Problem 348 - Calculus Volume 1

Question

For the following exercises, use logarithmic differentiation to find \(\frac{d y}{d x}\) $$y=(\ln x)^{\ln x}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = (\ln x)^{\ln x - 1} \cdot (\ln(\ln x) + 1)\)

Step 1: Take the Natural Logarithm of Both Sides

Start by taking the natural logarithm of both sides of the equation. Given that \(y = (\ln x)^{\ln x}\), apply the natural logarithm: \(\ln y = \ln((\ln x)^{\ln x})\). Using the power rule of logarithms, \(\ln y = \ln x \cdot \ln(\ln x)\).

Step 2: Differentiate Both Sides

Differentiate both sides of the equation with respect to \(x\). The left side becomes \(\frac{1}{y}\frac{dy}{dx}\). The right side requires the product rule for differentiation, as it is the product of \(\ln x\) and \(\ln(\ln x)\). Thus, the derivative is \((\ln(\ln x)) \cdot \frac{1}{x} + \ln x \cdot \frac{1}{\ln x} \cdot \frac{1}{x}\).

Step 3: Simplify the Derivative

Simplify the differentiated expression: \((\ln(\ln x)) \cdot \frac{1}{x} + \frac{1}{x}\). Combine terms to get \(\frac{\ln(\ln x) + 1}{x}\).

Step 4: Solve for \(\frac{dy}{dx}\)

Rearrange to solve for \(\frac{dy}{dx}\) by multiplying both sides by \(y\): \(\frac{dy}{dx} = y \cdot \frac{\ln(\ln x) + 1}{x}\). Since \(y = (\ln x)^{\ln x}\), substitute back: \(\frac{dy}{dx} = (\ln x)^{\ln x} \cdot \frac{\ln(\ln x) + 1}{x}\).

Step 5: Final Expression for \(\frac{dy}{dx}\)

Conclude with the final simplified form: \(\frac{dy}{dx} = (\ln x)^{\ln x - 1} \cdot (\ln(\ln x) + 1)\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a fundamental mathematical function, representing the logarithm to the base \( e \), where \( e \) is approximately 2.718.

• It is particularly useful in calculus for simplifying complex expressions involving exponential growth.
• Logarithms transform multiplicative operations into additive operations, which are easier to handle.

In this exercise, taking the natural logarithm of both sides of the given equation is the first step. This operation helps simplify the problem by using the power rule of logarithms, allowing us to manage the exponent more conveniently.

Power Rule of Logarithms

The power rule of logarithms is a property that states: \( \ln(a^b) = b \cdot \ln(a) \). This transformation turns an exponentiation problem into a multiplication problem.

• It is incredibly useful when differentiating functions where the exponent is not a constant.
• Applying this rule reduces the complexity of the function, making differentiation feasible.

In the provided solution, using the power rule of logarithms converts \( \ln((\ln x)^{\ln x}) \) into \( \ln x \cdot \ln(\ln x) \), simplifying the expression significantly and preparing it for differentiation.

Product Rule for Differentiation

The product rule in differentiation is essential when taking the derivative of the product of two functions, \( u \) and \( v \). It states that the derivative is \( \frac{d}{dx}(uv) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \).

• This rule helps differentiate composite functions without disentangling them first.
• In the current problem, \( \ln x \cdot \ln(\ln x) \) is differentiated using the product rule.

By recognizing \( \ln x \) and \( \ln(\ln x) \) as separate functions, we apply the product rule, leading to the expression \( (\ln(\ln x)) \cdot \frac{1}{x} + \ln x \cdot \frac{1}{\ln x} \cdot \frac{1}{x} \). This clears the way to simplify further.

Derivative Simplification

After differentiation, the next task is to simplify the derivative to its most efficient form. Simplification involves combining like terms and reducing complex fractions.

• In this context, simplification leads to combining \( (\ln(\ln x)) \cdot \frac{1}{x} + \frac{1}{x} \).
• This can be expressed as \( \frac{\ln(\ln x) + 1}{x} \), which highlights the key parts of the expression.

Finally, to find \( \frac{dy}{dx} \), we multiply by the original function \( y \) to reverse the natural logarithm applied initially. This results in the final simplified form \( \frac{dy}{dx} = (\ln x)^{\ln x} \cdot \frac{\ln(\ln x) + 1}{x} \), which combines all these simplification steps into a coherent answer.