Problem 1 - 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 \(f^{\prime}\). \(F(x, y) \equiv y^{3}+y-x^{2}=0 ;\left(x_{0}, y_{0}\right)=(0,0)\)

Step-by-Step Solution

Answer

After verifying the initial conditions and applying the Implicit Function Theorem, the derivative of the function \(f\) with respect to \(x\) is \(f'(x) = \frac{2x}{3y^2 + 1}\), and for \(x_0 = 0\), \(f'(0) = 0\).

Step 1: Verify the Initial Condition

Check whether the given point \(x_0, y_0\) satisfies the equation \(F(x, y) = y^3 + y - x^2 = 0\). Substituting \(x_0 = 0\) and \(y_0 = 0\) into \(F(x, y)\), we get \(0^3 + 0 - 0^2 = 0\), which satisfies the equation. Therefore, \(F(0, 0) = 0\).

Step 2: Apply the Implicit Function Theorem

To show that \(y\) is a function of \(x\) in an interval \(I\) about \(x_0\), we need to verify the conditions of the Implicit Function Theorem (IFT). The IFT states \(y\) can be expressed as a function of \(x\) if \(F_y = \frac{\partial F}{\partial y} eq 0\) at the point \(\left(x_{0}, y_{0}\right)\). Calculate \(F_y\) by differentiating \(F(x, y)\) with respect to \(y\): \(F_y = 3y^2 + 1\). At \(\left(x_{0}, y_{0}\right) = (0, 0)\), we have \(F_y(0, 0) = 3(0)^2 + 1 = 1 \eq 0\), hence satisfying the IFT.

Step 3: Differentiate the Relation Implicitly to Find \(f'(x)\)

To find \(f'(x)\), differentiate the given relation \(F(x, y)=0\) with respect to \(x\) implicitly assuming \(y\) is a function of \(x\), i.e., \(y = f(x)\). The derivative is given by \(F_x + F_y \cdot f'(x) = 0\). Differentiating \(F(x, y)\), we get \(F_x = -2x\) and we already found \(F_y = 3y^2 + 1\). Thus, we have \(-2x + (3y^2 + 1)f'(x) = 0\). Solving for \(f'(x)\), we get \(f'(x) = \frac{2x}{3y^2 + 1}\). Substituting \(x_0 = 0\) and \(y_0 = 0\), the derivative at the point \(x_0\) is \(f'(0) = \frac{2 \cdot 0}{3 \cdot 0^2 + 1} = 0\).

Key Concepts

Real Analysis

Real analysis is a significant branch of mathematical analysis dealing with the real numbers and the real value functions. Specifically, it encompasses theories related to limits, continuity, derivatives, and integrals of functions of real variables.

Understanding the underlying principles of real analysis is crucial when exploring the behavior of functions and their interactions. In our exercise, the concept is applied to determine if a function implicitly defined by an equation, such as the given relation \(F(x, y)=0\), can be expressed explicitly as a function with a variable dependency, i.e., \(y\) as a function of \(x\), and to determine the function's derivative.

Implicit Differentiation

Implicit differentiation is a technique to find the derivative of a function that is not expressed in the standard form \(y=f(x)\). Instead, the function is defined implicitly by an equation involving both \(x\) and \(y\).

This process requires treating \(y\) as an implicitly defined function of \(x\), which allows us to differentiate both sides of the equation with respect to \(x\) and then solve for the derivative \(dy/dx\). In our exercise, we use implicit differentiation to derive the formula for \(f'(x)\) from the equation \(y^3 + y - x^2 = 0\). The unique challenge here is to differentiate terms involving \(y\) with respect to \(x\), and this is where implicit differentiation shines. It simplifies the process by letting us directly differentiate with respect to \(x\), without explicitly solving for \(y\).

Function Derivative

The derivative of a function at a point provides the slope of the tangent line to the function's graph at that point, and it is a fundamental concept in calculus. It quantifies the rate at which the function's value changes with respect to changes in its input value.

In the context of our exercise, we're interested in finding the derivative of \(y\), denoted as \(f'(x)\), which represents how \(y\) varies with \(x\) in the implicitly defined function \(F(x, y)=0\). The function's derivative is crucial for understanding the behavior of the function near a given point. In the given exercise, after implicitly differentiating the relation \(F(x, y) = y^3 + y - x^2 = 0\), we find that \(f'(x) = \frac{2x}{3y^2 + 1}\), offering insight into the instantaneous rate of change of \(y\) with respect to \(x\) at any point along the curve defined by \(F\).