Problem 2 - Basic Analysis: Introduction to Real Analysis

Question

Suppose I, \(J\) are intervals and a monotone onto \(f: I \rightarrow J\) has an inverse \(g: J \rightarrow I\). Suppose you already know that both \(f\) and \(g\) are differentiable everywhere and \(f^{\prime}\) is never zero. Using chain rule but not Lemma 4.4.1 prove the formula \(g^{\prime}(y)=\frac{1}{f^{\prime}(g(y))} .\)

Step-by-Step Solution

Answer

Using the chain rule and the fact that the derivative of identity is 1, we find that \( g'(y) = \frac{1}{f'(g(y))} \).

Step 1: Understand Inverse and Differentiability

Given a function \( f: I \rightarrow J \) that is onto and monotone, it implies that \( f \) has an inverse function \( g: J \rightarrow I \). The problem states that both \( f \) and its inverse \( g \) are differentiable everywhere, meaning their derivatives exist at all points in their respective domains.

Step 2: Use the Chain Rule to Relate Derivatives

According to the chain rule in calculus, if two functions are differentiable, then the composition of those functions is also differentiable. For \( f \) and \( g \), we have:\[ f(g(y)) = y \] Differentiating both sides with respect to \( y \) using the chain rule:\[ \frac{d}{dy}[f(g(y))] = \frac{d}{dy}[y] \] This simplifies to:\[ f'(g(y)) imes g'(y) = 1 \]

Step 3: Solve for the Derivative of the Inverse Function

The resulting equation from Step 2 is:\[ f'(g(y)) imes g'(y) = 1 \]To isolate \( g'(y) \), divide both sides by \( f'(g(y)) \):\[ g'(y) = \frac{1}{f'(g(y))} \]This gives us the formula for the derivative of the inverse function.

Key Concepts

Understanding the Chain Rule

In calculus, the chain rule is a fundamental concept used to find the derivative of a composed function. This rule is particularly useful when dealing with functions where one function is nested within another. For example, if you have a function defined as \( h(x) = f(g(x)) \), the chain rule helps express the derivative \( h'(x) \) in terms of \( f'(x) \) and \( g'(x) \). The formula for the chain rule is:
\[ h'(x) = f'(g(x)) \cdot g'(x).\]
This means the derivative of the outer function, \( f \), is evaluated at the inner function, \( g(x) \), and then multiplied by the derivative of the inner function, \( g \).
To understand the problem given, the chain rule helps us differentiate the composition, \( f(g(y)) = y \). By applying the chain rule to differentiate both sides concerning \( y \), we arrive at the formula which shows how \( g'(y) \) relates inversely to \( f'(g(y)) \), proving \( g'(y) = \frac{1}{f'(g(y))} \).

Differentiability Explored

Differentiability is a critical mathematical concept that describes whether a function has a derivative at all points within its domain. If a function is differentiable, it is smooth and has no breaks, sharp points, or cusps at any point in its interval.
Both functions \( f \) and \( g \) in the problem are said to be differentiable everywhere. This implies that you can find their derivatives in their entire domains. Having this property ensures that their behavior is smooth and predictable, which is crucial for applying the chain rule properly.
Differentiability plays a significant role here, as the exercise asks us to use the chain rule. Since we're dealing with differentiable functions, we can safely assume that all the necessary derivatives exist and can be computed, allowing us to find the derivative of the inverse function as requested in the exercise.

Monotonic Functions and Inverses

Monotonic functions are functions that either never increase or never decrease as the input value increases. Such functions can be strictly increasing, strictly decreasing, or constant within an interval. Significantly, their nature gives them unique properties concerning inverses.
In the problem, the function \( f \) is monotone and onto, meaning each output value is uniquely covered without repetition. Because it is monotone, \( f \) is guaranteed to have an inverse, \( g \), that is also monotonic and differentiated in its range.
Understanding the monotonic property is vital as it assures that \( f \) is invertible and that any derivative concerns will not face complications due to overlaps or undefined regions. This step ensures that when the inverse function \( g(y) \) is differentiated, the inverse nature is preserved and respected concerning the formula \( g'(y) = \frac{1}{f'(g(y))} \). Such smooth and predictable behavior showcases the coherence and stability provided by monotonic functions in calculus.