Problem 27 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\ln \frac{1}{x \sqrt{x+1}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{-3x - 2}{2x(x+1)} \).

Step 1: Rewrite the Function

First, simplify the expression inside the natural logarithm. The given function is \( y = \ln \left( \frac{1}{x \sqrt{x+1}} \right) \). This can be rewritten using properties of logarithms as \( y = \ln(1) - \ln(x) - \ln(\sqrt{x+1}) \). Since \( \ln(1) = 0 \), the function simplifies to \( y = -\ln(x) - \frac{1}{2}\ln(x+1) \).

Step 2: Differentiate Using Logarithmic Properties

Differentiate each term separately. First, the derivative of \(-\ln(x)\) with respect to \(x\) is \(-\frac{1}{x}\). Next, the derivative of \(-\frac{1}{2} \ln(x+1)\) is \(-\frac{1}{2} \times \frac{1}{x+1}\), using the chain rule.

Step 3: Combine the Derivatives

Add the derivatives from the previous step. The derivative \( \frac{dy}{dx} \) is \(-\frac{1}{x} - \frac{1}{2(x+1)} \).

Step 4: Simplify the Expression

To combine the two fractions, find a common denominator. The fractions are \(-\frac{1}{x} \) and \(-\frac{1}{2(x+1)}\), with a common denominator of \(2x(x+1)\). Rewrite the derivatives as \(\frac{-2(x+1)}{2x(x+1)}\) and \(\frac{-x}{2x(x+1)}\). Combine them to get: \[\frac{-2(x+1) - x}{2x(x+1)} = \frac{-2x - 2 - x}{2x(x+1)} = \frac{-3x - 2}{2x(x+1)}\].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used when dealing with complex functions, especially those involving products, quotients, or exponents. In our exercise, we start with the function \( y = \ln \left( \frac{1}{x \sqrt{x+1}} \right) \). By applying logarithmic properties, the function can be simplified by breaking it into smaller parts.

These properties allow us to write:

• \( \ln(\frac{1}{x \sqrt{x+1}}) = \ln(1) - \ln(x) - \ln(\sqrt{x+1}) \)
• Since \( \ln(1) = 0 \), we simplify this to \( -\ln(x) - \frac{1}{2}\ln(x+1) \)

This simplification is key as it turns a complex logarithmic expression into manageable pieces.

Logarithmic differentiation not only makes differentiation easier but also ensures we efficiently tackle each part individually.

Chain Rule

The chain rule is fundamental in calculus when differentiating composite functions. It allows us to find the derivative of complicated expressions by breaking them down into their simpler components.

In our exercise, the function \( y = -\ln(x) - \frac{1}{2}\ln(x+1) \) involves terms that require the chain rule. When differentiating \(-\frac{1}{2} \ln(x+1)\), we apply the chain rule:

• The derivative of \( \ln(u) \) is \( \frac{1}{u} \), where \( u = x+1 \)
• We then need the derivative of \( u \), which is \( 1 \)
• So, we multiply \( \frac{1}{u} \) by the derivative of \( u \), resulting in \(-\frac{1}{2} \cdot \frac{1}{x+1} \)

This approach simplifies the differentiation process, ensuring each part of the function is treated correctly.

The chain rule thus enables us to handle derivatives efficiently, even when the function contains nested operations.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special logarithmic function with base \( e \), where \( e \approx 2.718 \). It appears frequently in calculus due to its smooth properties and practical applications in exponential growth and decay.

In our problem, the natural logarithm feature simplifies the process of differentiation. This is because the derivative of \( \ln(x) \) is simply \( \frac{1}{x} \).

Here are some core properties of the natural logarithm:

• \( \ln(ab) = \ln(a) + \ln(b) \)
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• \( \ln(a^b) = b \ln(a) \)

These properties make it easier to break down complex expressions before applying differentiation.

Using these properties carefully leads to smoother and more manageable computations when handling derivatives of functions involving the natural logarithm.