Problem 65 - Thomas Calculus

Question

In Exercises \(55-68,\) use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$ y=\frac{x \sqrt{x^{2}+1}}{(x+1)^{2 / 3}} $$

Step-by-Step Solution

Answer

The derivative \(\frac{dy}{dx}\) is \( \frac{x \sqrt{x^2 + 1}}{(x+1)^{2/3}} \left( \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3(x+1)} \right) \).

Step 1: Take the Natural Logarithm of Both Sides

Begin by taking the natural logarithm of both sides of the equation to simplify the process. This gives us: \[ \ln y = \ln \left( \frac{x \sqrt{x^2 + 1}}{(x+1)^{2/3}} \right). \] Using the property of logarithms, simplify this expression as:\[ \ln y = \ln x + \ln \sqrt{x^2 + 1} - \ln (x+1)^{2/3}. \]

Step 2: Simplify Using Logarithmic Identities

In this step, further simplify the logarithmic expression using identities. Recall that \(\ln (a^b) = b \ln a\) and \( \ln \sqrt{a} = \frac{1}{2} \ln a \):\[ \ln y = \ln x + \frac{1}{2} \ln (x^2 + 1) - \frac{2}{3} \ln (x+1). \]

Step 3: Differentiate Both Sides with Respect to x

Now, take the derivative of both sides with respect to \(x\). For the left side, recall that the derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y} \frac{dy}{dx}\):\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\ln x + \frac{1}{2} \ln (x^2 + 1) - \frac{2}{3} \ln (x+1)). \]

Step 4: Apply Derivative Rules to the Right Side

Using the derivatives of basic functions (\(\ln x\), constant multiples, and chain rule), find derivatives:- \(\frac{d}{dx} \ln x = \frac{1}{x}\)- \(\frac{d}{dx} \frac{1}{2} \ln (x^2 + 1) = \frac{1}{2} \cdot \frac{2x}{x^2 + 1} = \frac{x}{x^2 + 1}\) by the chain rule,- \(\frac{d}{dx} -\frac{2}{3} \ln (x+1) = -\frac{2}{3} \cdot \frac{1}{x+1} = -\frac{2}{3(x+1)}.\)So, the whole expression becomes:\[ \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3(x+1)}. \]

Step 5: Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y\) to solve for the derivative \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = y \left( \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3(x+1)} \right). \]Substitute back \(y = \frac{x \sqrt{x^2 + 1}}{(x+1)^{2/3}}\):\[ \frac{dy}{dx} = \frac{x \sqrt{x^2 + 1}}{(x+1)^{2/3}} \left( \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3(x+1)} \right). \]

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a powerful mathematical function particularly useful in simplifying complex expressions through differentiation or integration. It is defined as the logarithm to the base \( e \), where \( e \approx 2.71828 \) is an irrational constant known as Euler's number.

• A key property of natural logarithms is their ability to transform multiplication, division, and exponentiation into addition, subtraction, and multiplication, respectively. This comes from their basic properties: \( \ln(ab) = \ln(a) + \ln(b) \), \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \), and \( \ln(a^b) = b \cdot \ln(a) \).
• In our exercise, we first applied \( \ln \) to both sides of the equation, simplifying the expression into a more workable form.

Breaking down the original expression allowed us to separate the terms involving \( x \) and ensure each part could be differentiated more effortlessly later on. This simplification is essential in logarithmic differentiation as it leverages the rules of logarithms, which are easier to differentiate.

Derivatives of Logarithmic Functions

When dealing with derivatives, logarithmic functions offer specific guidelines. The derivative of the natural logarithm function \( \ln(x) \) is \( \frac{1}{x} \). This rule lays the groundwork for tackling more complex logarithmic expressions.

• For expressions like \( \ln(u) \), where \( u \) is a function of \( x \), the derivative is \( \frac{1}{u} \cdot \frac{du}{dx} \), in alignment with the chain rule.
• Additionally, if you have an expression composed of various terms, you can break it down using the sum and difference rules for derivatives.

In the exercise provided, after simplifying the equation, each term like \( \ln(x) \), \( \ln(x^2 + 1) \), and \( \ln(x + 1) \) undergoes differentiation. For example, leveraging the identity \( \ln(a^b) = b\ln(a) \), \( \frac{1}{2} \ln(x^2 + 1) \) is derived to \( \frac{x}{x^2 + 1} \) using basic differentiation combined with the chain rule.
Differentiating these logarithmic terms systematically helps achieve our final goal, which is finding \( \frac{dy}{dx} \) for the given expression. Each step effectively combines the fundamental differentiation rules to tackle an initially complex function.

Chain Rule

The chain rule is a pivotal concept in calculus, particularly when differentiating composite functions. It provides a method to differentiate a function that is nested within another function.

• The chain rule states that if \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• This rule is essential when differentiating expressions where a function is composed within another, helping to break down the derivative into manageable segments.

In the solution, the chain rule is applied to terms like \( \ln(x^2 + 1) \) and \( \ln(x+1) \). For instance, in finding the derivative of \( \frac{1}{2} \ln(x^2 + 1) \), the exterior function \( \ln(u) \) has \( u = x^2 + 1 \). By the chain rule, it results in \( \frac{1}{2} \cdot \frac{2x}{x^2 + 1} \).
The chain rule thus ensures that each part of a more intricate function gets its due treatment in the differentiation process. When combined with the derivatives of basic logarithmic functions, it provides a streamlined path to solving derivatives for complex equations like the original one in our exercise. By applying the chain rule, differentiation becomes systematic and more efficient, allowing you to handle more sophisticated mathematical models.