Problem 4 - 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 \(f^{\prime}\). \(F(x, y) \equiv \sin x+2 \cos y-\frac{1}{2}=0 ;\left(x_{0}, y_{0}\right)=(\pi / 6,3 \pi / 2)\)

Step-by-Step Solution

Answer

By verifying \( F(x_0, y_0) = 0 \) and \( \frac{\partial F}{\partial y}(x_0, y_0) eq 0 \) and then using implicit differentiation to calculate \( f'(x) \) from \( F(x, y) = 0 \) we can define \( y \) as a function of \( x \) and find its derivative.

Step 1: Verify the Initial Condition

First check whether the given point \( (x_0, y_0) = (\pi/6, 3\pi/2) \) satisfies the equation \( F(x, y) = 0 \) by plugging the values into the equation \( F(x, y) = \sin(x) + 2\cos(y) - 1/2 \) and ensuring the equality holds.

Step 2: Apply the Implicit Function Theorem

For \( F(x, y) \) to define \( y \) as a function of \( x \) in some interval \( I \) around \( x_0 \) based on the Implicit Function Theorem, we need to verify that \( \frac{\partial F}{\partial y} eq 0 \) at the point \( (x_0, y_0) \) using the partial derivative \( \frac{\partial F}{\partial y} \) evaluated at \( (x_0, y_0) \) is different from zero.

Step 3: Calculate the Derivative

Use the differentiation of implicit functions to find \( f'(x) \) by differentiating both sides of the equation \( F(x, y) = 0 \) with respect to \( x \) treating \( y \) as a function of \( x \) (\( y = f(x) \) and using the chain rule for derivatives).

Step 4: Solve for \( f'(x) \) Explicitly

After differentiation, solve the resulting equation for \( f'(x) \) to find the derivative of \( y \) with respect to \( x \) explicitly.

Key Concepts

Partial Derivative

When we talk about the partial derivative, we refer to the rate of change of a multivariable function with respect to one variable, holding the other variables constant. In the context of the Implicit Function Theorem, this concept is key for it determines whether a function of multiple variables can be simplified into a function of a single variable within a certain domain.

Understanding partial derivatives is crucial when dealing with complex functions where each variable's specific influence needs to be isolated. For instance, in our function F(x, y) = sin(x) + 2cos(y) - 1/2, to see how F changes as y varies while keeping x constant, we compute the partial derivative with respect to y denoted as \(\frac{\partial F}{\partial y}\). If this derivative is non-zero at a point, according to the Implicit Function Theorem, we can locally express y as a function of x, which allows us to proceed to find a more explicit relationship between x and y.

Function of a Function

The concept of a function of a function, often encountered in calculus as the chain rule, arises when we have one function nested within another. This idea is fundamental when differentiating composite functions—where we apply the derivative of the outer function to the inner function and multiply it by the derivative of the inner function itself.

For example, if y is implicitly defined as a function of x via y = f(x), and we are given another function g(y), then g(f(x)) is a function of a function. When we differentiate g(f(x)) with respect to x, we apply the chain rule to accommodate the nesting of functions. This principle allows us to manoeuvre through problems where direct differentiation isn't straightforward due to the implicit nature of the relationship between variables.

Differentiation of Implicit Functions

The differentiation of implicit functions comes into play when we're dealing with equations that define one variable implicitly in terms of another. Unlike explicit functions where we can directly apply differentiation rules, implicit functions require a different approach that often involves the chain rule for derivatives.

In our exercise, to find f'(x), we implicitly differentiate the equation sin(x) + 2cos(y) - 1/2 = 0 with respect to x, treating y as a function of x (y = f(x)). This allows us to calculate the derivative of y with respect to x, denoted as f'(x), by recognizing and respecting the interdependence of the variables. Once the derivatives are calculated, we rearrange the equation to solve for f'(x), giving us the rate of change of the implicitly defined function.