The derivative of the function at \( x = 0.25 \) is approximately \( y'(0.25) \approx -1.353 \).

Would you like more details on how this was calculated?

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Here are 5 related questions:

1. How does changing the value of \( x \) affect the derivative of the function?
2. What does the negative sign in the derivative imply about the behavior of the function at \( x = 0.25 \)?
3. How would you evaluate \( y'(x) \) for a different value of \( x \)?
4. What is the significance of finding the derivative at a specific point?
5. How does the exponential part of the function influence the rate of change?

**Tip:** When evaluating a derivative at a specific point, consider the behavior of each component function (e.g., exponential growth vs. oscillatory behavior of sine/cosine).

Understand how to calculate the derivative of an exponential function at x = 0.25 with a step-by-step explanation. This problem involves key calculus concepts, including derivatives and exponential functions, and applies the Mean Value Theorem. Suitable for advanced high school and college-level students.

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