Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\) by logarithmic differentiation \(y=\frac{\left(x^{2}+3\right)^{2 / 3}(3 x+2)^{2}}{\sqrt{x+1}}\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{(x^2+3)^{2/3}(3x+2)^2}{\sqrt{x+1}} \left( \frac{2x}{3(x^2+3)} + \frac{6}{3x+2} - \frac{1}{2(x+1)} \right)\)

Step 1: Take the natural logarithm of both sides

To simplify the differentiation process, start by applying the natural logarithm to both sides of the equation. We have: \[ \ln(y) = \ln\left( \frac{(x^2 + 3)^{2/3}(3x+2)^2}{\sqrt{x+1}} \right) \] Use the properties of logarithms to expand the expression: \[ \ln(y) = \frac{2}{3}\ln(x^2 + 3) + 2\ln(3x+2) - \frac{1}{2}\ln(x+1) \]

Step 2: Differentiate both sides with respect to x

Differentiate the equation with respect to \(x\). On the left, use the chain rule: \[ \frac{d}{dx}\left(\ln(y)\right) = \frac{1}{y}\frac{dy}{dx}\] For the right-hand side, differentiate term by term to get: \[ \frac{2}{3} \cdot \frac{2x}{x^2+3} + 2 \cdot \frac{3}{3x+2} - \frac{1}{2} \cdot \frac{1}{x+1} \] Simplify the expression to find: \[ \frac{2x}{3(x^2+3)} + \frac{6}{3x+2} - \frac{1}{2(x+1)} \]

Step 3: Solve for \(\frac{dy}{dx}\)

To find \(\frac{dy}{dx}\), multiply both sides by \(y\): \[ \frac{dy}{dx} = y \left( \frac{2x}{3(x^2+3)} + \frac{6}{3x+2} - \frac{1}{2(x+1)} \right) \] Substitute back \(y\) from the original function: \[ y = \frac{(x^2+3)^{2/3}(3x+2)^2}{\sqrt{x+1}} \]Thus, \[ \frac{dy}{dx} = \frac{(x^2+3)^{2/3}(3x+2)^2}{\sqrt{x+1}} \left( \frac{2x}{3(x^2+3)} + \frac{6}{3x+2} - \frac{1}{2(x+1)} \right) \]

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function with respect to a variable. When we talk about finding the derivative of a function, we are essentially looking to understand how changes in one variable lead to changes in the function's output. In the context of logarithmic differentiation, we aim to simplify complex expressions, making them more manageable for differentiation. By taking the natural logarithm of both sides of an equation, we can transform products and quotients into sums and differences, which are easier to differentiate.

Applying the Product Rule

The product rule is an essential derivative rule used when differentiating the product of two functions. If you have two functions, say \( f(x) \) and \( g(x) \), their product is represented as \( f(x) \cdot g(x) \). The product rule states that the derivative of this product is:

• \( \frac{d}{dx}[f(x) \cdot g(x)] = f(x) \cdot g'(x) + g(x) \cdot f'(x) \)

This rule is particularly useful when dealing with multiple function products, allowing us to break the differentiation problem into more straightforward parts. In logarithmic differentiation, the use of logarithms transforms the product into a sum, which simplifies the use of the product rule.

Utilizing the Chain Rule

The chain rule is another crucial technique in calculus, especially when differentiating composite functions. A composite function is when one function is nested inside another, such as \( f(g(x)) \). The chain rule states:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

This allows us to "chain" the differentiation of the outer function with the derivative of the inner function. In the exercise, the chain rule is used when differentiating each part of the expanded logarithmic expression, ensuring that all components are correctly accounted for as functions of \( x \). Understanding how and when to apply the chain rule is vital for successfully navigating complex differentiation tasks.

Mastering Calculus Techniques

In calculus, various techniques such as the product rule, chain rule, and logarithmic differentiation are often used in conjunction to solve intricate problems efficiently. Logarithmic differentiation is particularly beneficial for functions that involve products, quotients, and powers, as it leverages the properties of logarithms to transform the differentiation problem.
Each technique has its strengths:

• The product rule helps break down and manage multiplications of functions.
• The chain rule is key for handling composite functions with ease.
• Logarithmic differentiation provides a strategy for managing functions with multiple layers or complexities.

Mastering these techniques equips you with a powerful toolkit for tackling a broad spectrum of calculus problems, turning complexity into clarity.