Problem 69 - Calculus Single Variable

Question

Find \(f^{-1}(\gamma)\) for the given \(f\) and \(\gamma\) (but do not try to calculate \(f^{-1}(t)\) for a general value of \(t\) ). Then calculate \(\left(f^{-1}\right)^{\prime}(\gamma)\). $$ f(s)=s \ln (s), \gamma=0 $$

Step-by-Step Solution

Answer

The value of \( f^{-1}(0) \) is 1, and \((f^{-1})'(0) = 1\).

Step 1: Set Function Equal to Gamma

We want to find the inverse function \( f^{-1} \) evaluated at \( \gamma = 0 \). First, equate the function to \( \gamma \): \[ f(s) = s \ln(s) = 0 \].

Step 2: Solve for s

To solve \( s \ln(s) = 0 \), we recognize that this equation is satisfied when \( s = 1 \), since \( \ln(1) = 0 \) and therefore \( 1 \cdot \ln(1) = 0 \). Therefore, \( s = 1 \).

Step 3: Determine Inverse Value

Since \( s = 1 \) satisfies \( f(s) = \gamma = 0 \), we have \( f^{-1}(0) = 1 \).

Step 4: Use Inverse Derivative Formula

To find the derivative of the inverse function at \( \gamma = 0 \), use the formula \((f^{-1})'(y) = \frac{1}{f'(x)}\), where \( x = f^{-1}(y) \). In this case, \( y = \gamma = 0 \) and \( x = 1 \).

Step 5: Calculate Derivative of the Original Function

Find \( f'(s) \) by differentiating \( f(s) = s \ln(s) \) using the product rule. \[ f'(s) = 1 \cdot \ln(s) + s \cdot \frac{1}{s} = \ln(s) + 1 \].

Step 6: Evaluate Derivative at s=1

Substitute \( s = 1 \) into \( f'(s) \) to evaluate it: \[ f'(1) = \ln(1) + 1 = 0 + 1 = 1 \].

Step 7: Calculate Inverse Derivative

Using the formula for the derivative of the inverse, we have \((f^{-1})'(0) = \frac{1}{f'(1)} = \frac{1}{1} = 1\).

Key Concepts

Derivative of Inverse Function

Understanding the derivative of an inverse function is essential in calculus. Suppose you have a function \( f \) and its inverse \( f^{-1} \). The derivative of the inverse function at a particular point \( \gamma \) is found using a simple but powerful formula:

• \( (f^{-1})'(\gamma) = \frac{1}{f'(x)} \)

Here, \( x \) is the input to the original function such that \( f(x) = \gamma \). This formula essentially tells us that to find how the inverse function changes, we need the reciprocal of how the original function changes at the related point.

For example, if \( f(1) = 0 \), and we want \( (f^{-1})'(0) \), we first find \( f'(1) \) and then take its reciprocal. This approach highlights how calculus can simplify understanding the behavior of functions and their inverses.

Logarithmic Functions

Logarithms are a central topic in mathematics, particularly integral and differential calculus. A logarithmic function is the inverse of an exponential function. If \( y = \ln(x) \), then it means that \( e^y = x \), where \( e \) is the base of the natural logarithms, approximately equal to 2.718.

The natural logarithm, \( \ln(x) \), is especially significant because it arises naturally in many contexts beyond pure mathematics. In inverse function problems, understanding the properties of logarithmic functions helps us solve complex equations involving exponential growth and decay.

• \( \ln(1) = 0 \)
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \)

These properties are crucial for problems where we have products involving logarithmic terms, as seen in the expression \( s \ln(s) \). They also aid in solving equations where logarithms simplify expressions.

Product Rule

When dealing with the differentiation of products of functions, the product rule is your go-to tool. It states that if you have two differentiable functions \( u(x) \) and \( v(x) \), the derivative of their product is:

• \( (uv)' = u'v + uv' \)

This rule is incredibly helpful because it allows us to take derivatives of complex expressions by breaking them down into simpler parts.

Consider \( f(s) = s \ln(s) \). Here, \( u(s) = s \) and \( v(s) = \ln(s) \). Applying the product rule, we find:

• \( f'(s) = 1 \cdot \ln(s) + s \cdot \frac{1}{s} = \ln(s) + 1 \)

Using the product rule correctly ensures we capture all the nuances of how the functions interact, providing the accurate derivative needed to evaluate the inverse function's behavior.

Solving Equations

In calculus, solving equations is a fundamental operation to finding key values that satisfy given conditions. When we solve \( s \ln(s) = 0 \), we need to identify solutions by recognizing zeroing conditions.

Consider the equation \( f(s) = s \ln(s) = 0 \). To solve it:

• The product \( s \ln(s) \) is zero if either \( s = 0 \) or \( \ln(s) = 0 \).
• Since \( \ln(1) = 0 \), we discover that \( s = 1 \) satisfies the equation.

Recognizing that \( s = 1 \) works simplifies our understanding of solutions. Solving such equations often hinges on knowing properties of logarithms and strategic reasoning about zero values, crucial for finding inverse functions and their derivatives.