Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\) by logarithmic differentiation \(y=\frac{\sqrt{x+13}}{(x-4) \sqrt[3]{2 x+1}}\)

Step-by-Step Solution

Answer

Differentiate the function using logarithmic differentiation and simplify the result.

Step 1: Take ln of both sides

\(\ln y = \frac{1}{2}\ln(x+13) - \ln(x-4) - \frac{1}{3}\ln(2x+1)\)

Step 2: Differentiate

\(\frac{y'}{y} = \frac{1}{2(x+13)} - \frac{1}{x-4} - \frac{2}{3(2x+1)}\)

\(y' = y\left[\frac{1}{2(x+13)} - \frac{1}{x-4} - \frac{2}{3(2x+1)}\right]\)

Key Concepts

Natural Logarithm

The natural logarithm is a crucial tool in calculus, especially when dealing with complex functions like the one in our problem. The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.718. It converts multiplicative relationships into additive ones, simplifying processes like differentiation.

• When applying the natural logarithm to both sides of an equation, it can make otherwise complicated fractions and products more manageable.
• This simplification is invaluable in finding derivatives of products, quotients, or powers of functions.

In the provided problem, applying the natural logarithm helps decompose the initial complex fraction into a more straightforward sum of logarithmic terms.

Derivative of a Function

The derivative of a function represents the rate at which the function's value changes at any given point. In the context of logarithmic differentiation, the primary goal is to uncover the derivative of complex functions that contain products, quotients, or powers.

• Using logarithmic differentiation, we can derive such functions by applying differentiation to the logarithmic form rather than directly on the original function.
• This technique is particularly useful for functions where variables are in the base or the exponent of a term.

For the function \( y = \frac{\sqrt{x+13}}{(x-4)\sqrt[3]{2x+1}} \), logarithmic differentiation makes it more straightforward to find \( \frac{dy}{dx} \) by converting multiplicative elements into additive ones.

Logarithmic Properties

Understanding logarithmic properties is fundamental when applying logarithmic differentiation. These properties allow us to transform and simplify expressions efficiently. Some key properties that were used in the solution are:

• \( \ln(ab) = \ln a + \ln b \): This property helps to split the logarithm of a product into the sum of the logarithms.
• \( \ln(\frac{a}{b}) = \ln a - \ln b \): In the original solution, this property is used to separate the logarithm of a quotient into a difference.
• \( \ln(a^b) = b \cdot \ln a \): Allows us to bring powers down, which simplifies differentiation.

These properties enable the simplification of complex expressions, making taking derivatives through logarithmic differentiation manageable.

Calculus Problem Solving

In calculus, problem-solving often requires applying various strategies and understanding multiple concepts simultaneously. Logarithmic differentiation is one such strategy that combines skills in algebra, calculus, and logarithms to tackle otherwise difficult differentiation problems.

• It provides a methodical approach, turning multiplicative and complex power-based expressions into simpler additive forms.
• Success in calculus problem solving comes from the ability to recognize which technique or approach will simplify the task, such as leveraging logarithmic properties or natural logarithms in differentiation.

In practice, using logarithmic differentiation not only simplifies the differentiation process but also in many cases makes it the most viable approach for obtaining the result in a clear and structured manner.