Problem 10 - Basic Analysis: Introduction to Real Analysis

Question

Let \(I_{1}, I_{2}\) be intervals. Let \(f: I_{1} \rightarrow I_{2}\) be a bijective function and \(g: I_{2} \rightarrow I_{1}\) be the inverse. Suppose that both \(f\) is differentiable at \(c \in I_{1}\) and \(f^{\prime}(c) \neq 0\) and \(g\) is differentiable at \(f(c) .\) Use the chain rule to find a formula for \(g^{\prime}(f(c))\) (in terms of \(\left.f^{\prime}(c)\right)\).

Step-by-Step Solution

Answer

The derivative of the inverse function is \\(g'(f(c)) = \frac{1}{f'(c)}\\).

Step 1: Understanding the Problem

We need to find the derivative of the inverse function, \(g\), at \(f(c)\) using the chain rule. Given that \(f: I_{1} \rightarrow I_{2}\) is a bijective and differentiable function with non-zero derivative \(f'(c)\), and \(g: I_{2} \rightarrow I_{1}\) is its inverse and differentiable at \(f(c)\).

Step 2: Applying the Chain Rule

Since \(g\) is the inverse of \(f\), we have the identity \(g(f(x)) = x\) for \(x \in I_{1}\). Differentiating both sides with respect to \(x\), we use the chain rule: \((g(f(x)))' = g'(f(x)) \cdot f'(x) = 1\).

Step 3: Substituting Known Values

Substituting \(x = c\) into the differentiated identity gives \(g'(f(c)) \cdot f'(c) = 1\). This is because \(f(x) = f(c)\) when \(x = c\).

Step 4: Solving for the Derivative of the Inverse

Solving the equation \(g'(f(c)) \cdot f'(c) = 1\) for \(g'(f(c))\) gives \(g'(f(c)) = \frac{1}{f'(c)}\). This relates the derivative of the inverse function at \(f(c)\) to the derivative of \(f\) at \(c\).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions. When you have two functions, such as \(f\) and \(g\), where \(g\) is a function of \(f(x)\), the chain rule helps find the derivative of this composition. It's formulated as follows: if you have \(y = f(u)\) and \(u = g(x)\), then the derivative \(\frac{dy}{dx}\) is the product of the derivatives of each function: \(\frac{dy}{du} \cdot \frac{du}{dx}\).

For the problem at hand, we consider the functions \(g\) and \(f\) such that \(g\) is the inverse of \(f\), i.e., \(g(f(x)) = x\). By applying the chain rule to this identity, we differentiate both sides with respect to \(x\).

• The derivative of the left side becomes \(g'(f(x)) \cdot f'(x)\), where \(g'(f(x))\) is the rate of change of \(g\) at \(f(x)\), and \(f'(x)\) is the rate of change of \(f\) at \(x\).
• The derivative of the right side, \(x\), simplifies to 1, since the derivative of \(x\) with respect to \(x\) is 1.

So, the chain rule directly connects the rates of change of \(f\) and its inverse \(g\), allowing us to solve for \(g'(f(x))\).

Differentiable Functions

Differentiable functions are those that have a derivative at all points in their domain. This means you can find a rate of change at any given point for such functions. Understanding whether a function is differentiable helps predict its behavior and smoothness.

For a function to be differentiable at a point, it must be continuous there, and its derivative must exist. In this exercise, it is crucial that both \(f\) and \(g\), the inverse of \(f\), are differentiable.

• When \(f\) is differentiable at a point \(c\) and \(f'(c) eq 0\), it ensures that the tangent line is not horizontal, indicating a predictable behavior of \(f\) at \(c\).
• Similarly, \(g\) being differentiable at \(f(c)\) means that \(g\) too has a well-defined tangent at \(f(c)\).

This differentiability is the foundation upon which the chain rule applied, enabling us to relate the derivatives of \(f\) and \(g\).

Bijective Function

A bijective function is a special kind of function that is both injective (one-to-one) and surjective (onto). This means every element in the domain maps to a unique element in the codomain and covers the codomain entirely. For a function to have an inverse, it must be bijective.

In the given problem, the function \(f: I_{1} \rightarrow I_{2}\) is stated to be bijective. This property is crucial for the existence of the inverse function \(g\), as only bijective functions have inverses that map back perfectly.

• An injective function ensures that \(f\) doesn't map two different elements from \(I_1\) to the same element in \(I_2\).
• A surjective function ensures every \(y\) in \(I_2\) is mapped to by some \(x\) in \(I_1\), confirming \(f\)'s range is exactly \(I_2\).

This bijectivity ensures that there is a one-to-one correspondence between the input and output, an essential feature that allows both \(f\) and \(g\) to be differentiable and their derivatives to be related.

Derivatives

Derivatives represent the rate of change of a function with respect to its variable. It's the mathematical essence of motion and change, showing how a function behaves as its input changes slightly.

In calculus, understanding derivatives is key to solving problems involving optimization, motion, and change, such as finding the slope of a tangent line to a curve at a point.

• The derivative of \(f\) at a point \(c\), written as \(f'(c)\), captures how \(f\) changes as you move away slightly from \(c\).
• In this exercise, knowing \(f'(c) eq 0\) is crucial because if \(f'(c) = 0\), the function's rate of change would be zero, impairing our ability to relate it to \(g'(f(c))\).

By finding \(g'(f(c)) = \frac{1}{f'(c)}\), we use the power of derivatives to understand how the inverse function \(g\) changes at the point \(f(c)\), showcasing the interconnectedness of \(f\) and its inverse through differentiation.