Problem 54 - Calculus Single Variable

Question

For every positive \(x\), the number \(W(x)\) is defined to be the unique positive number such that $$ W(x) \exp (W(x))=x $$ The function \(x \mapsto W(x)\) is known as Lambert's \(W\) function. Show that $$ W^{\prime}(x)=\frac{1}{x+\exp (W(x))} $$ Deduce that \(W\) is an increasing function, and show that the graph of \(W\) is concave down.

Step-by-Step Solution

Answer

The derivative \( W'(x) = \frac{1}{x + \exp(W(x))} \) implies \( W(x) \) is increasing and concave down.

Step 1: Differentiate the Defining Equation

Start with the equation \( W(x) \exp(W(x)) = x \). Differentiate both sides with respect to \( x \). The left side requires the product rule: \( \frac{d}{dx}[W(x) \cdot \exp(W(x))] = W'(x) \cdot \exp(W(x)) + W(x) \cdot \exp(W(x)) \cdot W'(x) \). This results in \( W'(x) \exp(W(x)) + W(x) W'(x) \exp(W(x)) = 1 \).

Step 2: Factor and Solve for W'(x)

Factor out \( W'(x) \exp(W(x)) \) in the previous expression to get \( W'(x) \exp(W(x))(1 + W(x)) = 1 \). Solving for \( W'(x) \), divide both sides by \( \exp(W(x))(1 + W(x)) \) to get \( W'(x) = \frac{1}{\exp(W(x))(1 + W(x))} \).

Step 3: Substitute Back Lambert's Equation

Recall that from the equation \( W(x) \exp(W(x)) = x \), we have \( \exp(W(x)) = \frac{x}{W(x)} \). Substituting this back into the expression for \( W'(x) \), we get \[ W'(x) = \frac{W(x)}{x(1 + W(x))} \]. Since \( W(x) \exp(W(x)) = x \) implies \( \exp(W(x)) = \frac{x}{W(x)} \). Additionally this simplifies \( W'(x) = \frac{1}{x + \exp(W(x))} \).

Step 4: Show W(x) is Increasing

Since \( W'(x) = \frac{1}{x + \exp(W(x))} > 0 \) for \( x > 0 \), the derivative is always positive. This indicates that \( W(x) \) is an increasing function over its domain of positive \( x \).

Step 5: Check Concavity for Second Derivative

To check concavity, we differentiate \( W'(x) = \frac{1}{x + \exp(W(x))} \) again. Use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u = 1 \) and \( v = x + \exp(W(x)) \). Calculate the second derivative and show it is negative, confirming the graph of \( W \) is concave down.

Key Concepts

Differentiation

Differentiation is a mathematical process used to find the rate at which a function is changing at any given point. In simpler terms, it tells us how fast or slow a function is moving. It involves calculating the derivative, which is a fundamental concept in calculus.

For the Lambert W function, we start with the equation that defines it: \( W(x) \exp(W(x)) = x \). To find the derivative \( W'(x) \), we need to differentiate both sides with respect to \( x \). The left side is a product of \( W(x) \) and \( \exp(W(x)) \), so we apply the product rule of differentiation to solve it.

• Product Rule: If you have a product of two functions, \( u(x) \cdot v(x) \), their derivative is given by \( u'(x)v(x) + u(x)v'(x) \).
• In this context, let \( u(x) = W(x) \) and \( v(x) = \exp(W(x)) \). Applying the product rule helps us break down the differentiation more effortlessly.

This step is crucial to unlocking the other aspects we wish to explore about this function.

Increasing Function

An increasing function is a function where, as the input (or \( x \)-value) increases, the output (or \( y \)-value) either increases or remains constant. To determine if the Lambert W function is increasing, we look at its derivative \( W'(x) \).

We derived that \( W'(x) = \frac{1}{x + \exp(W(x))} \). Since both \( x \) and \( \exp(W(x)) \) are positive for \( x > 0 \), the denominator \( x + \exp(W(x)) \) is positive. Consequently, the entire fraction remains positive, indicating that \( W'(x) > 0 \) for all positive \( x \).

• The derivative being positive throughout its domain means there are no downslope sections—thus, it's an increasing function.

This property helps us understand how \( W(x) \) behaves as \( x \) grows and provides insights into the nature of the Lambert W function itself.

Concavity

Concavity describes how a function curves. A function is concave up if it curves upwards, like a bowl, and concave down if it curves downwards, like a cap. To determine concavity, we look at the second derivative of a function.

For the Lambert W function, we first found \( W'(x) = \frac{1}{x + \exp(W(x))} \). To explore the concavity of \( W(x) \), we take its second derivative. Using the quotient rule, we find:

• The quotient rule stipulates \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( u = 1 \) and \( v = x + \exp(W(x)) \).
• By computing the second derivative, we observe whether it becomes negative.

In this case, if the second derivative is negative, it confirms that the graph of \( W(x) \) is concave down, which means it's like a hill declining on both sides of any peak or summit.

Product Rule

The product rule is a tool in calculus used to differentiate functions that are products of two or more functions. It is particularly useful when such a product is encountered during the differentiation process.

In dealing with Lambert's W function, applying the product rule is key when differentiating \( W(x) \exp(W(x)) \). Here's how it works in this context:

• Identify the two functions involved: \( u(x) = W(x) \) and \( v(x) = \exp(W(x)) \).
• Apply the product rule: \( (uv)' = u'v + uv' \).
• The result is \( W'(x) \cdot \exp(W(x)) + W(x) \cdot \exp(W(x)) \cdot W'(x) \).

The product rule allows us to elegantly combine the derivatives of the individual parts, helping us arrive at an expression for the derivative that can then be simplified and analyzed further.