Problem 2 - A First Course in Real Analysis

Question

Show that the relation \(F(x, y)=0\) yields \(y\) as a function of \(x\) in an interval \(I\) about \(x_{0}\) where \(F\left(x_{0}, y_{0}\right)=0\). Denote the function by \(f\) and compute \(f^{\prime}\). \(F(x, y) \equiv x^{2 / 3}+y^{2 / 3}-4=0 ;\left(x_{0}, y_{0}\right)=(1,3 \sqrt{3})\)

Step-by-Step Solution

Answer

After verifying that \(F(1, 3\sqrt{3})=0\) and checking \(\frac{\partial F}{\partial y}(1, 3\sqrt{3}) eq 0\), the Implicit Function Theorem guarantees the existence of a function \(f\) in an interval about 1. Differentiating \(F(x, y)=0\) implicitly yields \(f'(x)\), which at \(x_{0}=1\), provides the slope of \(y=f(x)\) at that point.

Step 1: Verify the initial condition

First, verify that the given point \(x_{0}, y_{0})=(1,3\sqrt{3}\) satisfies the equation. Substitute \(x_{0}\) and \(y_{0}\) into \(F(x, y)\) and check if \(F(1, 3\sqrt{3})=0\).

Step 2: Apply the Implicit Function Theorem

To show that \(y\) can be expressed as a function of \(x\) in an interval \(I\) about \(x_{0}\), check if the conditions of the Implicit Function Theorem are met, which require \(\frac{\partial F}{\partial y}(x_{0}, y_{0}) eq 0\). Compute \(\frac{\partial F}{\partial y}(1, 3\sqrt{3})\) to verify this condition.

Step 3: Differentiate implicitly to find \(f'(x)\)

Differentiate both sides of \(F(x, y)=0\) implicitly with respect to \(x\) to find \(\frac{dy}{dx}=f'(x)\). In this process, apply the chain rule to derivatives involving \(y\), treating \(y\) as an implicit function of \(x\).

Step 4: Solve for \(f'(x)\) at the point \(x_{0}\)

With the derivative obtained in step 3, solve for \(f'(x)\) at the point \(x=x_{0}=1\) to find the slope of the function \(y=f(x)\) at that point.

Key Concepts

Real Analysis

Real analysis is the branch of mathematics that deals with the set of real numbers and the functions defined on them. It encompasses various topics including limits, continuity, differentiability, and integration. Fundamental to real analysis is the concept of a function and how small changes in its input, or 'domain', can affect its output, termed as the 'range'. The explicit and implicit relationships between variables are scrupulously studied, employing rigorous proofs to ensure understanding of these relationships.

For instance, in the context of the exercise provided, real analysis helps us understand the behavior of the function described implicitly by the equation \( F(x, y) = 0 \). By exploring how changes in the variable \( x \) affect the variable \( y \), and vice versa, the subject helps articulate the conditions under which one can be expressed as a function of the other in a particular interval.

Partial Derivatives

Within the realm of real analysis, partial derivatives are crucial when dealing with functions of multiple variables. A partial derivative measures how a function changes as one of its variables is varied while the others are held constant. This concept is represented symbolically as \( \frac{\partial f}{\partial x} \) or \( \frac{\partial f}{\partial y} \), for example, depending on which variable is being considered.

The exercise employs partial derivatives to use the Implicit Function Theorem. Here, the existence of the derivative \( \frac{\partial F}{\partial y} \) at the point \( (x_0, y_0) \) is pivotal. If this derivative isn't zero, it suggests that at a local level, the function behaves regularly enough for us to treat \( y \) as a function of \( x \) near that point. Calculating and understanding partial derivatives are foundational for analyzing how functions behave in a multivariable context.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not given in the standard form \( y=f(x) \), but rather is defined by an equation involving both \( x \) and \( y \) that constrains their relationship. To find the derivative of \( y \) with respect to \( x \), we differentiate both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \). This technique requires the chain rule, a fundamental rule in calculus that computes the derivative of composite functions.

In the provided problem, one applies implicit differentiation to the equation \( F(x, y) = 0 \) to find \( f'(x) \), or \( \frac{dy}{dx} \). It involves finding the derivative of each term with respect of \( x \), and solving for \( \frac{dy}{dx} \) thereby providing insights into how \( y \) changes in response to changes in \( x \).

Function Differentiability

Function differentiability is a measure of how smoothly a function's output changes with respect to its input. If a function is differentiable at a point, it not only means that the function has a derivative there, but also that it can be locally approximated by a linear function. This is a stringent requirement, implying continuity and the lack of sharp corners or cusps at that point.

In the context of the exercise, verifying the differentiability of function \( f \) that implicitly describes \( y \) in terms of \( x \) is part of what the Implicit Function Theorem assures. But the theorem also needs the partial derivative \( \frac{\partial F}{\partial y} \) to be non-zero to apply. Through implicit differentiation, we can then find the derivative function \( f'(x) \), further studying the function's differentiability. It is worth noting that differentiability implies continuity, meaning if a function is differentiable at a point, it must also be continuous there. This deepens our understanding of a function's behavior in real analysis.