Question

Derivatives a. Use limits to find the derivative function \(f^{\prime}\) for the following functions \(f\) b. Evaluate \(f^{\prime}(a)\) for the given values of \(a\) $$f(t)=\frac{1}{\sqrt{t}} ; a=9, \frac{1}{4}$$

Short answer

Based on the given function, we found that the derivative function \(f^{\prime}(t) = -\frac{1}{2t\sqrt{t}}\). Then, we evaluated the derivative function at the given values: \(f^{\prime}(9) = -\frac{1}{54}\) and \(f^{\prime}\left(\frac{1}{4}\right) = -4\).

Step-by-step solution

Rewrite the function in terms of exponentiation

We begin by rewriting the given function \(f(t) = \frac{1}{\sqrt{t}}\) in terms of exponentiation, as this will make it easier to work with when we compute its derivative. $$ f(t) = \frac{1}{\sqrt{t}} = t^{-\frac{1}{2}} $$

Use the limit definition of the derivative

We will now use the limit definition of the derivative to find the derivative function \(f^{\prime}(t)\). The limit definition of the derivative is given by: $$ f^{\prime}(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} $$ We will apply this definition to our function \(f(t) = t^{-\frac{1}{2}}\).

Substitute the function and simplify

Now, we will substitute our function into the limit definition and simplify as much as possible: $$ f^{\prime}(t) = \lim_{h \to 0} \frac{ (t+h)^{-\frac{1}{2}} - t^{-\frac{1}{2}}}{h} $$

Find the common denominator

In order to simplify this expression further, we need to find a common denominator for the two terms inside the limit. We can use \((t+h)^{\frac{1}{2}}t^{\frac{1}{2}}\) as the common denominator for the sum in the numerator: $$ f^{\prime}(t) = \lim_{h \to 0} \frac{t^{\frac{1}{2}} - (t+h)^{\frac{1}{2}}}{h (t+h)^{\frac{1}{2}}t^{\frac{1}{2}}} $$

Multiply the numerator and denominator by the conjugate

To continue simplifying the expression, we will multiply both the numerator and denominator by the conjugate of the numerator, which is \((t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}})\): $$ f^{\prime}(t) = \lim_{h \to 0} \frac{(t^{\frac{1}{2}} - (t+h)^{\frac{1}{2}})(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}})}{h (t+h)^{\frac{1}{2}}t^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}})} $$ Now, simplify the numerator using the difference of squares formula, and simplify the denominator: $$ f^{\prime}(t) = \lim_{h \to 0} \frac{t - (t+h)}{h(t^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}}))} $$ Simplifying further: $$ f^{\prime}(t) = \lim_{h \to 0} \frac{-h}{h(t^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}}))} $$ We can now cancel out the \(h\) in the numerator and denominator: $$ f^{\prime}(t) = \lim_{h \to 0} \frac{-1}{t^{\frac{1}{2}}(t^{\frac{1}{2}} + (t+h)^{\frac{1}{2}})} $$

Take the limit as h approaches 0

Finally, we will take the limit as \(h\) approaches 0: $$ f^{\prime}(t) = \frac{-1}{t^{\frac{1}{2}}(t^{\frac{1}{2}} + (t)^{\frac{1}{2}})} = \frac{-1}{2t^{\frac{1}{2}}(t^{\frac{1}{2}})} = -\frac{1}{2t\sqrt{t}} $$ Thus, the derivative function \(f^{\prime}(t) = -\frac{1}{2t\sqrt{t}}\)

Evaluate the derivative for the given values of a

We are now asked to evaluate the derivative function \(f^{\prime}(t)\) for the given values of \(a\), \(a=9\) and \(a=\frac{1}{4}\). For \(a=9\): $$ f^{\prime}(9) = -\frac{1}{2\cdot9\sqrt{9}} = -\frac{1}{2\cdot9\cdot3} = -\frac{1}{54} $$ For \(a=\frac{1}{4}\): $$ f^{\prime}\left(\frac{1}{4}\right) = -\frac{1}{2\left(\frac{1}{4}\right)\sqrt{\frac{1}{4}}} = -\frac{1}{2\left(\frac{1}{4}\right)\frac{1}{2}} = -\frac{1}{\frac{1}{4}} = -4 $$ So, \(f^{\prime}(9) = -\frac{1}{54}\) and \(f^{\prime}\left(\frac{1}{4}\right) = -4\).

Key concepts

Understanding the Limit Definition of a Derivative

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: A Step-By-Step Approach

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.

Calculus Problem Solving: Bringing It All Together

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.