Answer
$\frac{dy}{dt} = -2\pi e^{\cos^{2}(\pi t-1)} \cdot \cos(\pi t-1) \cdot \sin(\pi t-1)$.
Work Step by Step
To find the derivative of $y=e^{cos^{2}(\pi t-1)}$ with respect to $t$, we can use the chain rule.
The derivative of $y$ with respect to $t$ is given by:
$\frac{dy}{dt} = \frac{d}{dt} (e^{\cos^{2}(\pi t-1)})$
Applying the chain rule, we get:
$\frac{dy}{dt} = e^{\cos^{2}(\pi t-1)} \cdot \frac{d}{dt} (\cos^{2}(\pi t-1))$
Now, let's find the derivative of $\cos^{2}(\pi t-1)$ with respect to $t$:
$\frac{d}{dt} (\cos^{2}(\pi t-1)) = -2\pi \cos(\pi t-1) \cdot \sin(\pi t-1)$
Substituting this derivative back into the original equation, we get:
$\frac{dy}{dt} = e^{cos^{2}(\pi t-1)} \cdot (-2\pi cos(\pi t-1) \cdot sin(\pi t-1))$
Therefore, the derivative of $y=e^{cos^{2}(\pi t-1)}$ with respect to $t$ is $\frac{dy}{dt} = -2\pi e^{cos^{2}(\pi t-1)} \cdot cos(\pi t-1) \cdot sin(\pi t-1)$.
