Problem 77 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \begin{equation}y=\log _{3}\left(\left(\frac{x+1}{x-1}\right)^{\ln 3}\right)\end{equation}

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{-2}{(x+1)(x-1)} \).

Step 1: Simplify the Expression

Start by simplifying the given function. Since the logarithm is involved, recognize that \( y = \log_{3}\left(\left(\frac{x+1}{x-1}\right)^{\ln 3}\right) \) can be rewritten. Utilize the change of base formula: \( \log_{a}(b^c) = c \log_{a}(b) \). Therefore, \( y = \ln 3 \cdot \log_{3}\left(\frac{x+1}{x-1}\right) \).

Step 2: Apply Change of Base Formula

Use the change of base formula to further simplify: \( \log_{3}\left(\frac{x+1}{x-1}\right) = \frac{\ln\left(\frac{x+1}{x-1}\right)}{\ln 3} \). Now substitute back into \( y \): \( y = \ln 3 \cdot \frac{\ln\left(\frac{x+1}{x-1}\right)}{\ln 3} = \ln\left(\frac{x+1}{x-1}\right) \).

Step 3: Differentiate the Simplified Expression

Now that the function \( y = \ln\left(\frac{x+1}{x-1}\right) \) is simplified, differentiate it. Use the derivative of the natural logarithm: \( \frac{d}{dx}\ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \frac{x+1}{x-1} \).

Step 4: Find Derivative of Inner Function

Find the derivative \( \frac{du}{dx} \) of \( u = \frac{x+1}{x-1} \). Use the quotient rule: If \( u = \frac{f(x)}{g(x)} \), then \( \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \). Here, \( f(x) = x+1 \) and \( g(x) = x-1 \). Calculate \( f'(x) = 1 \), \( g'(x) = 1 \). Therefore, \( \frac{du}{dx} = \frac{1(x-1) - (x+1)1}{(x-1)^2} = \frac{-2}{(x-1)^2} \).

Step 5: Combine Results

Combine the results from the previous steps to find \( \frac{dy}{dx} \). Substitute back \( u = \frac{x+1}{x-1} \) and \( \frac{du}{dx} = \frac{-2}{(x-1)^2} \) into the derivative of the logarithmic function: \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{\frac{x+1}{x-1}} \cdot \frac{-2}{(x-1)^2} = \left(\frac{x-1}{x+1}\right) \cdot \frac{-2}{(x-1)^2} = \frac{-2}{(x+1)(x-1)} \).

Step 6: Simplify Final Derivative

The simplified derivative is: \( \frac{dy}{dx} = \frac{-2}{(x+1)(x-1)} \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a technique used to find the derivative of a function that has a form making direct differentiation complicated. This approach is particularly handy for functions involving products, quotients, or powers. By taking the natural logarithm of both sides of such an equation, you can often simplify the differentiation process significantly.
In essence, logarithmic differentiation allows you to convert higher computational operations, like multiplication and exponentiation, into simpler ones, like addition and multiplication. This is why the logarithmic properties, such as the power property (\( \log_b(a^c) = c \cdot \log_b(a) \)), are very useful. This property simplifies the process here by letting you bring down exponents, ultimately preparing the equation for further steps using basic derivative rules.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions. A composite function is essentially a function within another function, like \( y = \ln\left(\frac{x+1}{x-1}\right) \).
The chain rule states that the derivative of a composite function \( f(g(x)) \) is the derivative of the outer function \( f'(g(x)) \) times the derivative of the inner function \( g'(x) \). In our example, the outer function is the natural logarithm, and the inner function is the fraction \( \frac{x+1}{x-1} \).

• First, find the derivative of the outer function, applied to the inner function.
• Secondly, multiply it by the derivative of the inner function.

This step ensures that any complexities within the function are properly accounted for, leading to an accurate calculation of the derivative.

Quotient Rule

Quotient rule is applied when differentiating functions that are divisions of two other functions. For a function \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule is expressed as:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
This means the derivative of the quotient is the difference between the product of the derivative of the numerator and the denominator and the product of the numerator and the derivative of the denominator, all over the square of the denominator.
In our context, because \( u = x + 1 \) and \( v = x - 1 \), applying the quotient rule simplifies the expression, allowing for straightforward differentiation. This process ultimately contributes to simplifying the resulting derivative needed to apply the chain rule effectively.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with base \( e \), where \( e \) is approximately 2.71828. It is commonly used in mathematics due to its simple derivation properties. The derivative of \( \ln(x) \) is straightforward and is given by:
\[ \frac{d}{dx}\ln(x) = \frac{1}{x} \]
This derivative is particularly convenient when dealing with functions under a logarithmic form, as it breaks down more complex components into simpler linear components. In the given problem, the natural logarithm allows us to peal back the layers of complex expressions in the function, making the differentiation more manageable.
The simplification through this step reduces complicated expressions into forms that are readily differentiable by other rules like the chain and quotient rule, thus easing the entire differentiation process.