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The limit definition of the derivative is a fundamental concept in calculus. It allows us to determine the rate at which a function changes at a particular point. This is essentially what the derivative represents—a function's instant rate of change. Using limits, we can express the derivative of a function \( f(t) \) as:\[f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}\]This formula might seem complex at first, but let's break it down.

• \( h \) represents an infinitesimally small change in the input.
• \( f(t+h) - f(t) \) represents the change in the function's output when the input changes by \( h \).
• By dividing the change in function by \( h \), we find the average rate of change over an interval.

As \( h \) approaches zero, the average rate of change over a finite interval becomes the instantaneous rate of change, or the derivative, at point \( t \). This is why derivatives are such powerful tools in calculus—they provide precise insights into how functions behave at every tiny point.

Function differentiation is the process of computing the derivative of a function. In the example provided, we start with the function \( f(t) = \frac{1}{\sqrt{t}} \). To make it easier to apply the limit definition, rewrite the function with an exponent: \( f(t) = t^{-\frac{1}{2}} \). Next, substitute this back into the limit definition of the derivative:\[f'(t) = \lim_{h \to 0} \frac{(t+h)^{-\frac{1}{2}} - t^{-\frac{1}{2}}}{h}\]The next step is simplifying the expression inside the limit. To do this:

• Seek a common denominator for the terms in the numerator.
• Use the identity: \((a-b)(a+b) = a^2-b^2\) to help simplify expressions.
• Multiply by the conjugate to further simplify the complex fractions.

Finally, upon canceling terms and taking the limit as \( h \rightarrow 0 \), you will arrive at the derivative function: \( f'(t) = -\frac{1}{2t\sqrt{t}} \).This result is crucial because it represents the rate of change of the original function \( f(t) \) at any point \( t \). This process highlights how derivatives allow us to understand a function's behavior in mathematical models.

Solving calculus problems involves a clear understanding of the derivative and its interpretations. The problem presented involved evaluating the derivative at certain points. Having derived \( f'(t) = -\frac{1}{2t\sqrt{t}} \), we need to evaluate it for given values like \( a = 9 \) and \( a = \frac{1}{4} \).For \( a = 9 \):

• Substitute \( t = 9 \) into \( f'(t) \).
• Calculate: \(-\frac{1}{2 \cdot 9 \cdot 3} = -\frac{1}{54}\).

For \( a = \frac{1}{4} \):

• Substitute \( t = \frac{1}{4} \) into \( f'(t) \).
• Calculate: \(-\frac{1}{2 \cdot \frac{1}{4} \cdot \frac{1}{2}} = -4\).

Understanding this process demonstrates how calculated derivatives can provide insights into the rates of change for different function values. Successfully solving calculus problems often involves applying theoretical concepts into practical applications—here, grasping how the derivative's structure allows for these evaluations.