find \\( \\frac { d y } { d t } \\).\n\\( y = e ^ { \\cos ^ { 2 } ( \\pi t - 2 ) } \\)\n\\( \\frac { d y } {…

Explanation:

Step1: Let (u = \cos^{2}(\pi t - 2))

So (y = e^{u}). By the chain - rule (\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}). First, find (\frac{dy}{du}). Since (y = e^{u}), then (\frac{dy}{du}=e^{u}).

Step2: Find (\frac{du}{dt})

Let (v=\cos(\pi t - 2)), then (u = v^{2}). By the chain - rule (\frac{du}{dt}=\frac{du}{dv}\cdot\frac{dv}{dt}). Since (u = v^{2}), (\frac{du}{dv}=2v). And since (v=\cos(\pi t - 2)), (\frac{dv}{dt}=-\sin(\pi t - 2)\cdot\pi) (using the chain - rule for (y = \cos(f(t))) where (f(t)=\pi t - 2) and (\frac{d}{dt}\cos(f(t))=-\sin(f(t))\cdot f^{\prime}(t))). So (\frac{du}{dt}=2\cos(\pi t - 2)\cdot(-\pi\sin(\pi t - 2))=-\pi\sin(2(\pi t - 2))) (using the double - angle formula (\sin(2\alpha)=2\sin\alpha\cos\alpha)).

Step3: Combine results

Since (\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}) and (y = e^{u}), (u=\cos^{2}(\pi t - 2)), (\frac{du}{dt}=-\pi\sin(2(\pi t - 2))) (\frac{dy}{dt}=e^{\cos^{2}(\pi t - 2)}\cdot(-\pi\sin(2(\pi t - 2))))

Answer:

(-\pi\sin(2(\pi t - 2))e^{\cos^{2}(\pi t - 2)})