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The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. For instance, when you have a function like \( f(t) = 25\sqrt{t-1} \), it's a composition involving an outer function \( g(u) = 25u^{1/2} \) and an inner function \( u(t) = t-1 \). The chain rule provides a method to find the derivative of such composite functions effectively. To differentiate using the chain rule, follow these steps:

• Differentiate the outer function with respect to the inner function.
• Multiply it by the derivative of the inner function.

In our example:- The derivative of \( 25u^{1/2} \) with respect to \( u \) is \( 25 \cdot \frac{1}{2}u^{-1/2} \).- The derivative of \( u = t-1 \) with respect to \( t \) is \( 1 \).So, the derivative \( f'(t) \) becomes \( \frac{25}{2\sqrt{t-1}} \), which combines both parts.

Finding the derivative of a function is a key step in understanding how the function behaves over time. Derivatives represent the rate of change or the slope of a function at any given point. For the function \( f(t) = 25\sqrt{t-1} \), differentiating helps us understand the behavior of \( f(t) \) regarding its change with respect to \( t \).Using the chain rule, we determined that the derivative, \( f'(t) \), is \( \frac{25}{2\sqrt{t-1}} \). This expression tells us how sensitive the function is to changes in \( t \). A greater derivative means the function changes "faster" at that point.

Evaluating derivatives involves substituting specific values of \( t \) into the derivative function to find the rate of change at particular points. For the problem given, after finding the derivative \( f'(t) = \frac{25}{2\sqrt{t-1}} \), we evaluate it at \( t = 6 \).This step involves:

• Substituting \( t = 6 \) into \( f'(t) \).
• Performing simplifications to find the result.

Substituting gives \( \frac{1}{2(6-1)} = \frac{1}{10} \). Thus, at \( t = 6 \), the function's change rate is \( \frac{1}{10} \), providing a clear view of how quickly \( f(t) \) is changing relative to its value.

Calculus is not just about solving equations; it's about understanding how to apply concepts to interpret real-world scenarios. In this exercise, we found the relative rate of change for the function \( f(t) = 25\sqrt{t-1} \), which required a methodical approach to use calculus tools effectively.Here's a brief look at the steps involved:

• Identify the function with a composite structure and apply the chain rule.
• Calculate both the derivative \( f'(t) \) and the relative rate of change.
• Substitute specific values to derive meaningful insights from the mathematics.

The interpretation of the results provides us insight into how the function behaves, especially at specific points, revealing the interconnectedness of mathematical theory and practical application.