Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$$$ D_{x} \ln (x-4)^{3} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{3}{x-4} \).

Step 1: Rewrite the Function

To simplify taking the derivative of the function, first rewrite \( \ln(x-4)^3 \) as \( 3\ln(x-4) \). This is possible due to the logarithmic property \( \ln(a^b) = b\ln(a) \).

Step 2: Determine the Derivative Using Constant Multiple Rule

Apply the derivative rule for a constant multiple, which states \( \frac{d}{dx}[c\cdot f(x)] = c \cdot f'(x) \). Here, \( c = 3 \), so the derivative is \( 3 \cdot \frac{d}{dx}[\ln(x-4)] \).

Step 3: Apply the Derivative of the Logarithmic Function

The derivative of \( \ln(u) \), where \( u \) is a differentiable function, is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = x-4 \), so \( \frac{d}{dx}[\ln(x-4)] = \frac{1}{x-4} \).

Step 4: Combine the Results

Multiply the results to obtain the final derivative: \( 3 \cdot \frac{1}{x-4} = \frac{3}{x-4} \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used for finding the derivatives of products or quotients of functions. It proves especially helpful in cases where normal differentiation rules are tricky to apply. In the exercise, the original function is \( \ln(x-4)^3 \). Simplifying this involves a key property of logarithms: \( \ln(a^b) = b \ln(a) \). By applying this property, you can rewrite \( \ln(x-4)^3 \) as \( 3 \ln(x-4) \).
Logarithmic differentiation is particularly useful when:

• You have functions raised to a power, as in this exercise.
• The relationship involves products or quotients, where breaking down into simpler parts aids in differentiation.

It essentially transforms complex multiplication or exponent problems into simpler addition or linear multiplication problems, making differentiation straightforward.

Chain Rule in Calculus

The chain rule is a fundamental differentiation rule in calculus. It helps when differentiating composite functions, i.e., functions nested inside one another. The rule states that the derivative of a composite function \( y = f(g(x)) \) is given by \( f'(g(x)) \cdot g'(x) \). In this case, consider the function \( f(u) = \ln u \) and the inner function \( u = x-4 \).
Applying the chain rule means identifying:

• An outer function — here it's \( \ln(u) \).
• An inner differentiable function — here it's \( x-4 \).

By substituting and applying the chain rule, you find that the derivative of \( \ln(x-4) \) is \( \frac{1}{x-4} \). Remember, the derivative of \( \ln(u) \) involves multiplying by the derivative of \( u \), ensuring the rule's application is correct.

Derivative Rules

Derivative rules are the foundational tools in calculus for finding rates of change or slopes of curves. Key rules include the constant rule, power rule, product rule, quotient rule, and, notably, the logarithmic and chain rules you've seen applied here.
When differentiating \( 3 \ln(x-4) \), the constant multiple rule is applied, which says the derivative of \( c \cdot f(x) \) is \( c \cdot f'(x) \). The constant multiple in this case is 3, so you multiply the derivative of the logarithmic part by 3.
The final derivative of the function \( (x-4)^3 \) represented in logarithmic form is \( \frac{3}{x-4} \). This outcome was achieved through understanding:

• The combination and application of various derivative rules.
• Breaking down complex expressions into more manageable components.

This makes it easier to apply the derivative rules efficiently.