we multiply all the derivatives together: $$ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dt} $$ $$ \frac{dy}{dt} = (-3 \cos^2 (\sec^2 2t) \sin (\sec^2 2t)) \cdot (2 \sec^2 (2t) \tan (2t)) \cdot (2) $$ $$ \frac{dy}{dt} = -12 \cos^2 (\sec^2 2t) \sin (\sec^2 2t) \sec^2 (2t) \tan (2t) $$
Final Answer: $\frac{dy}{dt} = -12 \cos^2 (\sec^2 2t) \sin (\sec^2 2t) \sec^2 (2t) \tan (2t)$