Problem 79 - Calculus Single Variable

Question

Show that the function \(y \mapsto y \exp (y)\) is increasing for \(y>0 .\) Deduce that for every positive \(x,\) there is a unique \(y\) such that \(y\) exp \((y)=x\). This relationship inplicitly determines a function that is often denoted by \(W\) and is called Lambert's \(W\) function: $$ W(x) \exp (W(x))=x \quad(x>0) $$ Use implicit differentiation to show that \(W\) is an increasing function. Show that $$ W^{\prime}(x)=\frac{W(x)}{x(1+W(x))} $$

Step-by-Step Solution

Answer

The function \( y \mapsto y \exp(y) \) is increasing for \( y > 0 \). There is a unique \( y \) such that \( y \exp(y) = x \) for each \( x > 0 \). The Lambert W function is increasing, with \( W'(x) = \frac{W(x)}{x(1+W(x))} \).

Step 1: Differentiate the Function

To show that the function \( y \mapsto y \exp(y) \) is increasing for \( y > 0 \), we first differentiate it with respect to \( y \). This gives us:\[ \frac{d}{dy}[y \exp(y)] = \exp(y) + y \exp(y) = \exp(y) (1 + y)\]

Step 2: Analyze the Derivative

Since the derivative of \( y \exp(y) \) is \( \exp(y)(1 + y) \), and both \( \exp(y) > 0 \) and \( 1 + y > 0 \) for \( y > 0 \), it follows that the product is positive. Hence, the function \( y \exp(y) \) is increasing for \( y > 0 \).

Step 3: Show Unique Existence

For every positive \( x \), we deduce using the increasing nature of \( y \exp(y) \) that there is a unique \( y \) such that \( y \exp(y) = x \). This is because an increasing function has exactly one input for every output in its range.

Step 4: Differentiate W Implicitly

Given the equation for the Lambert W function, \( W(x) \exp(W(x)) = x \), differentiate both sides with respect to \( x \): \[ \frac{d}{dx}[W(x) \exp(W(x))] = 1 \]Use the product rule on the left side:\[ W'(x) \exp(W(x)) + W(x) W'(x) \exp(W(x)) = 1 \]

Step 5: Solve for W Prime

Factor out \( W'(x) \) on the left side:\[ W'(x)(\exp(W(x)) + W(x) \exp(W(x))) = 1 \]Simplify by factoring:\[ W'(x) \exp(W(x))(1 + W(x)) = 1 \]Solve for \( W'(x) \):\[ W'(x) = \frac{1}{\exp(W(x))(1 + W(x))} \]Since \( x = W(x) \exp(W(x)) \), we have \( \exp(W(x)) = \frac{x}{W(x)} \). Substitute this back into the expression:\[ W'(x) = \frac{W(x)}{x(1 + W(x))} \]

Step 6: Verify W(x) is Increasing

From the expression of \( W'(x) = \frac{W(x)}{x(1 + W(x))} \), observe that all components are positive for \( x > 0 \). Therefore, \( W'(x) > 0 \), indicating that the Lambert W function is increasing for \( x > 0 \).

Key Concepts

Understanding Increasing Functions

An increasing function, in simple terms, means that as you move along the x-axis, the value of the function goes up. This concept makes it easier to understand changes and predict outcomes in graphs.
To be precise, a function \(f(x)\) is considered increasing if for any two values \(a\) and \(b\), such that \(a < b\), it holds that \(f(a) < f(b)\).
In calculus, this is often shown using derivatives, which help in determining whether a function is increasing or decreasing.

• If \(f'(x) > 0\), the function is increasing.
• If \(f'(x) < 0\), the function is decreasing.

In our original exercise, the function \(y \to y \exp(y)\) is shown to be increasing for \(y > 0\) because its derivative, \(\exp(y)(1 + y)\), is positive for those values.

Exploring Implicit Differentiation

Implicit differentiation is a technique used when a function is not simply expressed as \(y = f(x)\), but rather in a more complex relationship where \(x\) and \(y\) are intertwined.
For example, instead of \(y = x^2\), you might deal with something like \(x^2 + y^2 = 1\), where you want to differentiate with respect to a specific variable.
This involves differentiating both sides of the equation with respect to \(x\) and often involves implicitly differentiated terms, such as \(\frac{dy}{dx}\).

• Apply the differentiation rules systematically.
• Keep track of both direct and indirect derivatives using the chain rule.

In our exercise, implicit differentiation is used to differentiate \(W(x) \exp(W(x)) = x\), which leads us to find the derivative of the Lambert W function, \(W'(x)\).

The Nature of Exponential Functions

Exponential functions are prevalent in calculus and represent a constant base raised to a variable exponent, typically expressed as \(a^x\) or \(\exp(x)\) if the base is \(e\). Such functions grow rapidly and are key in various mathematical concepts.
In an exponential function, any change in the variable results in a proportional change in the function’s value.

• Exponential growth refers to increases at a consistent percentage rate over time.
• Exponential decay is the reverse, where it decreases at a constant decay percentage.

In the exercise, the exponential function \(\exp(y)\) appears frequently, and its property of being always positive for real numbers is crucial when analyzing the derivative of \(y \exp(y)\). This property ensures the outcome is positive, supporting the notion that the function is increasing.

Navigating Calculus Problem Solving

Calculus problem-solving often involves understanding the relationships and changes between values. It often requires tools like derivatives to find rates of change, and integrals to find area under curves.
In tackling calculus problems:

• Break down the problem into manageable parts.
• Identify the applicable calculus concepts—like differentiation or integration.
• Use known formulas and rules, such as the chain rule or product rule, to solve specific problems.

In our scenario with the Lambert W function, the process involves using implicit differentiation and knowledge of exponential functions to show that \(W(x)\) is an increasing function. Each step builds on fundamental calculus rules to simplify and solve the problem.