Problem 46 - Thomas Calculus

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. \(y=(\ln x)^{\ln x}\)

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = (\ln x)^{\ln x} \left( \frac{\ln(\ln x)}{x} + \frac{1}{x \ln x} \right)\).

Step 1: Take the Natural Logarithm of Both Sides

To use logarithmic differentiation, start by taking the natural logarithm of both sides of the equation. If \(y = (\ln x)^{\ln x}\), then we have \(\ln y = \ln((\ln x)^{\ln x})\). Using the power rule of logarithms, this simplifies to \(\ln y = \ln x \cdot \ln(\ln x)\).

Step 2: Differentiate Both Sides with Respect to x

Differentiate both sides of the equation with respect to \(x\). For the left side, \(\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}\). For the right side, use the product rule: \(\frac{d}{dx}(\ln x \cdot \ln(\ln x)) = \ln(\ln x) \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(\ln(\ln x))\).

Step 3: Compute Derivatives

Compute each required derivative from Step 2. The derivative of \(\ln x\) is \(\frac{1}{x}\). For \(\ln(\ln x)\), the derivative is \(\frac{1}{\ln x} \cdot \frac{1}{x}\). Thus, the right side becomes \(\frac{\ln(\ln x)}{x} + \frac{1}{x} \cdot \frac{1}{\ln x}\).

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

Combine the terms on the right: \(\frac{\ln(\ln x)}{x} + \frac{1}{x \ln x}\). Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y \left( \frac{\ln(\ln x)}{x} + \frac{1}{x \ln x} \right)\). Substitute back \(y = (\ln x)^{\ln x}\) to get \(\frac{dy}{dx} = (\ln x)^{\ln x} \left( \frac{\ln(\ln x)}{x} + \frac{1}{x \ln x} \right)\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \()\), refers to the logarithm base \(e\). \(e\) is an irrational constant approximately equal to 2.71828. Natural logarithms are widely used in mathematics due to their simple rules and properties, particularly in calculus. Taking the natural logarithm of a number \(x\) gives the exponent to which \(e\) must be raised to obtain \(x\). The expression \(()x)\), which appears in our problem, represents the natural logarithm of itself. Such expressions are common when simplifying complex exponential forms. The natural logarithm has the following noteworthy properties:

• The natural logarithm of 1 is 0: \(()1 = 0\).
• For numbers greater than 1, the natural logarithm is positive.
• For numbers between 0 and 1, the natural logarithm is negative.
• \((ab) = ((a) + ((b))\), known as the product rule for logarithms.
• \(((a^b) = b ((a)\), used for expressions like \((((x))^{ln(x)})\).

By using the natural logarithm, we simplify the differentiation of power-like expressions, as seen in our exercise.

Product Rule

The product rule is a fundamental differentiation technique used when differentiating products of two or more functions. If you have a function \(f(x)\) and \(g(x)\), then the product rule states that: \[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\]Using the product rule allows us to differentiate more complex expressions step-by-step. In our solved exercise, the right-side expression \(((x) \cdot (((x)))\) requires the use of the product rule. First, treat \(((x)\) as \(f(x)\) and \((((x)))\) as \(g(x)\). Applying the product rule, we get:

• Derivative of \(((x))\) is \(\frac{1}{x}\).
• Derivative of \((((x)))\) is \(\frac{1}{((x))} \cdot \frac{1}{x}\), using the chain rule.

The product rule helps simplify the process, ensuring each part of the product is handled correctly.

Derivative of Logarithmic Functions

Differentiating logarithmic functions is a key skill in calculus. The basic rule is: \[ \frac{d}{dx}(((x))) = \frac{1}{x}\]This rule forms the foundation for differentiating more complex logarithmic expressions like \((((x)))\), which appears in this exercise.To differentiate \((((x)))\), use the chain rule:

• You take the derivative of the outer function \((((x)))\) with respect to \(((x))\): \(\frac{1}{((x))}\).
• Then, differentiate the inner function \(((x))\) with respect to \(x\), which is \(\frac{1}{x}\).

Putting these together, the complete derivative is \(\frac{1}{((x))} \cdot \frac{1}{x}\).Understanding these differentiation rules is crucial for solving logarithmic differentiation problems, ensuring you use the correct method for each part of an expression.

Differentiation Techniques

Differentiation techniques are methods to find the derivative of a function. In this exercise, we used several important calculus concepts:

• Logarithmic differentiation: A method useful for differentiating functions where variables appear as exponents. By taking the natural log of both sides, complicated expressions often become easier to differentiate.
• Chain rule: Used for differentiating nested functions. For example, to find the derivative of \((((x)))\), you must apply the chain rule to handle the composition of \(((n)\) and \(x\).
• Product rule: Employed when differentiating products of two functions, like \(((x) \cdot (((x)))\) in our problem.

Each technique has its own specific use, but they all allow us to break down complicated expressions into manageable parts.By mastering these methods, you will have a solid toolkit for tackling diverse and challenging calculus problems, enhancing your understanding and ability to solve real-world math problems efficiently.