Problem 17 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{x(1 + 2\log x)}{(1 + \log x)^2}\)

Step 1: Identify the Rule to Use

To find \(\frac{dy}{dx}\) of the function \(y = \frac{x^{2}}{1+\log x}\), we need the quotient rule because the function is a fraction with a numerator and a denominator that are both functions of \(x\). The quotient rule states that for \(y = \frac{u}{v}\), then \(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\).

Step 2: Differentiate the Numerator

The numerator of our function is \(u = x^2\). Differentiating it gives \(\frac{du}{dx} = 2x\).

Step 3: Differentiate the Denominator

The denominator of our function is \(v = 1 + \log x\). Differentiating it gives \(\frac{dv}{dx} = \frac{1}{x}\) since the derivative of \(\log x\) is \(\frac{1}{x}\).

Step 4: Apply the Quotient Rule

Using the quotient rule, substitute \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the formula: \[ \frac{dy}{dx} = \frac{(1 + \log x)(2x) - (x^2)\left( \frac{1}{x} \right)}{(1 + \log x)^2} \] Simplifying inside the numerator: \[ \frac{dy}{dx} = \frac{2x + 2x \log x - x}{(1 + \log x)^2} \] Combine like terms: \[ \frac{dy}{dx} = \frac{x(1 + 2\log x)}{(1 + \log x)^2} \]

Step 5: Simplify the Expression (if needed)

The expression is already simplified. However, it never hurts to check your work to ensure every term has been calculated and combined correctly. The final expression is: \[ \frac{dy}{dx} = \frac{x(1 + 2\log x)}{(1 + \log x)^2} \]

Key Concepts

Quotient Rule

When you encounter a function that is a fraction with both numerator and denominator involving variables, you typically use the quotient rule to find its derivative. This rule is essential when the expressions are too complex to separate easily. The quotient rule can be envisioned like this:

• Consider a function in the form \( y = \frac{u}{v} \).
• The derivative \( \frac{dy}{dx} \) is given by the formula \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).

In simple terms, the rule tells us that to differentiate, you:

• Multiply the function at the bottom (denominator) by the derivative of the function at the top (numerator).
• Subtract the product of the function at the top and the derivative of the function at the bottom.
• Finally, divide by the square of the function at the bottom.

This systematic method helps in ensuring that all parts of the function are correctly differentiated and assembled together.

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm, play a fundamental role in calculus and the differentiation process. The natural logarithm, denoted as \( \log x \) or \( \ln x \), has a very straightforward derivative:

• The derivative of \( \log x \) with respect to \( x \) is \( \frac{1}{x} \).

Understanding this property is crucial when dealing with expressions that involve logarithms in differentiation exercises.
Logarithmic functions are commonly used to simplify complex multiplication problems into manageable addition ones, which are easier to differentiate. This feature makes them very versatile, not just in theoretical mathematics but also in practical applications like computer science and engineering.
Mastering derivatives involving logarithmic functions will provide a substantial advantage in solving various calculus problems.

Derivatives of Functions

The concept of derivatives is central to calculus and essential understanding for students tackling calculus problems. A derivative essentially measures the "rate of change" or how one quantity changes with respect to another. For instance:

• If you have a function \( f(x) \), its derivative \( \frac{df}{dx} \) tells you the rate at which \( f(x) \) changes as \( x \) changes.

Different rules exist for different types of functions, like polynomial, exponential, and trigonometric functions. Each function type has its unique derivative rule that you can apply directly. For example, the derivative of \( x^n \) is \( nx^{n-1} \).
Grasping these rules allows for the combining and creative manipulation necessary to differentiate more complicated functions. Knowing when and how to apply rules like the product rule, chain rule, or quotient rule is crucial.
Becoming comfortable with derivatives of functions will enable you to tackle calculus problems with confidence, opening up a deeper understanding of the mathematical changes of our world.