Problem 22 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The derivative of the inverse function is \((f^{-1})'(t) = \frac{1}{2\sqrt{\ln(t)}t}\).

Step 1: Find General Form of Inverse Derivative

To find the derivative of the inverse function, we use the inverse function derivative rule: if \(y = f(x)\), then \( \frac{d}{dx} [f^{-1}(y)] = \frac{1}{f'(x)} \). Therefore, \( (f^{-1})'(t) = \frac{1}{f'(s)} \) where \(s = f^{-1}(t)\).

Step 2: Differentiate the Function

First, we calculate the derivative of \(f(s) = \exp(s^2)\) with respect to \(s\). Using the chain rule, we have: \( f'(s) = \frac{d}{ds}[\exp(s^2)] = \exp(s^2) \cdot 2s = 2s \exp(s^2) \).

Step 3: Find the Expression for the Inverse Derivative

Substitute the expression for \(f'(s)\) obtained in Step 2 into the formula from Step 1. We have: \( (f^{-1})'(t) = \frac{1}{2s \exp(s^2)} \) where \( s = f^{-1}(t) \).

Step 4: Express in terms of t

Since \(t = f(s) = \exp(s^2)\), it follows that \(s = \sqrt{\ln(t)}\). Thus, substitute \(s = \sqrt{\ln(t)}\) into the expression for \( (f^{-1})'(t)\). We get: \((f^{-1})'(t) = \frac{1}{2 \sqrt{\ln(t)} \cdot t} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. Imagine you have two functions, one nested inside the other, like layers of an onion. The chain rule lets you peel back these layers by differentiating each one in steps.

To apply the chain rule, you multiply the derivative of the outer function by the derivative of the inner function. In mathematical terms, if you have a function \( y = g(f(x)) \), the derivative \( \frac{dy}{dx} \) is calculated as: \[ \frac{dy}{dx} = g'(f(x)) \cdot f'(x) \] For our problem, where \( f(s) = \exp(s^2) \), the outer function is the exponential and the inner function is \( s^2 \). Using the chain rule, we differentiate:

• The outer function \( \exp(u) \), derivative is \( \exp(u) \).
• The inner function \( s^2 \), derivative is \( 2s \).

Thus, the whole derivative is \( f'(s) = \exp(s^2) \cdot 2s \). This method makes it straightforward to tackle more complex functions step-by-step.

Exponential Function

An exponential function is characterized by having a constant base raised to a variable exponent. It is one of the most important functions in mathematics. In our exercise, we are working with the exponential function \( f(s) = \exp(s^2) \), implying \( e \) is the constant base, and the exponent is \( s^2 \).

Its behavior is special because it grows rapidly, and it has the unique property that its rate of increase is proportional to its current value. We say that:

• \( y = \exp(x) \) is its basic form.
• \( \exp(s^2) \) indicates every output is an exponential of \( s^2 \).

This unique nature makes the exponential function a perfect model for situations involving growth and decay.
Half the power in using exponential functions lies in the simplicity of differentiating them. The derivative \( \frac{d}{dx}[\exp(x)] = \exp(x) \), which forms the basis of finding \( f'(s) \) in the exercise.

Derivative of Inverse Functions

The derivative of inverse functions is a useful concept when dealing with functions that are invertible. If a function \( y = f(x) \) is one-to-one, it has an inverse function \( f^{-1}(y) \). Calculating the derivative of this inverse function can be simplified using the inverse function derivative rule.

This rule tells us that if \( y = f(x) \), then \[ \frac{d}{dx} [f^{-1}(y)] = \frac{1}{f'(x)} \] Basically, this means you can find the derivative of the inverse function by taking the reciprocal of the derivative of the original function.

• If \( f'(s) = 2s \exp(s^2) \), then \( (f^{-1})'(t) = \frac{1}{2s \exp(s^2)} \).

In application, you first differentiate the function and use it to find the reciprocal, giving the inverse’s derivative.

Inverse Functions

Inverse functions are functions that "reverse" another function. If you have a function \( f(x) \), its inverse \( f^{-1}(x) \) finds the original input given the output. For an inverse to exist, the original function must be one-to-one, meaning each \( y \) comes from only one \( x \).

Understanding inverse functions boils down to finding the correspondence between inputs and outputs such that:

• \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) for all values within their domains.

In the problem we tackled, given \( f(s) = \exp(s^2) \), we find \( f^{-1}(t) \) by setting \( t = \exp(s^2) \) and rearranging to find \( s \). This gives: \[ s = \sqrt{\ln(t)} \] This form allows us to express \( (f^{-1})'(t) \) in terms of \( t \), resulting in: \[ \frac{1}{2 \sqrt{\ln(t)} \cdot t} \] Grasping the concept of inverse functions opens the door to solving complex problems involving reverse calculations.