Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ \frac{d r}{d x} \text { if } r=\frac{\ln x}{x^{2} \ln x^{2}}+\left(\ln \frac{1}{x}\right)^{3} $$

Step-by-Step Solution

Answer

\( \frac{dr}{dx} = -\frac{1}{x^3} - \frac{3(\ln x)^2}{x} \)

Step 1: Simplify the Expression

First, simplify the expression for \( r \). The expression \( r = \frac{\ln x}{x^{2} \ln x^{2}} + \left(\ln \frac{1}{x}\right)^{3} \) can be rewritten by simplifying each part. Notice that \( \ln x^2 = 2 \ln x \). Hence, the first term is \( \frac{\ln x}{x^2 \cdot 2 \ln x} = \frac{1}{2x^2} \). The second term becomes \((\ln x^{-1})^3 = (-\ln x)^3 = - (\ln x)^3 \). So, \( r = \frac{1}{2x^2} - (\ln x)^3 \).

Step 2: Derive the Simplified Function

Now find the derivative \( \frac{dr}{dx} \) of the simplified expression \( r = \frac{1}{2x^2} - (\ln x)^3 \). Use the power and chain rules here. For \( \frac{1}{2x^2} \), rewrite it as \( \frac{1}{2} x^{-2} \). Its derivative is \( -\frac{2}{2}x^{-3} = -\frac{1}{x^3} \). For \( (\ln x)^3 \), use the chain rule: the derivative is \( 3(\ln x)^2 \cdot \frac{1}{x} \).

Step 3: Combine the Derivatives

Combine the derivatives of the two terms. The derivative of \( r \) is \( \frac{dr}{dx} = -\frac{1}{x^3} - \frac{3(\ln x)^2}{x} \). Simplify this to \[ \frac{dr}{dx} = -\frac{1}{x^3} - \frac{3(\ln x)^2}{x} \].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a clever technique used when finding derivatives of functions that involve logarithmic expressions which are difficult to differentiate using standard rules. This can include expressions where the variable appears both in the base and the exponent, or products, and quotients of functions that can be simplified using logarithms.

To apply logarithmic differentiation:

• Take the natural logarithm of both sides of the equation.
• Differentiation of both sides using the chain rule.
• Simplify and solve for the derivative.

In the given exercise, although the expressions are already simplified, you initially encounter logarithmic terms like \(\ln x\) and \(\left( \ln \frac{1}{x} \right)^3\). The logarithmic identities help in the simplification process, for instance, transforming \(\ln x^2 = 2 \ln x\), and \(\ln \frac{1}{x} = -\ln x\). These transformations make it easier to see how each component's derivative can be individually addressed, paving the way for straightforward application of derivative rules.

Chain Rule

The chain rule is a fundamental tool in calculus, essential when differentiating composite functions. It dictates that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

To understand the chain rule, consider a function \(f(g(x))\), where \(f\) is the outer function and \(g\) is the inner one:

• Differentiate the outer function while keeping the inner function unchanged.
• Multiply by the derivative of the inner function.

In our exercise, this rule finds its place particularly in differentiating \((\ln x)^3\). Here, we treated \((\ln x)^3\) as \(f(u)\) where \(u = \ln x\). The differentiation follows as: derive \(f(u) = u^3\), giving \(3u^2\) and multiplying by the derivative of \(u = \ln x\), which is \(\frac{1}{x}\). As a result, the complete derivative for that term is \(3(\ln x)^2 \cdot \frac{1}{x}\). This shows the seamless harmony of the chain rule's application.

Power Rule

The power rule is often one of the first derivative rules learned and it provides a straightforward method for differentiating functions of the form \(x^n\). The rule is quite simple: the derivative of \(x^n\) is \(nx^{n-1}\). This rule is immensely helpful in various calculus problems, including those which involve simplifying complex expressions.

In the provided exercise, the power rule is applied to terms that have expressions in the form of \(x\) raised to a power. For example:

• To derive \(\frac{1}{2x^2}\), first express it as \( \frac{1}{2} x^{-2} \). The derivative is calculated as \( -2 \cdot \frac{1}{2} x^{-3} = -\frac{1}{x^3} \).

This method makes it simple to handle complex fractions or negative exponent problems by allowing us to systematically determine their derivative. By grasping the power rule, students can efficiently handle the derivatives of polynomials and rational functions.