Problem 68 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative is obtained by implicit differentiation and substitution.

Step 1: Express as a Logarithm

Given the function \(y = \sqrt[3]{\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}}\), we first express it using logarithms. Rewrite the expression as an exponential form: \(y = \left(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\right)^{\frac{1}{3}}\). Taking the natural logarithm on both sides, we have: \(\ln y = \frac{1}{3} \ln \left(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\right)\).

Step 2: Apply Logarithm Properties

Use the logarithm rules to simplify the expression: \(\ln y = \frac{1}{3} [\ln(x \cdot (x+1) \cdot (x-2)) - \ln((x^2+1)(2x+3))]\). Apply the fact that \(\ln(ab) = \ln a + \ln b\) and \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\). This gives us: \(\ln y = \frac{1}{3} [(\ln x + \ln(x+1) + \ln(x-2)) - (\ln(x^2+1) + \ln(2x+3))]\).

Step 3: Differentiate Implicitly with Respect to x

Differentiate both sides of the equation with respect to \(x\). On the left side, use the derivative of \(\ln y\): \(\frac{d}{dx}[\ln y] = \frac{1}{y} \cdot \frac{dy}{dx}\). On the right, differentiate each term individually: \(\frac{d}{dx}(\ln y) = \) \(\frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right)\).

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

Equating the derivatives obtained, we have: \(\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right)\). Solving for \(\frac{dy}{dx}\), we get: \(\frac{dy}{dx} = y \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right)\).

Step 5: Substitute Back for y

Replace \(y\) back with its original expression: \(y = \sqrt[3]{\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}}\). Substituting this in gives \(\frac{dy}{dx} = \sqrt[3]{\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}} \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right)\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. In this type of differentiation, we assume that both variables are related in some way and differentiate both sides of the given equation with respect to the independent variable. In our problem, we start by expressing the given function in logarithmic form, which has both sides of the equation in terms of functions involving x. This prepares the equation for differentiation.

• Let's consider the function: \( \ln y = \frac{1}{3} \left[\ln(x(x+1)(x-2)) - \ln((x^2+1)(2x+3))\right] \).
• We apply the derivative to both sides: \( \frac{d}{dx}[\ln y] = \frac{1}{y} \cdot \frac{dy}{dx} \).

Implicit differentiation allows us to find \( \frac{dy}{dx} \) in terms of y and x by differentiating as if y were a function of x, solving for the derivative.

Derivative Calculation

Derivative calculation is the process of finding the rate at which a function is changing at any given point. In logarithmic differentiation, this becomes crucial after the logarithms are applied, as they simplify the product quotient forms in the original function. By differentiating each term separately, it becomes simpler compared to directly differentiating a complex fraction raised to a power.

• The left side when differentiated is \( \frac{d}{dx}[\ln y] = \frac{1}{y} \cdot \frac{dy}{dx} \).
• The right-hand terms are differentiated individually. For example, \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).

The final step involves solving for \( \frac{dy}{dx} \). This gives us the overall rate of change of the original function in terms of y and x.

Logarithmic Properties

Understanding logarithmic properties is essential when it comes to simplifying the expressions in logarithmic differentiation. These properties help in transforming complex expressions into manageable terms through basic rules such as \( \ln(ab) = \ln a + \ln b \) and \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \).For the given function:

• The product \(x(x+1)(x-2)\) is expressed as \(\ln x + \ln(x+1) + \ln(x-2)\).
• Similarly, \((x^2+1)(2x+3)\) is split into \( \ln(x^2+1) + \ln(2x+3) \).

These properties simplify the differentiation process by breaking down complex multiplicative terms into additive ones, making the calculus operations more straightforward.

Chain Rule

The chain rule is a fundamental concept in calculus for finding derivatives of composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.In this problem, the chain rule is applied implicitly when differentiating \( \ln y = \frac{1}{3} [\ln(x(x+1)(x-2)) - \ln((x^2+1)(2x+3))] \). Here,

• Each term such as \( \ln(x+1) \) uses the chain rule since it involves a composite function structure. Its derivative is \( \frac{1}{x+1} \).
• When tackling \( \ln(x^2+1) \), the derivative requires the derivative of the inside \(2x\), following the chain rule.

By applying the chain rule, we ensure that the differentiation process is correctly executed for complex nested functions, empowering us to solve for all necessary derivatives.