Problem 19 - Calculus Single Variable

Question

Use the Inverse Function Derivative Rule to calculate \(\left(f^{-1}\right)^{\prime}(t)\). $$ f:(1,2) \rightarrow(2,5 / 2), f(s)=\left(s^{2}+1\right) / s $$

Step-by-Step Solution

Answer

\( \left(f^{-1}\right)'(t) = \frac{s^2}{s^2 - 1} \) where \( s = \frac{t + \sqrt{t^2 - 4}}{2} \).

Step 1: Understand the function

The given function is \( f(s) = \frac{s^2 + 1}{s} \), which is defined on the interval \((1,2)\). We need to find the derivative of its inverse at a certain point \( t \).

Step 2: Apply the inverse function derivative rule

The Inverse Function Derivative Rule states that if \( f : (a, b) \rightarrow (c, d) \) is a differentiable function with an inverse \( f^{-1} \), then the derivative of the inverse at a point \( t \) is given by \( \left(f^{-1}\right)'(t) = \frac{1}{f'(f^{-1}(t))} \).

Step 3: Identify the inverse function and the point \( t \)

We need the derivative of the inverse function at a specific point \( t \). We assume that \( t \) is related to a point where \( f(s) = t \). Let's find such \( s \) that \( f(s) = t \).

Step 4: Differentiate \( f(s) \)

Calculate \( f'(s) \) using the quotient rule. If \( f(s) = \frac{s^2 + 1}{s} \), then \( f(s) = s + \frac{1}{s} \). The derivative is \( f'(s) = 1 - \frac{1}{s^2} \).

Step 5: Solve for \( s \) at \( f(s) = t \)

Given the function \( f(s) = s + \frac{1}{s} = t \), rearrange to find \( s \) as a function of \( t \). Solve \( s^2 - ts + 1 = 0 \) using the quadratic formula: \( s = \frac{t \pm \sqrt{t^2 - 4}}{2} \).

Step 6: Apply the inverse function derivative rule

Use \( \left(f^{-1}\right)'(t) = \frac{1}{f'(s)} \) where \( s = f^{-1}(t) \). Choose the meaningful \( s \) from the quadratic formula solution based on the interval for \( s \). If \( s_1 = \frac{t + \sqrt{t^2 - 4}}{2} \), then \( \left(f^{-1}\right)'(t) = \frac{1}{1 - \frac{1}{s^2}} \) evaluated at this \( s_1 \).

Step 7: Evaluate \( \left(f^{-1}\right)'(t) \)

Final answer: \( \left(f^{-1}\right)'(t) = \frac{s^2}{s^2 - 1} \) where \( s = \frac{t + \sqrt{t^2 - 4}}{2} \). Choose \( s_1 \) correctly according to the domain constraints.

Key Concepts

Function Derivative

A function derivative is a central concept in calculus, representing the rate at which a function changes at any given point. It provides a way to understand how the function behaves in the neighborhood of that point.Derivatives are critical for analyzing and understanding the local behavior of functions. For a given function like \( f(s) \), the derivative \( f'(s) \) is calculated by considering small changes in \( s \) and how they impact \( f(s) \). In this exercise's context, the derivative helps in applying the inverse function derivative rule.Calculating derivatives involves applying specific rules based on the function type, such as the power rule, product rule, or quotient rule, which we will explore further in another section. Understanding these rules allows you to determine the derivative efficiently and accurately.

Inverse Function

An inverse function \( f^{-1}(t) \) essentially reverses the action of the original function \( f(s) \). If \( f(s) \) takes \( s \) to \( t \), then \( f^{-1}(t) \) takes \( t \) back to \( s \).In simpler terms:

• If \( f(a) = b \), then \( f^{-1}(b) = a \).
• The existence of an inverse requires \( f(s) \) to be one-to-one (injective) and onto (surjective).

The inverse function is crucial because it allows us to navigate between variables and understand relationships in reverse. For the problem at hand, we utilize the inverse function to find \( s \) such that \( f(s) = t \).The inverse function derivative rule, which will be discussed later, leverages this relationship to find the derivative of the inverse function.

Quotient Rule

The quotient rule is fundamental for finding the derivative of functions expressed as one function divided by another. It's represented mathematically as:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]This rule applies when dealing with a function of the form \( f(s) = \frac{u(s)}{v(s)} \). For the function given in the exercise, \( f(s) = \frac{s^2 + 1}{s} \), applying the quotient rule involves:

• Identifying the numerator \( u(s) = s^2 + 1 \), and denominator \( v(s) = s \).
• Differentiate both: \( u'(s) = 2s \) and \( v'(s) = 1 \).

Plug these into the quotient rule formula to find the derivative:\[ f'(s) = \frac{(2s)(s) - (s^2 + 1)(1)}{s^2} \]Simplifying yields \( f'(s) = 1 - \frac{1}{s^2} \). This derivative is vital for applying the inverse function derivative rule, as it helps evaluate the rate of change at any given \( s \).

Derivative Calculation

Derivative calculation involves applying the appropriate differentiation rules to find the rate of change of a function. In the case of this exercise, the calculation focuses on finding the inverse function derivative at a given \( t \).Start by recognizing that you need to solve for \( s \) such that \( f(s) = t \). This is done using the quadratic equation:\[ s^2 - ts + 1 = 0 \]Employ the quadratic formula \( s = \frac{t \pm \sqrt{t^2 - 4}}{2} \) to find the possible values of \( s \). Choose the value that falls within the appropriate interval \((1,2)\). Once \( s \) is selected, it provides the necessary input to use in the inverse function derivative rule:\[ \left(f^{-1}\right)'(t) = \frac{1}{f'(s)} \]Substituting the correct \( s \) into the formula, follow through by using the previously calculated \( f'(s) \). This will give:\[ \left(f^{-1}\right)'(t) = \frac{s^2}{s^2 - 1} \]This calculation process allows understanding the derivative in the context of inverse functions, demonstrating how function behavior reverses, and changes still can be measured.