Question

, find dy/dx by logarithmic differentiation. $$ y=\frac{\sqrt{x+13}}{(x-4) \sqrt[3]{2 x+1}} $$

Short answer

\(\frac{dy}{dx} = \frac{\sqrt{x+13}}{(x-4) \sqrt[3]{2x+1}} \left[ \frac{1}{2(x+13)} - \frac{1}{x-4} - \frac{2}{3(2x+1)} \right]\)

Step-by-step solution

Take the Natural Logarithm of Both Sides

Start by taking the natural logarithm on both sides of the equation to simplify the differentiation process. So we have:\[\ln(y) = \ln\left(\frac{\sqrt{x+13}}{(x-4) \sqrt[3]{2x+1}}\right)\]Using logarithmic properties, split the expression:\[\ln(y) = \ln(\sqrt{x+13}) - \ln(x-4) - \ln(\sqrt[3]{2x+1})\]

Simplify the Logarithmic Expression

Use the logarithm power property, \(\ln(a^b) = b\ln(a)\), to simplify each term:\[\ln(y) = \frac{1}{2}\ln(x+13) - \ln(x-4) - \frac{1}{3}\ln(2x+1)\]

Differentiate Both Sides with Respect to x

Differentiate both sides of the logarithmic equation with respect to \(x\). Remember that the derivative of \(\ln(u)\) is \(\frac{1}{u} \frac{du}{dx}\):\[\frac{1}{y} \frac{dy}{dx} = \frac{1}{2(x+13)} - \frac{1}{x-4} - \frac{1}{3(2x+1)} \cdot 2\]Simplify the expression:\[\frac{dy}{dx} = y \left[ \frac{1}{2(x+13)} - \frac{1}{x-4} - \frac{2}{3(2x+1)} \right]\]

Substitute Back for y

Replace \(y\) with the original expression to finish the derivative calculation:\[\frac{dy}{dx} = \frac{\sqrt{x+13}}{(x-4) \sqrt[3]{2x+1}} \left[ \frac{1}{2(x+13)} - \frac{1}{x-4} - \frac{2}{3(2x+1)} \right]\]

Key concepts

Derivative Calculation

When dealing with functions that are complex and involve products or quotients, calculating their derivatives can be challenging. Logarithmic differentiation is a powerful tool that simplifies this process. Instead of directly deriving the function, we first take the natural logarithm of both sides. This allows us to break down products and quotients into sums and differences, making the derivative easier to handle.

For the given function, \[y=\frac{\sqrt{x+13}}{(x-4) \sqrt[3]{2x+1}},\]we begin by introducing the natural logarithm to both sides. This step converts the function into a form that is far simpler:
\[\ln(y) = \ln\left(\frac{\sqrt{x+13}}{(x-4) \sqrt[3]{2x+1}}\right).\]

This transformation sets the stage for differentiation by exploiting logarithmic properties below, simplifying our calculation significantly.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), acts as a bridge between complex expressions and their manageable components. By applying \(\ln\) to a function, we can utilize its inherent properties to simplify terms into additive and subtractive forms. This is incredibly useful in differentiation.

For example, the natural logarithm helped break down the initial complex expression into simpler fragments:
\[\ln(y) = \ln(\sqrt{x+13}) - \ln(x-4) - \ln(\sqrt[3]{2x+1}).\]

Utilizing the natural logarithm's attribute of converting products into sums and quotients into differences, it turns what might seem like an impossible derivative task into a straightforward application of algebra and calculus.
Its powerful relationship to exponentials allows us to wield logarithms in various mathematical manipulations, amplifying our ability to tackle sophisticated equations with ease.

Logarithmic Properties

Logarithmic properties are key to manipulating logarithmic expressions efficiently. Three main properties are particularly useful:

• **Product Property**: \(\ln(ab) = \ln(a) + \ln(b)\)
• **Quotient Property**: \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• **Power Property**: \(\ln(a^b) = b\ln(a)\)

In our example, the quotient property allowed us to split the initial function into a series of subtractions, enabling us to focus on each part individually. The power property further simplified expressions like \(\ln(\sqrt{x+13})\) by transforming it to \(\frac{1}{2}\ln(x+13)\).

These properties empower us to reshape complicated expressions into more workable components. They play a pivotal role in making logarithmic differentiation a practical and favored technique in calculus.

Chain Rule

The chain rule is essential for differentiating composite functions within our exercise. It provides a systematic approach to finding derivatives of nested functions. When differentiating \(\ln(y)\), you recognize that \(y\) itself is a function of \(x\). The chain rule expresses this as:
\[\frac{1}{y} \frac{dy}{dx}\]

The practice of applying the chain rule becomes evident when differentiating terms such as \(\ln(\sqrt{x+13})\). Here, the external function is the natural logarithm, and the internal function is \(\sqrt{x+13}\). By the chain rule, we compute the derivative of the external function, multiply it by the derivative of the internal one, and adhere these products across the whole equation.

This rule acts as a guiding tool in unraveling layers of a function to reveal derivatives with precision. Understanding how to apply the rule effectively is a critical skill in mastering calculus.