Problem 53 - University Calculus: Early Transcendentals

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt[3]{\frac{x(x-2)}{x^{2}+1}}$$

Step-by-Step Solution

Answer

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

Step 1: Express the Function with Exponent

Start by rewriting the given expression using exponential notation. Note that the cube root can be expressed as an exponent of \( \frac{1}{3} \). The function becomes:\[y = \left( \frac{x(x-2)}{x^{2}+1} \right)^{\frac{1}{3}}\]

Step 2: Take Natural Logarithm of Both Sides

Apply the natural logarithm to both sides of the equation to enable logarithmic differentiation:\[\ln y = \ln \left( \left( \frac{x(x-2)}{x^{2}+1} \right)^{\frac{1}{3}} \right)\]Simplify using the logarithm power rule \( \ln(a^b) = b \ln a \):\[\ln y = \frac{1}{3} \ln \left( \frac{x(x-2)}{x^{2}+1} \right)\]

Step 3: Use Logarithm Quotient Rule

Apply the logarithm quotient rule \( \ln \left( \frac{a}{b} \right) = \ln a - \ln b \):\[\ln y = \frac{1}{3} \left( \ln (x(x-2)) - \ln (x^2 + 1) \right)\]

Step 4: Expand Using Logarithm Product Rule

Expand \( \ln(x(x-2)) \) using the logarithm product rule \( \ln(ab) = \ln a + \ln b \):\[\ln y = \frac{1}{3} \left( \ln x + \ln (x-2) - \ln (x^2+1) \right)\]

Step 5: Differentiate Both Sides

Differentiate both sides with respect to \( x \). For the left side:\[\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}\]For the right side:\[\frac{1}{3} \left( \frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln(x-2)) - \frac{d}{dx}(\ln(x^2+1)) \right)\]Calculate each derivative:- \( \frac{d}{dx}(\ln x) = \frac{1}{x} \)- \( \frac{d}{dx}(\ln (x-2)) = \frac{1}{x-2} \)- \( \frac{d}{dx}(\ln (x^2+1)) = \frac{2x}{x^2+1} \)Thus the equation becomes:\[\frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{1}{x} + \frac{1}{x-2} - \frac{2x}{x^2+1} \right)\]

Step 6: Solve for the Derivative

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

Key Concepts

Natural Logarithm

In calculus, we often use logarithms to simplify complex differentiation problems. The natural logarithm, denoted as \( \ln \), is particularly useful because of its unique properties.
Natural logarithms are based on the constant \( e \), where \( e \approx 2.71828 \), which arises naturally in various mathematical contexts.

• The natural log of a number is the power to which \( e \) must be raised to obtain that number.
• Using \( \ln \) can transform products, quotients, and powers into simpler sums and differences, making differentiation easier.

In our use case, we applied the natural logarithm to both sides of the equation \( y = \left( \frac{x(x-2)}{x^2+1} \right)^{1/3} \) in order to facilitate differentiation by breaking down the components of this expression.

Product Rule

When dealing with derivatives of products of functions, the product rule is your go-to tool. The rule states that:
\[ (fg)' = f'g + fg' \]
where \( f \) and \( g \) are functions of \( x \).
This rule is particularly useful when you have an expression that consists of multiple factors, each being a function of \( x \).

• This was applied in our solution when breaking down the product within the logarithmic expression: \( \ln(x(x-2)) \).
• We used the product rule conceptually to expand \( \ln \) of a product into a sum, i.e., \( \ln(x) + \ln(x-2) \).

This simplifies the differentiation, as each term \( \ln(x) \) and \( \ln(x-2) \) can be independently derived.

Quotient Rule

The quotient rule is a powerful tool for differentiating expressions that consist of one function divided by another. The rule is given by:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
where \( u \) and \( v \) are functions of \( x \).
This rule was used in our solution to handle the differentiation of a function divided by another within a logarithmic expression.

• Specifically, we applied \( \ln \left( \frac{x(x-2)}{x^2+1} \right) = \ln(x(x-2)) - \ln(x^2+1) \).
• By breaking the expression into subtraction, differentiation became more straightforward since derivatives of \( \ln a \) and \( \ln b \) have well-known forms.

Remember, it's crucial to carefully apply the quotient rule to ensure all components are correctly differentiated.

Chain Rule

The chain rule is essential for differentiating composite functions—functions within functions. The rule is defined as follows:
If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
This enables tackling derivatives of complicated compositions by treating them layer by layer.

• In our example, we encountered composites in \( \ln(x-2) \) and \( \ln(x^2+1) \).
• Each required applying the chain rule to differentiate correctly: first determining the derivative of the inner function, and then the derivative of the outer one, multiplying them after.

Without the chain rule, such differentiations would be challenging since direct derivations would miss the complexities of internal functions. Therefore, mastering the chain rule equips you to handle derivatives of nested functions efficiently.