Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(\frac{d r}{d x}\) if \(r=\frac{\ln x}{x^{2} \ln x^{2}}+\left(\ln \frac{1}{x}\right)^{3}\)

Step-by-Step Solution

Answer

\(\frac{d r}{d x} = -\frac{1}{x^3} - \frac{3(-\ln x)^2}{x}\).

Step 1: Simplify the Function

The given function is \(r = \frac{\ln x}{x^{2} \ln x^{2}} + \left(\ln \frac{1}{x}\right)^{3}\). Simplify the second term: \(\ln \frac{1}{x} = -\ln x\). Therefore, the function becomes \(r = \frac{\ln x}{x^{2} \cdot 2\ln x} + (-\ln x)^{3}\), which further simplifies to \(r = \frac{1}{2x^{2}} + (-\ln x)^{3}\).

Step 2: Differentiate the Simplified Function

Differentiate each term separately. Start with \(\frac{1}{2x^{2}}\). Use the power rule, so \(\frac{d}{dx}\left(\frac{1}{2x^{2}}\right) = \frac{d}{dx}\left(\frac{1}{2}x^{-2}\right) = -x^{-3}\), so it becomes \( -\frac{1}{x^{3}}\).

Step 3: Differentiate the Second Term

Differentiate \((-\ln x)^{3}\) using the chain rule. Let \(u = -\ln x\), so \(u(x) = -\ln x\) and \(u^3 = (-\ln x)^3\). Therefore, \(\frac{d}{dx}(-\ln x)^3 = 3(-\ln x)^2 \cdot \frac{d}{dx}(-\ln x)\). Since \(\frac{d}{dx}(-\ln x) = -\frac{1}{x}\), the derivative of the second term is \(-\frac{3(-\ln x)^2}{x}\).

Step 4: Combine the Derivatives

Combine the derivatives from Step 2 and Step 3. This gives us the derivative \(\frac{d r}{d x} = -\frac{1}{x^3} - \frac{3(-\ln x)^2}{x}\).

Key Concepts

Chain Rule

The chain rule is a method used in calculus to find the derivative of a composite function. When you have a function nested inside another function, the chain rule helps you perform differentiation efficiently. This rule is essential when dealing with functions like \(f(g(x))\), where both \(f\) and \(g\) are functions of \(x\).

• To apply the chain rule, you first differentiate the outer function, keeping the inner function unchanged.
• Next, you multiply by the derivative of the inner function.

This approach simplifies the differentiation of complex structures.
In the problem, when differentiating \((-\ln x)^3\), the chain rule is applied by first differentiating \(u^3\) with \(u = -\ln x\) and then multiplying by the derivative of \(-\ln x\). This ensures each part is appropriately accounted for.

Power Rule

The power rule is an efficient method for differentiating polynomial functions. It's particularly useful for expressions of the form \(x^n\), where \(n\) is any real number. The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\).

• This rule is straightforward: you bring down the exponent as a coefficient and reduce the exponent by one.
• It works smoothly for any power, including negative and fractional exponents.

In the given exercise, the power rule is used when differentiating \(\frac{1}{2x^2}\).
By expressing \(\frac{1}{2x^2}\) as \(\frac{1}{2}x^{-2}\), we can apply the power rule directly to find the derivative: \(-x^{-3}\). This transformation simplifies the differentiation process and leads to an efficient solution.

Logarithmic Functions

Logarithmic functions are critical in calculus due to their unique properties relating to growth rates and their behavior inverse to exponential functions. The natural logarithm, denoted as \(\ln x\), is particularly significant due to its base \(e\).

• The derivative of \(\ln x\) is straightforward: \(\frac{1}{x}\).
• Logarithms transform multiplication into addition, which is handy in calculus.

In this exercise, logarithmic properties simplify the expression before differentiation.
The substitution \(\ln \frac{1}{x} = -\ln x\) allows the function to be rewritten more conveniently. This manipulation is essential as it prepares the function for applying rules like the chain and power rules, showing how logarithms tie algebraic steps to calculus.

Differentiation

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It is the fundamental tool for understanding instantaneous rates of change in calculus.

• The derivative provides insights into the function's behavior, such as its slope and whether it is increasing or decreasing.
• It consists of various techniques such as the chain, power, and product rules to tackle all kinds of functions.

In this problem, differentiation is used to find \(\frac{d r}{d x}\) by applying multiple rules: the power rule for simple polynomials, and the chain rule for more complex expressions.
By systematically applying these differentiation techniques, one can break down even complicated functions into simpler parts, allowing for precise and accurate derivation of the function's change rate.