Problem 10 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The derivative of the inverse function is \( \left(f^{-1}\right)'(t) = 2t \) for \( t > 0 \).

Step 1: Understand the Problem

We are tasked with finding the derivative of the inverse function of \( f(s) = \sqrt{s} \). The function \( f \) maps from \((0, \infty)\) to \((0, \infty)\), and we need to find the derivative of its inverse, \( \left(f^{-1}\right)'(t) \), using the Inverse Function Derivative Rule.

Step 2: Identify the Inverse Function

To apply the Inverse Function Derivative Rule, we first need to find the inverse function of \( f(s) = \sqrt{s} \). Set \( y = \sqrt{s} \), which implies \( s = y^2 \). Thus, the inverse function is \( f^{-1}(t) = t^2 \), where \( t = \sqrt{s} \).

Step 3: Apply the Inverse Function Derivative Rule

The Inverse Function Derivative Rule is given by \( \left(f^{-1}\right)'(t) = \frac{1}{f'(f^{-1}(t))} \). We need to find \( f'(s) \) first.

Step 4: Differentiate the Original Function

Find the derivative of \( f(s) = \sqrt{s} \). This is \( f'(s) = \frac{d}{ds}(s^{1/2}) = \frac{1}{2}s^{-1/2} = \frac{1}{2\sqrt{s}} \).

Step 5: Substitute and Simplify

Substitute \( f^{-1}(t) = t^2 \) into the derivative of \( f \). We find \( f'(t^2) = \frac{1}{2\sqrt{t^2}} = \frac{1}{2t} \). Therefore, \( \left(f^{-1}\right)'(t) = \frac{1}{f'(t^2)} = \frac{1}{\frac{1}{2t}} = 2t \).

Step 6: Verify the Domain

Double-check the domain restrictions. Since the original function \( f(s) = \sqrt{s} \) maps \((0, \infty) \rightarrow (0, \infty)\), its inverse \( f^{-1}(t) = t^2 \) is indeed valid for \( t > 0 \). Our result \( \left(f^{-1}\right)^{\prime}(t) = 2t \) applies for all \( t > 0 \).

Key Concepts

Inverse Function

An inverse function essentially "undoes" what the original function does. For a function \( f \), its inverse, denoted as \( f^{-1} \), will map outputs from \( f \) back to their original inputs. If \( f(a) = b \), then \( f^{-1}(b) = a \).

To find the inverse of a function, you swap the roles of the independent and dependent variables and solve for the new dependent variable. In this exercise, we started with the function \( f(s) = \sqrt{s} \). By setting \( y = \sqrt{s} \), we can rearrange to find \( s = y^2 \), which implies \( f^{-1}(t) = t^2 \). This shows the inverse of the square root function is essentially squaring the value.

Remember, inverse functions are not defined everywhere. They require the original function to be one-to-one, which means it doesn’t repeat output values. The given function \( f(s) = \sqrt{s} \) is naturally one-to-one because the square root function is always increasing.

Derivative Rule

The Derivative Rule refers specifically in this context to the Inverse Function Derivative Rule. It's a powerful tool that allows us to find the derivative of an inverse function without having to explicitly differentiate the inverse. This rule states:

• \( \left(f^{-1}\right)'(t) = \frac{1}{f'(f^{-1}(t))} \)

When applying this rule, the idea is to first differentiate the original function \( f(s) \). Then you use the derivative in the rule by substituting the inverse function \( f^{-1}(t) \) back into it. The result gives the derivative of the inverse's output with respect to its input. In this problem, using the rule simplified our calculations.

Square Root Function

The square root function is a fundamental operation in mathematics. For any positive number \( s \), \( \sqrt{s} \) is the non-negative number which, when squared, equals \( s \). This function is defined only for non-negative inputs when dealing with real numbers, as the square root of a negative number is not real.

In this exercise, the square root function \( f(s) = \sqrt{s} \) is mapped over the domain \((0, \infty)\), covering all positive real numbers. This ensures it remains a function since it will not hit any undefined points or require complex numbers. Furthermore, the square root is specifically useful for its property that every positive number has exactly one positive square root, making it inherently suited for inverse operations like this one.

Derivative Calculation

Finding the derivative of a function provides the rate at which the function's output changes as its input changes. For the function \( f(s) = \sqrt{s} \), its derivative \( f'(s) \) tells us how the square root of \( s \) changes as \( s \) itself changes.

• Begin by expressing \( \sqrt{s} \) as \( s^{1/2} \).
• The power rule, \( \frac{d}{ds}(s^n) = ns^{n-1} \), guides us, resulting in \( f'(s) = \frac{1}{2}s^{-1/2} \).
• Simplify this to \( f'(s) = \frac{1}{2\sqrt{s}} \), giving the rate of change of the square root function.

When applying the inverse function derivative rule, we substitute into this derivative, leading to the conclusion that the derivative of the inverse, \( f^{-1}(t) = t^2 \), is \( 2t \), emphasizing this step's vital role in understanding function behaviors.