Problem 66 - Calculus and its applications

Question

Consider the function \(y=x^{x},\) with \(x>0\). a) Find \(\frac{d y}{d x}\). (Hint: Take the natural logarithm of both sides and differentiate implicitly.) b) Find the minimum value of \(y\) on \((0, \infty)\).

Step-by-Step Solution

Answer

a) \(\frac{dy}{dx} = x^x(\ln x + 1)\). b) The minimum value of \(y\) is \(\left(\frac{1}{e}\right)^{\frac{1}{e}}\).

Step 1: Take the Natural Logarithm

First, take the natural logarithm of both sides of the function to simplify the expression for differentiation. Start with the equation: \(y = x^x\). Take the natural logarithm: \(\ln y = \ln(x^x)\). This simplifies to: \(\ln y = x \ln x\).

Step 2: Differentiate Implicitly

Differentiate both sides of the equation \(\ln y = x \ln x\) with respect to \(x\). The left-hand side gives \(\frac{1}{y}\frac{dy}{dx}\), using the chain rule for differentiation, and the right-hand side gives \(1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\). So, we have \(\frac{1}{y}\frac{dy}{dx} = \ln x + 1\).

Step 3: Solve for \(\frac{d y}{d x}\)

Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\). This gives \(\frac{dy}{dx} = y (\ln x + 1)\). Substitute back \(y = x^x\) to get \(\frac{dy}{dx} = x^x(\ln x + 1)\).

Step 4: Find Critical Points for Minimum Value

To find where the minimum occurs, we need the derivative \(\frac{dy}{dx} = 0\). Set \(x^x(\ln x + 1) = 0\). Since \(x^x > 0\) for \(x > 0\), we need \(\ln x + 1 = 0\), which implies \(\ln x = -1\) or \(x = e^{-1} = \frac{1}{e}\).

Step 5: Analyze for Minimum

The critical point found is \(x = \frac{1}{e}\). To ensure it is a minimum, check the second derivative or consider the behavior of \(\ln x + 1\) around \(x = \frac{1}{e}\). As \(x\) decreases from 1, \(\ln x + 1\) changes sign only at \(x = \frac{1}{e}\), indicating a change in slope direction, confirming a minimum.

Step 6: Calculate the Minimum Value of \(y\)

Substitute \(x = \frac{1}{e}\) back into \(y = x^x\) to find the minimum value. So, \(y = \left(\frac{1}{e}\right)^{\frac{1}{e}}\). This is the minimum value of the function on \((0, \infty)\).

Key Concepts

Natural Logarithm

A natural logarithm, denoted by \(\ln\), is a powerful tool in calculus that helps simplify exponential expressions. In our function \(y = x^x\), direct differentiation is complex. By taking the natural logarithm of both sides, we apply:

• \(\ln y = \ln(x^x)\)
• Using logarithmic identity: \(\ln(x^x) = x \ln x\)

This transformation leverages the property \(\ln(a^b) = b \ln a\), making the differentiation straightforward. After this step, the expression becomes linear in terms of \(x\), which is easier to handle in further calculus operations. The use of natural logarithm in expressions like \(x^x\) is crucial because it allows us to differentiate functions we otherwise could not tackle directly.

Critical Points

Critical points in calculus are where the derivative of a function equals zero or is undefined. These points are vital because they help us identify where a function might reach its minimum or maximum values.

In solving for the critical points of \(y = x^x\), we first derive:

• \(\frac{dy}{dx} = x^x(\ln x + 1)\)

To find when the function reaches a minimum, set \(\frac{dy}{dx} = 0\):

• This breaks down to \(\ln x + 1 = 0\), which simplifies to \(\ln x = -1\), or \(x = \frac{1}{e}\)

In this context:

• \(x = \frac{1}{e}\) is the critical point necessary to find the function's minimum value.

Analyzing these points helps us understand the nuances of the function's behavior and ensure we locate the minimum along the given domain \((0, \infty)\).

Chain Rule

The Chain Rule is a fundamental differentiation technique in calculus. It is utilized when dealing with composite functions and allows us to differentiate using the formula \((f(g(x)))' = f'(g(x))g'(x)\). In the function \(y = x^x\), taking the natural log simplifies the problem:

• \(\ln y = x \ln x\)
• Differentiate both sides with respect to \(x\): \(\frac{1}{y}\frac{dy}{dx} = \ln x + 1\)

Here, we apply the Chain Rule to differentiate \(\ln y\). We multiply the derivative \(\frac{1}{y}\frac{dy}{dx}\) by \(y\) to get \(\frac{dy}{dx}\), ensuring we can express \(y\) in terms of \(x\) again:

• \(\frac{dy}{dx} = y(\ln x + 1)\)
• Substitute \(y = x^x\) to get the result in the original function's terms.

Thus, using the Chain Rule enables us to handle implicit differentiation, allowing for the manipulation and solution of more complex functions like \(x^x\).