Problem 18 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\frac{\log x}{1+\log x}$$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = \frac{1}{x(1+\log x)^2} \).

Step 1: Apply the Quotient Rule

To differentiate the function \( y = \frac{\log x}{1+\log x} \), we will use the quotient rule. The quotient rule states \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \). Here, \( u = \log x \) and \( v = 1 + \log x \).

Step 2: Differentiate the Numerator and Denominator

Differentiate \( u = \log x \) to get \( \frac{du}{dx} = \frac{1}{x} \).Differentiate \( v = 1 + \log x \) to get \( \frac{dv}{dx} = \frac{1}{x} \).

Step 3: Substitute into the Quotient Rule Formula

Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule:\[ \frac{d}{dx} \left( \frac{\log x}{1+\log x} \right) = \frac{(1 + \log x) \left( \frac{1}{x} \right) - \log x \left( \frac{1}{x} \right)}{(1 + \log x)^2} \].

Step 4: Simplify the Expression

Simplify the numerator:\[ (1 + \log x) \frac{1}{x} - \log x \frac{1}{x} = \frac{1}{x} + \frac{\log x}{x} - \frac{\log x}{x} = \frac{1}{x} \].Thus, the differentiated function becomes:\[ \frac{1}{x(1+\log x)^2} \].

Key Concepts

Quotient Rule in Derivative Calculus

The quotient rule is a powerful tool in calculus, useful for differentiating ratios of functions. When you have a function expressed as a fraction, where the numerator and denominator are both differentiable functions of the same variable, the quotient rule becomes essential. This rule states:\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]Here, \( u \) represents the numerator and \( v \) the denominator of the fraction. This formula is particularly helpful when the fraction cannot be easily simplified or when both the numerator and the denominator are complex.

• Ensure both \( u \) and \( v \) are differentiable.
• Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), the derivatives of \( u \) and \( v \).
• Substitute these into the quotient rule formula and simplify.

Using the quotient rule correctly can simplify your derivative operations, making it less error-prone, especially when dealing with logarithmic or other complex functions.

Logarithmic Differentiation

Logarithmic differentiation is an alternative method useful for finding the derivative of functions that contain logarithms or complex products, quotients, and powers. It's particularly handy when the function is a product or quotient of terms where each term has a variable base raised to some power.To use logarithmic differentiation, follow these steps:

• Take the natural logarithm of both sides of the equation \( y = f(x) \).
• Use logarithmic properties to simplify the equation, converting products into sums and quotients into differences.
• Differentiate implicitly with respect to \( x \), utilizing the derivative of the natural log function, which is \( \frac{d}{dx}[\log x] = \frac{1}{x} \).
• Solve for \( \frac{dy}{dx} \).

This method can simplify differentiation, especially for complex functions, by reducing the potential for algebraic mistakes made when handling complicated expressions directly.

Differentiation Techniques in Calculus

Differentiation is a core concept in calculus, and mastering various techniques is essential for tackling a wide range of functions. Techniques are strategic approaches to find the derivative of a function, which is the rate at which the function's value changes.Key differentiation techniques include:

• Power Rule: If \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).
• Product Rule: If \( y = uv \), where both \( u \) and \( v \) are functions of \( x \), then \( \frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx} \).
• Chain Rule: Applied when a function is composed of another function, \( y = f(g(x)) \), gives \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• Quotient Rule: Used for differentiating divisions of functions, already detailed above.

Each technique has its ideal application scenario, and often, complex problems require a combination of these techniques. Understanding when and how to apply each method is crucial for efficient problem-solving in calculus.