Problem 68 - Thomas Calculus in SI Units

Question

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 \( \frac{dy}{dx} = \frac{1}{3} y \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right) \), where \( y = \sqrt[3]{\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}} \).

Step 1: Apply Logarithm

Start by taking the natural logarithm of both sides of the given equation to facilitate differentiation. This helps to simplify the expression where the product and quotient can be transformed into sums and differences: \[ \ln y = \ln \left( \sqrt[3]{\frac{x(x+1)(x-2)}{(x^{2}+1)(2x+3)}} \right) = \frac{1}{3} \ln \left( \frac{x(x+1)(x-2)}{(x^{2}+1)(2x+3)} \right) \] Using the properties of logarithms, we can separate the expression: \[ \ln y = \frac{1}{3} \left( \ln(x) + \ln(x+1) + \ln(x-2) - \ln(x^2+1) - \ln(2x+3) \right) \]

Step 2: Differentiate with respect to x

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

Step 3: Solve for dy/dx

Now solve for \( \frac{dy}{dx} \):\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right) \] Multiply both sides by \(y\) to isolate \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = y \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right) \]

Step 4: Substitute back for y

Substitute the expression for \( y \) back into the equation to express the derivative in terms of \(x\): \[ \frac{dy}{dx} = \left(\sqrt[3]{\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}} \right) \cdot \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-2} - \frac{2x}{x^2+1} - \frac{2}{2x+3} \right) \] This is the derivative of \( y \) with respect to \( x \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a method used when dealing with equations that are not easily solved for one variable in terms of another. In scenarios where a function is not explicitly defined as "\(y = f(x)\)", implicit differentiation allows us to find derivatives nonetheless. To apply this technique, we differentiate both sides of the equation with respect to the independent variable (like \(x\)), treating the dependent variable (like \(y\)) as a function of the independent variable. This often leads to deriving a function in terms of both variables. Then we can solve for the derivative of the function.

• Treat \(y\) as a function of \(x\), so \(y\) behaves as an "unknown function".
• Differentiate both sides with respect to \(x\), which means we apply the chain rule to \(y\).
• The differentiation for \(\ln(y)\) uses the chain rule giving us \(\frac{1}{y} \cdot \frac{dy}{dx}\).

Using implicit differentiation is particularly useful here because it simplifies managing fractions and powers in complex equations.

Chain Rule

The chain rule is a fundamental concept for finding the derivative of composite functions. It is especially useful when dealing with functions nested within functions – a common occurrence in logarithmic differentiation.When a function \(g(x)\) is wrapped inside another function \(f\), forming \(f(g(x))\), we use the chain rule to differentiate it: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]* For example, we see the chain rule in action when differentiating \(\ln(y)\). Here, \(y\) is a function of \(x\), so: * Differentiate \(\ln(y)\) using the outer derivative \(\frac{1}{y}\), * Then multiply by the derivative of the inner function \(\frac{dy}{dx}\).When applying the chain rule:

• Recognize the inner and outer functions involved.
• Differentiate the outer function first.
• Multiply by the derivative of the inner function.

In our case, it helps manage the complexity of multiple layered functions.

Derivative of a Function

The derivative of a function provides us with the rate at which the function changes with respect to the change in its input variable. Derivatives are crucial in calculus as they help in understanding how functions behave and change. With the given function, applying derivatives helps in pinpointing where these changes occur for a specified variable.

• The derivative of a power function follows the rule: \(\frac{d}{dx}[x^n] = nx^{n-1}\).
• For a natural logarithm function \(\ln(x)\), the derivative is \(\frac{1}{x}\).
• Using these rules together enables us to find derivatives even within complex expressions.

In our original exercise, breaking down the given function into its components allowed each part to be differentiated separately, utilizing the properties of derivatives to find the overall derivative.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a key element in logarithmic differentiation. Its primary advantage comes from its ability to convert products and quotients into sums and differences. This transformation simplifies the differentiation process, especially when dealing with complex expressions.Properties of natural logarithms that are widely used include:

• \(\ln(ab) = \ln(a) + \ln(b)\)
• \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• \(\ln(a^b) = b \cdot \ln(a)\)

By leveraging these properties, each component of the exercise was broken down into manageable parts. Hence, logarithmic differentiation can be performed more straightforwardly, saving the time and effort we’d otherwise spend managing the complex terms. In the solution, separating the function into multiple logs led us to an easier pathway for differentiation.