Problem 23 - Calculus Single Variable

Question

Use the Inverse Function Derivative Rule to calculate \(\left(f^{-1}\right)^{\prime}(t)\). $$ f:(0, \infty) \rightarrow(0, \infty), f(s)=\log _{2}(1+s) $$

Step-by-Step Solution

Answer

\((f^{-1})'(t) = \ln(2) \, 2^t.\)

Step 1: Understand the Inverse Function Derivative Rule

The Inverse Function Derivative Rule states that if \(f\) is a one-to-one differentiable function with a derivative that is non-zero on its domain, then the inverse function \(f^{-1}\) is also differentiable at any point \(t = f(s)\), and its derivative is given by \( (f^{-1})'(t) = \frac{1}{f'(s)}\) where \(t = f(s)\).

Step 2: Find the Expression of the Inverse Function

Given \(f(s) = \log_2(1+s)\), its inverse function \(f^{-1}(t)\) is obtained by rewriting the expression for \(s\) in terms of \(t\):1. Rewrite \(y = \log_2(1+s)\) as \(2^y = 1+s\) because \(\log_2(x)\) is equivalent to the exponential \(2^x\).2. Solve for \(s\): \(s = 2^y - 1\). This implies the inverse function \(s = f^{-1}(t) = 2^t - 1\).

Step 3: Differentiate the Original Function

Differentiate the original function \(f(s) = \log_2(1+s)\) with respect to \(s\). Use the change of base formula \(\log_b(x) = \frac{\ln(x)}{\ln(b)}\):\[f'(s) = \frac{d}{ds} \left( \frac{\ln(1+s)}{\ln(2)} \right) = \frac{1}{\ln(2)} \cdot \frac{1}{1+s}.\]

Step 4: Apply the Inverse Function Derivative Rule

Use the Inverse Function Derivative Rule to find \((f^{-1})'(t)\). Since \(f(s) = t\) and \(s = f^{-1}(t) = 2^t - 1\), the derivative of the inverse function is given by:\[(f^{-1})'(t) = \frac{1}{f'(s)} = \frac{1}{\frac{1}{\ln(2)(1+s)}} = \ln(2)(1+s).\]Substitute \(s = 2^t - 1\) into the expression:\[(f^{-1})'(t) = \ln(2)(1 + (2^t - 1)) = \ln(2) \, 2^t.\]

Key Concepts

Logarithmic Functions

Logarithmic functions are special mathematical functions that help us reverse exponential functions. This means if you have the exponential form and you want to find the original base number, logarithms can show you the way. For instance, \(\log_b(x)\) translates to asking which power you'd raise \(b\) to, to get \(x\). Think of it as undoing an exponential operation!

In our exercise, the function \(f(s) = \log_2(1+s)\) involves a logarithm with base 2. This particular function takes a number \(1+s\), asks for its power needed to reach that number starting from 2. When dealing with logarithms, familiarity with their rules, like the change of base formula \(\log_b(x) = \frac{\ln(x)}{\ln(b)}\), is crucial. This rule helps convert between different logarithm bases using natural logs (\(\ln\)). Understanding this allows you to differentiate logarithms easily or switch bases without much hassle.

Differentiation

Differentiation is a core concept in calculus which examines how functions change. Specifically, it involves calculating the derivative, which represents the rate of change of a function's output with respect to its input. Think of it as finding the slope at any given point on a curve.

To differentiate the given function \(f(s) = \log_2(1+s)\), we applied the change of base formula for logarithms and differentiated \(\frac{\ln(1+s)}{\ln(2)}\) with respect to \(s\). This operation required applying basic differentiation rules to \(\ln(1+s)\), leading to the function's derivative: \(f'(s) = \frac{1}{\ln(2)} \cdot \frac{1}{1+s}\). This expression reveals how our function grows at each point – very insightful for calculating changes!

The key here is understanding each differentiation step, such as using the chain rule or power rule, which require identifying how different components of a function contribute to its change. These are essential skill sets for handling varying functions, whether they are polynomial, exponential, or logarithmic.

Exponential Functions

Exponential functions are powerful mathematical constructs where a constant base is raised to a variable exponent. They describe rapid growth or decay, common in populations, finances, and radioactive substances. An example is \(2^x\), where 2 is the base exponentially influenced by \(x\).

In our exercise, understanding \(^2y = 1+s\) simplifies to \(1+s\) represented as an exponential function. This linkage enables rewriting the function in terms of its inverse, essential for inverse derivatives: \(s = 2^y - 1\).

The role of exponential functions, especially when represented inversely, is critical in calculus. They allow you to determine meaningful rates of change through their derivatives. For instance, knowing this inverse form was pivotal to calculating our final result for \((f^{-1})'(t)\) as \(\ln(2) \cdot 2^t\).Understanding both exponential and inverse functions provides multiple perspectives on how functions can transform and guide predictions on behavior over time.