Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\). $$ y=\frac{x+11}{\sqrt[3]{x^{2}-4}} $$

Short answer

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

Step-by-step solution

Take Natural Logarithms on Both Sides

To use logarithmic differentiation, first take the natural logarithm of both sides of the equation. This gives: \[\ln y = \ln \left( \frac{x + 11}{(x^2 - 4)^{1/3}} \right) \]Use the properties of logarithms to simplify. This results in: \[\ln y = \ln(x + 11) - \frac{1}{3} \ln(x^2 - 4)\]

Differentiate Both Sides

Differentiate each side of the equation with respect to \(x\). Using implicit differentiation for the left side, we have: \[\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[\ln(x+11)] - \frac{1}{3} \frac{d}{dx}[\ln(x^2-4)]\]The differential of the right side gives: \[\frac{1}{x + 11} - \frac{1}{3} \left( \frac{2x}{x^2 - 4} \right)\]

Solve for \( \frac{dy}{dx} \)

Reorganizing the equation to solve for \( \frac{dy}{dx} \), we have: \[\frac{1}{y} \frac{dy}{dx} = \frac{1}{x+11} - \frac{2x}{3(x^2 - 4)}\]Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):\[\frac{dy}{dx} = y \left( \frac{1}{x+11} - \frac{2x}{3(x^2-4)} \right)\]

Substitute Original Function for \( y \)

Recall that \( y = \frac{x+11}{(x^2-4)^{1/3}} \). Substitute this back into the equation to find \( \frac{dy}{dx} \):\[\frac{dy}{dx} = \frac{x+11}{(x^2-4)^{1/3}} \left( \frac{1}{x+11} - \frac{2x}{3(x^2-4)} \right)\]Simplify this expression:

Simplify the Expression

Simplify the expression:\[\frac{dy}{dx} = \frac{1}{(x^2-4)^{1/3}} - \frac{2x(x+11)}{3(x^2-4)^{4/3}}\]

Key concepts

Calculus Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any point. It forms the basis for understanding how functions behave and is crucial in various fields, such as physics, engineering, and economics.
Differentiation fundamentally involves calculating the derivative of a function. The derivative provides information about the rate of change of the dependent variable with respect to the independent variable.

• The derivative of a function is denoted as \( \frac{dy}{dx} \) when \( y \) is a function of \( x \).
• The process involves applying various differentiation rules, such as the power rule, product rule, and chain rule.
• Understanding differentiation allows for solving problems related to slopes of tangents, velocities, and optimization.

By understanding these basic principles of differentiation, we can delve deeper into specialized techniques such as logarithmic and implicit differentiation.

Implicit Differentiation

Implicit differentiation comes into play when dealing with equations that aren't easily expressed as a function of one variable in terms of another. In such scenarios, the function is "implicit," meaning not explicitly solved for a particular variable. This technique is useful for differentiating more complex equations.
Consider an equation involving two variables, such as \( y = \frac{x+11}{(x^2-4)^{1/3}} \). This equation could be difficult to solve directly for \( y \) in terms of \( x \) before differentiating.

• Implicit differentiation allows us to differentiate both sides of an equation with respect to \( x \) directly.
• This involves taking the derivative of each term in the equation separately, using the chain rule wherever necessary.
• The result often involves both \( dy/dx \) and \( y \), and solving for \( dy/dx \) could require some algebraic manipulation.

In the logarithmic differentiation technique, implicit differentiation simplifies the process and allows solving derivative problems that would otherwise be complex.

Natural Logarithms

Natural logarithms are logarithms with the base \( e \), where \( e \) is approximately 2.718. They are used extensively in calculus for simplifying complex differentiation problems, especially those involving products or quotients of functions.
By taking the natural logarithm of both sides of an equation, we can leverage their properties to transform products into sums and quotients into differences, drastically simplifying differentiation.

• The natural logarithm of a product \( \ln(ab) \) translates to \( \ln(a) + \ln(b) \).
• The natural logarithm of a quotient \( \ln(\frac{a}{b}) \) becomes \( \ln(a) - \ln(b) \).
• This transformation facilitates breaking down complex expressions into manageable parts, making differentiation straightforward.

Using these properties, logarithmic differentiation becomes a powerful tool for tackling otherwise challenging or cumbersome derivative calculations.