Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find dy/dx by logarithmic differentiation. $$ y=\frac{x+11}{\sqrt{x^{3}-4}} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1}{\sqrt{x^3 - 4}} - \frac{3x^2(x + 11)}{2(x^3 - 4)^{3/2}} \).

Step 1: Take Natural Logarithm on Both Sides

Given the function \( y = \frac{x+11}{\sqrt{x^3 - 4}} \), take the natural logarithm of both sides to simplify the expression: \( \ln y = \ln \left( \frac{x+11}{\sqrt{x^3 - 4}} \right) \).

Step 2: Simplify Using Logarithm Properties

Use the properties of logarithms to separate the terms: \( \ln y = \ln (x + 11) - \ln (\sqrt{x^3 - 4}) \). Recall that \( \ln(a/b) = \ln a - \ln b \) and \( \ln(\sqrt{a}) = \frac{1}{2} \ln a \).

Step 3: Differentiate Both Sides with Respect to x

Differentiate the equation \( \ln y = \ln (x + 11) - \frac{1}{2} \ln (x^3 - 4) \) with respect to \( x \). Use the chain rule for \( \ln y \): \( \frac{1}{y} \frac{dy}{dx} \), and for the right side: \( \frac{d}{dx}\ln(x + 11) = \frac{1}{x + 11} \) and \( \frac{d}{dx}\left(\frac{1}{2} \ln(x^3 - 4)\right) = \frac{1}{2} \cdot \frac{3x^2}{x^3 - 4} \).

Step 4: Solve for dy/dx

Equate and rearrange the derivatives to solve for \( \frac{dy}{dx} \): \( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x+11} - \frac{3x^2}{2(x^3 - 4)} \). Thus, \( \frac{dy}{dx} = y \left( \frac{1}{x+11} - \frac{3x^2}{2(x^3 - 4)} \right) \).

Step 5: Substitute y Back into the Expression

Since \( y = \frac{x+11}{\sqrt{x^3 - 4}} \), substitute back into the equation: \( \frac{dy}{dx} = \frac{x+11}{\sqrt{x^3 - 4}} \left( \frac{1}{x+11} - \frac{3x^2}{2(x^3 - 4)} \right) \).

Step 6: Simplify the Expression

Simplify the expression by canceling the terms where possible: \( \frac{dy}{dx} = \frac{1}{\sqrt{x^3 - 4}} - \frac{(x+11)\cdot 3x^2}{2(x^3 - 4)^{3/2}} \).

Key Concepts

Calculus Differentiation

Calculus differentiation is a key concept in mathematics, used to find the rate at which a quantity changes. It is essential for understanding the behavior of functions.
Differentiation involves calculating the derivative, represented as \( \frac{dy}{dx} \), which shows how the function \( y \) changes with respect to the variable \( x \).
In this context, we perform differentiation using logarithmic differentiation, which simplifies complex functions.

• Standard differentiation rules apply, like the power, product, and quotient rules.
• Logarithmic differentiation is especially useful for functions that are a product or quotient of exponentials and polynomials.

By using calculus differentiation, you can understand and predict the behavior of dynamic systems in natural sciences, engineering, and economics.

Chain Rule

The chain rule is a fundamental technique for finding derivatives of composite functions, where one function is nested inside another.
This rule allows us to differentiate functions by breaking them into simpler parts. For example, if you have a function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).
In logarithmic differentiation, we apply the chain rule when differentiating the natural logarithm of a function.

• This involves taking the derivative of the logarithm, and then multiplying it by the derivative of the internal function.
• In our problem, this helps us transition from \( \ln y \) to the actual calculation of \( \frac{dy}{dx} \).

Understanding the chain rule is crucial to solving complex differentiation problems, especially when functions are composed of multiple layers.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with base \( e \), where \( e \approx 2.718 \.\).
It is a powerful tool in differentiation and integration because of its unique properties.
When applying logarithmic differentiation, the natural logarithm helps by turning multiplication and division inside a function into addition and subtraction, thanks to properties like \( \ln(ab) = \ln a + \ln b \).

• It simplifies the differentiation process, especially for functions that are challenging to differentiate directly.
• Using \( \ln \) can transform the problem into a more manageable form, making subsequent steps of the solution easier.

In solving the original problem, we used \( \ln \) to effectively manage the complexities of the given function.

Derivative Simplification

After calculating the derivative, simplifying it is crucial for obtaining a clearer and more concise result.
Derivative simplification often involves canceling terms or reducing expressions to their simplest forms.
In the given exercise, once \( \frac{dy}{dx} \) is found, simplification helps in achieving the final clean form of the derivative.

• This step involves algebraic manipulation, like factoring and canceling out common terms.
• It ensures clarity and helps in analyzing and interpreting the derivative more efficiently.

Effective simplification not only provides insight into the underlying function but also aids in predicting how changes in \( x \) affect \( y \).