Problem 11 - Calculus Single Variable

Question

Use the Inverse Function Derivative Rule to calculate \(\left(f^{-1}\right)^{\prime}(t)\). $$ f:(1, \infty) \rightarrow(0,1 / 2), f(s)=1 / \sqrt{s+3} $$

Step-by-Step Solution

Answer

\((f^{-1})'(t) = -\frac{2}{t^3}\).

Step 1: Understanding the given function and its domain

We are given the function \( f(s) = \frac{1}{\sqrt{s+3}} \) with the domain \( s > 1 \). This means that \( f \) maps from the interval \((1, \infty)\) to \((0, \frac{1}{2})\). Our goal is to find the derivative of its inverse function \( (f^{-1})'(t) \) using the Inverse Function Derivative Rule.

Step 2: Recall the Inverse Function Derivative Rule

The inverse function derivative rule states that if \( f \) is a function that is continuous and differentiable, and it has an inverse \( f^{-1} \), then the derivative of the inverse function is given by: \( (f^{-1})'(t) = \frac{1}{f'(f^{-1}(t))} \). This formula requires us to find the derivative of \( f(s) \) and evaluate it at the inverse function \( f^{-1}(t) \).

Step 3: Differentiate the given function

Differentiate \( f(s) = \frac{1}{\sqrt{s+3}} \) with respect to \( s \). Using the chain rule, we find the derivative:\[f'(s) = \frac{d}{ds}\left(s+3\right)^{-1/2} = -\frac{1}{2}(s+3)^{-3/2}\]This is the derivative \( f'(s) \) that we will use in the next step.

Step 4: Solve for the inverse function

To find \( f^{-1}(t) \), we start with the equation \( t = \frac{1}{\sqrt{s+3}} \). Rearrange to express \( s \) in terms of \( t \):1. \( \sqrt{s+3} = \frac{1}{t} \)2. \( s+3 = \frac{1}{t^2} \)3. \( s = \frac{1}{t^2} - 3 \)Thus, \( f^{-1}(t) = \frac{1}{t^2} - 3 \).

Step 5: Substitute into the Inverse Function Derivative Rule

Replace \( s \) with \( f^{-1}(t) = \frac{1}{t^2} - 3 \) in the derivative \( f'(s) \):\[f'(f^{-1}(t)) = -\frac{1}{2}\left(\left(\frac{1}{t^2} - 3\right) + 3\right)^{-3/2}\]Simplify the expression to get:\[f'(f^{-1}(t)) = -\frac{1}{2}\left(\frac{1}{t^2}\right)^{-3/2} = -\frac{1}{2} t^3 \].

Step 6: Calculate the derivative of the inverse function

Finally, use the Inverse Function Derivative Rule:\[(f^{-1})'(t) = \frac{1}{f'(f^{-1}(t))} = \frac{1}{-\frac{1}{2} t^3} = -\frac{2}{t^3}\]This result gives the derivative of the inverse function \( (f^{-1})'(t) \).

Key Concepts

Inverse Functions

Inverse functions are a valuable concept in calculus as they allow us to "reverse" a function. If you have a function that maps an input to an output, the inverse function does the opposite, mapping each output back to its original input. Formally, if you have a function \( f(s) \), and it has an inverse \( f^{-1}(t) \), then \( f(f^{-1}(t)) = t \) and \( f^{-1}(f(s)) = s \). In our exercise, we are working with a function \( f(s) = \frac{1}{\sqrt{s+3}} \). We found its inverse as \( f^{-1}(t) = \frac{1}{t^2} - 3 \). The ability to find an inverse relationship is crucial for solving many calculus problems, especially when we need to revert outputs to inputs in calculations.

Function Differentiation

Differentiating a function is all about finding how the function's output changes as its input changes, which is precisely what gives us information about its rate of change or slope. The differentiation process is one of the cornerstones of calculus. For the given function:

• We started with \( f(s) = \frac{1}{\sqrt{s+3}} \).
• The differentiation involves using the power rule alongside a negative exponent: \((s+3)^{-1/2}\).
• The derivative turns out to be \( f'(s) = -\frac{1}{2}(s+3)^{-3/2} \).

Differentiation allows us to gauge the rate at which one quantity changes with respect to another. In this exercise, the derivative informs us about how \( f(s) \) changes as \( s \) changes.

Chain Rule

The chain rule is a vital tool in differentiation, especially when dealing with composite functions. It allows us to differentiate functions embedded within other functions. Consider our original function form used:
The chain rule states if you have a function \( g(x) \) inside another function \( h(g(x)) \), then the derivative is \( h'(g(x)) \cdot g'(x) \). In our case, since \( f(s) = \frac{1}{\sqrt{s+3}} \) can be seen as a composite function, applying the chain rule is essential.

• First, differentiate the outer layer: \( (s+3)^{-1/2} \) which gives us the expression \(-\frac{1}{2}(s+3)^{-3/2} \).
• Notice here the appearance of \( (s+3) \), signaling a composite nature that requires chain rule intervention.

Ultimately, the chain rule helps "unpack" the layers in functions like the one in our inverse function problem, revealing the interactions between the inner and outer function components.