Problem 25 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

\(\left(f^{-1}\right)'(t) = \frac{1}{\ln(2) \cdot t}\)

Step 1: Understand the Inverse Function Derivative Rule

The Inverse Function Derivative Rule states that if \( y = f(x) \) has an inverse function \( x = f^{-1}(y) \), then the derivative of the inverse function is given by \( \left(f^{-1}\right)^{\prime}(t) = \frac{1}{f^{\prime}(x)} \), where \( x = f^{-1}(t) \).

Step 2: Find \( f^{\prime}(s) \)

To use the inverse function derivative rule, we must first compute the derivative of the original function \( f(s) = 2^s \). The derivative with respect to \( s \) is:\[ f^{\prime}(s) = \ln(2) \cdot 2^s \]

Step 3: Find \( x = f^{-1}(t) \)

Since the original function \( f(s) = 2^s \) is exponential, its inverse is a logarithmic function. Thus, \( f^{-1}(t) = \log_2(t) \), where base 2 logarithm is used because of the base in the exponential function.

Step 4: Calculate \( \left(f^{-1}\right)^{\prime}(t) \) using the Inverse Function Derivative Rule

We substitute into the inverse derivative formula. Since \( x = f^{-1}(t) = \log_2(t) \), we have for \( t = 2^x \), \( x = \log_2(t) \). Thus, \( f^{\prime}(x) = \ln(2) \cdot 2^x = \ln(2) \cdot t \). Applying the rule:\[ \left(f^{-1}\right)^{\prime}(t) = \frac{1}{f^{\prime}(x)} = \frac{1}{\ln(2) \cdot t} \]

Key Concepts

Derivative of Exponential Functions

Exponential functions are powerful mathematical constructs that appear frequently in various fields, such as physics and finance. The derivative of an exponential function is noteworthy for its simplicity. If we have an exponential function in the form of \[ f(s) = a^s \],its derivative with respect to the variable \( s \) is \[ f^{\prime}(s) = \ln(a) \cdot a^s \].Here, \( a \) is the base of the exponential function, and \( \ln(a) \) is the natural logarithm of \( a \). This means that the rate at which the exponential function changes is proportional to both the function itself and the natural logarithm of the base.

• For \( f(s) = 2^s \), the derivative \( f^{\prime}(s) = \ln(2) \cdot 2^s \) illustrates that this function's rate of growth depends on \( 2^s \) multiplied by \( \ln(2) \).

Because of this behaviour, exponential functions grow rapidly, which is integral to their applications in modeling compound interest, population growth, and radioactive decay.

Inverse Functions

Inverse functions play a crucial role in mathematics by effectively 'undoing' the effect of a function. Given a function \( y = f(x) \), the inverse function \( x = f^{-1}(y) \) returns the original input \( x \) when given \( y \). In simple terms, the function and its inverse function 'cancel' each other out, such that \[ f(f^{-1}(y)) = y \] and \[ f^{-1}(f(x)) = x \].
For exponential functions like \( f(s) = 2^s \), the inverse is a logarithmic function, specifically \( f^{-1}(t) = \log_2(t) \). This is because the exponential function outputs powers of 2, and the logarithm base 2 effectively reverses this operation by determining what power of 2 yields \( t \).

• Understanding inverse functions allows us to solve equations involving exponential and logarithmic expressions efficiently.
• They are pivotal in calculus, especially when applying the Inverse Function Derivative Rule to find derivatives of inverse functions.

This concept is especially important in calculus to help facilitate understanding the relationships between different types of functions and their derivatives.

Logarithms

Logarithms are the inverse functions of exponential functions, providing the exponent needed to achieve a certain number. The logarithm with base \( a \) of a number \( t \) is written as \( \log_a(t) \) and is defined such that \[ a^{\log_a(t)} = t \].
In the given exercise, where \( f(s) = 2^s \), the inverse function \( f^{-1}(t) \) corresponds to the logarithm base 2: \( \log_2(t) \). This means solving for the exponent that 2 is raised to, yielding \( t \).
Logarithms are valuable due to numerous properties:

• They transform multiplication into addition, aiding in complex calculations and simplifying expressions.
• They are essential in solving equations involving exponential growth and decay.
• Natural logarithms, with base \( e \), are frequently used in calculus due to their smooth properties across real numbers.

By turning multiplicative processes into additive ones, logarithms help manage and analyze exponential relationships in diverse mathematical contexts.