Problem 33 - University Calculus: Early Transcendentals

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 \( \frac{dy}{dx} = -\frac{1}{x} - \frac{1}{2(x+1)} \).

Step 1: Simplify the Given Expression

First, simplify the expression inside the logarithm for easier differentiation. We have: \( y = \ln \frac{1}{x \sqrt{x+1}} \). This can be rewritten using properties of logarithms: \( y = \ln 1 - (\ln x + \ln \sqrt{x+1}) \). Since \( \ln 1 = 0 \) and using the power rule \( \ln \sqrt{x+1} = \frac{1}{2} \ln (x+1) \), the expression becomes: \( y = - (\ln x + \frac{1}{2} \ln (x+1)) = - \ln x - \frac{1}{2} \ln (x+1) \).

Step 2: Differentiate Each Term Separately

Now differentiate each term with respect to \(x\). - **First term**: For \(-\ln x\), the derivative is \(-\frac{1}{x}\).- **Second term**: For \(-\frac{1}{2} \ln (x+1)\), use the chain rule. The derivative is \(-\frac{1}{2} \cdot \frac{1}{x+1}\).

Step 3: Combine the Derivatives

Combine the derivatives of the individual terms from Step 2 to find the derivative of the entire expression:\[ \frac{dy}{dx} = -\frac{1}{x} - \frac{1}{2(x+1)} \].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used to find derivatives of functions that are tricky to differentiate directly. This method is particularly useful for products or quotients of functions, especially when they involve powers or are complex in nature. In this process, you take the natural logarithm of both sides of the given function. This helps in simplifying the differentiation by turning products into sums and quotients into differences thanks to properties of logarithms.

For example, in the exercise provided, the function is initially a quotient: \( y = \ln \frac{1}{x \sqrt{x+1}} \). By applying the property \( \ln \frac{a}{b} = \ln a - \ln b \), this expression simplifies considerably before applying the derivative. Logarithmic differentiation is often combined with other differentiation rules such as the chain rule, which you will see shortly. This method is especially helpful when dealing with variables raised to the power of other variables.

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of a composite function. A composite function is essentially a function wrapped inside another function, like \( f(g(x)) \). The chain rule helps in differentiating such functions by considering the derivative of the outer function while multiplying by the derivative of the inner function.

In the context of the given solution, we used the chain rule in calculating the derivative of \( -\frac{1}{2} \ln (x+1) \). The outer function \( f(u) = \ln u \) has a derivative of \( \frac{1}{u} \). The inner function \( g(x) = x+1 \) has a derivative of 1. According to the chain rule, the derivative of the composite \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \), which is \( \frac{1}{x+1} \cdot 1 \). Don't forget the constant multiplier of \(-\frac{1}{2}\), giving us a final derivative of \(-\frac{1}{2} \cdot \frac{1}{x+1}\).

• Remember: Differentiate the outer function first, holding the inner function in place.
• Then multiply by the derivative of the inner function to complete the process.

Properties of Logarithms

The properties of logarithms are crucial in simplifying expressions, especially when dealing with derivatives. Understanding these properties aids immensely in converting multiplication into addition and division into subtraction in logarithmic terms.

Key properties include:

• Product Rule: \( \ln(ab) = \ln a + \ln b \)
• Quotient Rule: \( \ln \frac{a}{b} = \ln a - \ln b \)
• Power Rule: \( \ln (a^n) = n \ln a \)

Applying these properties simplifies the function \( y = \ln \frac{1}{x \sqrt{x+1}} \) into a form that is easier to differentiate, leading to the equation \( y = - \ln x - \frac{1}{2} \ln (x+1) \) in the exercise solution. By knowing these properties, a seemingly complex expression becomes more manageable and reduces the potential for errors in differentiation.

Mastering these properties can greatly enhance your ability to work with logarithmic functions and their derivatives efficiently.