Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y / d x\) 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 the natural logarithm of both sides

We start by taking the natural logarithm of both sides of our equation to facilitate differentiation. This gives us: \[ \ln(y) = \ln\left(\frac{x+11}{\sqrt{x^3-4}}\right) \]

Step 2: Simplify the logarithmic expression

Using the properties of logarithms, we rewrite the right-hand side: \[ \ln(y) = \ln(x + 11) - \ln(x^3 - 4)^{1/2} \] The expression simplifies to: \[ \ln(y) = \ln(x + 11) - \frac{1}{2} \ln(x^3 - 4) \]

Step 3: Differentiate both sides

Differentiate both sides of the equation with respect to \(x\). The left-hand side gives \[ \frac{1}{y} \frac{dy}{dx} \] And the right-hand side becomes: \[ \frac{1}{x+11} - \frac{1}{2} \cdot \frac{3x^2}{x^3-4} \] This results in: \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x+11} - \frac{3x^2}{2(x^3-4)} \]

Step 4: Solve for \(\frac{dy}{dx}\)

Multiply through by \(y\) to solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = y \left( \frac{1}{x+11} - \frac{3x^2}{2(x^3-4)} \right) \]

Step 5: Substitute back the expression for \(y\)

Remember that \(y = \frac{x+11}{\sqrt{x^3-4}}\). Substitute this back into our expression: \[ \frac{dy}{dx} = \frac{x+11}{\sqrt{x^3-4}} \left( \frac{1}{x+11} - \frac{3x^2}{2(x^3-4)} \right) \] This simplifies to: \[ \frac{dy}{dx} = \frac{1}{\sqrt{x^3-4}} - \frac{3x^2(x+11)}{2(x^3-4)^{3/2}} \]

Key Concepts

Natural Logarithm Properties

Natural logarithms have unique properties that can simplify differentiation. One important property is that the natural log of a quotient is the difference of the logs: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). This property is helpful when dealing with complex expressions.
Another critical property is the power rule for logarithms: if you have \( \ln(x^b) \), it can be rewritten as \( b \ln(x) \). This makes it easier to differentiate functions expressed in exponential terms.
In the given exercise, these properties simplify the function before differentiation. This allows us to focus directly on finding the derivative efficiently and accurately. Make sure to always simplify expressions as much as possible before differentiating; it can save time and reduce complexity.

Differentiation Techniques

Differentiation techniques are essential tools for solving calculus problems. In this exercise, we utilize logarithmic differentiation—a technique that involves taking the logarithm of both sides of an equation to simplify the differentiation process. This is especially useful for products, quotients, or complex exponentials.
Consider the function: \( y = \frac{x+11}{\sqrt{x^{3}-4}} \). Taking the natural logarithm first and using logarithmic differentiation allows to convert multiplicative processes into additive ones. This greatly eases the differentiation process.
Here are key differentiation components applied in the solution:

• Differentiate \( \ln(y) \) as \( \frac{1}{y} \frac{dy}{dx} \), multiplying by \( y \) later to isolate \( \frac{dy}{dx} \).
• Apply quotient and chain rules on the right side for terms like \( \ln(x+11) \) and \( \ln(x^3 - 4) \).

By mastering these techniques, you'll be equipped to handle a wide range of differentiation problems.

Implicit Differentiation

Implicit differentiation is a technique used when it is difficult or impossible to explicitly solve an equation for one variable in terms of another. Instead of isolating \( y \), as in explicit differentiation, we differentiate each side of the equation with respect to \( x \), treating \( y \) as a dependent variable.
When you encounter complex expressions, such as those involving products or quotients of functions, implicit differentiation can simplify the process. In our exercise, after taking the logarithm of both sides, we use implicit differentiation by setting \( y \) as a function of \( x \) implicitly within the log expression.
This requires applying the chain rule: differentiate \( \ln(y) \) using \( \frac{d}{dx} \ln(y) = \frac{1}{y} \frac{dy}{dx} \), while respecting all other differentiation rules applicable to \( x \) and any nested functions.
Such a strategic use of implicit differentiation integrates seamlessly with our logarithmic approach in this problem to effectively arrive at the derivative \( \frac{dy}{dx} \).