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The derivative of a product of two functions requires a special rule known as the Product Rule. This rule helps us differentiate expressions where two functions are being multiplied. Imagine two functions, say \( u(t) \) and \( v(t) \). To find the derivative of their product, we take each function's derivative separately and then apply:

• Differentiate the first function \( u(t) \) and multiply it by the second function \( v(t) \).
• Then, differentiate the second function \( v(t) \) and multiply it by the first function \( u(t) \).
• Add these results together to get the derivative of the product.

This means, mathematically, you express this as: \[(uv)' = u'v + uv'\]In the exercise, we identified \( u = e^{-at} \) and \( v = \cos t \). Their derivatives were found as \( u' = -ae^{-at} \) and \( v' = -\sin t \). Applying the product rule, the derivative \( y'(t) \) becomes: \[y' = (-ae^{-at})(\cos t) + (e^{-at})(-\sin t)\]This technique is essential for efficiently dealing with products of functions in calculus.

Differential equations can often be categorized based on the variables and parameters they contain. One significant category is the nonautonomous differential equation. A differential equation is considered nonautonomous if it explicitly involves the independent variable, usually time \( t \), in its formulation.
In the given exercise, our differential equation is:\[y' = -e^{-at}(a \cos t + \sin t)\]Here, you can see the presence of the independent variable \( t \) in several terms, like \( e^{-at} \), \( \cos t \), and \( \sin t \). This explicit appearance indicates that the system's behavior directly depends on time, rather than being solely a function of the dependent variable \( y \) and its derivatives.
This contrasts with autonomous equations, where the system's behavior is only related to \( y \) and its derivatives, not the independent variable. Recognizing a differential equation as nonautonomous is crucial for understanding how a system evolves over time and under varying conditions.

Calculating derivatives is a cornerstone of calculus, providing the rate at which one quantity changes with respect to another. In this exercise, we compute the derivative of the function \( y = e^{-at} \cos t \) relative to time \( t \). To do this effectively, we use the Product Rule, considering functions \( u \) and \( v \) previously identified.
### Step-by-Step Derivative Process1. **Differentiate Individually:** - Compute the derivative of \( u(t) = e^{-at} \), yielding \( u'(t) = -ae^{-at} \). - Compute the derivative of \( v(t) = \cos t \), yielding \( v'(t) = -\sin t \).
2. **Apply the Product Rule:** - Insert the derivatives into the product rule formula: \[y' = u'(t)v(t) + u(t)v'(t) \] - Substituting the known derivatives and functions results in: \[y' = (-ae^{-at})(\cos t) + (e^{-at})(-\sin t) \]
3. **Simplify the Expression:** - Finally, simplify the result to confirm it aligns with the differential equation provided: \[y' = -ae^{-at}\cos t - e^{-at}\sin t \]
The accurate calculation and simplification of derivatives is essential for solving differential equations, confirming that solutions satisfy given conditions, and understanding change in mathematical and physical contexts.