Problem 29 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative is \(-\frac{2}{x^2-1}\).

Step 1: Simplify the Expression

First, recognize that the expression inside the logarithm can be simplified using the power rule of logarithms. The given function is \[ y = \log_3 \left( \left( \frac{x+1}{x-1} \right)^{\ln 3} \right) \]. Using the rule \( \log_b(a^c) = c \cdot \log_b(a) \), rewrite the expression as:\[ y = \ln 3 \cdot \log_3 \left( \frac{x+1}{x-1} \right) \].

Step 2: Change of Base Formula for Logarithms

Now, use the change of base formula to convert the base-3 logarithm to natural logarithms:\[ \log_3 \left( \frac{x+1}{x-1} \right) = \frac{\ln \left( \frac{x+1}{x-1} \right)}{\ln 3} \]. Substitute back into the expression for \( y \):\[ y = \ln 3 \cdot \frac{\ln \left( \frac{x+1}{x-1} \right)}{\ln 3} = \ln \left( \frac{x+1}{x-1} \right) \].

Step 3: Differentiate the Natural Logarithm Expression

To find the derivative of \( y \) with respect to \( x \), differentiate the expression \( y = \ln \left( \frac{x+1}{x-1} \right) \). Use the chain rule to differentiate:The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here \( u = \frac{x+1}{x-1} \),\( \frac{du}{dx} = \frac{(x-1) - (x+1)}{(x-1)^2} = \frac{-2}{(x-1)^2} \).Therefore, the derivative is:\[ \frac{dy}{dx} = \frac{1}{\frac{x+1}{x-1}} \cdot \frac{-2}{(x-1)^2} = -\frac{2}{(x+1)(x-1)}.\]

Step 4: Simplify the Result

Thus, the final simplified expression for the derivative of \( y \) with respect to \( x \) is:\[ \frac{dy}{dx} = -\frac{2}{x^2-1} \].This is the derivative of the given function with respect to the independent variable \( x \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is an incredibly handy tool when dealing with complex functions, especially those involving exponents and logarithms. It allows you to take derivatives more easily by first applying logarithms to both sides of an equation. This transforms complicated expressions into simpler ones. So, if you have a function where variables appear in the base, exponent, or both, logarithmic differentiation can be your best friend. It's like peeling an onion layer by layer until you reach the core. During differentiation, remember to bring the exponent down using the logarithmic identity and differentiate as you normally would. By simplifying the process, you're left with an expression that's much easier to handle zigzagging through the extensive calculus terrain.

Change of Base Formula

The change of base formula is an essential trick in logarithms that allows you to switch from one base to another. This is particularly useful when integrating or differentiating expressions because natural logarithms (base e) are easier to manipulate mathematically. The formula is:

• If you have \( \log_b a \), it can be rewritten as \( \frac{\ln a}{\ln b} \).

In the case of solving for derivatives, use this formula to transform logarithms into forms that work better with your differentiation toolkit. This method streamlines the process by converting base-3 logarithms into natural logarithms, as seen in the example, providing a straightforward pathway to differentiating the expression.

Chain Rule

The chain rule is an all-star player in differentiation, especially when tackling 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 formula terms:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

In simpler terms, think of it as peeling away layers of an onion: you tackle the external layer first and work your way down. In the step-by-step solution, the chain rule is employed to differentiate the natural logarithm expression efficiently. By breaking it down into the outer function \( \ln(u) \) and the inner function \( \frac{x+1}{x-1} \), you slice through the problem in manageable, logical steps.

Simplifying Logarithmic Expressions

Simplification is a fundamental step in dealing with logarithmic expressions, especially when finding derivatives. It’s all about making things as straightforward as possible before jumping into calculations. For logarithmic expressions, use identities such as the power rule,

• \( \log_b(a^c) = c \cdot \log_b(a) \).

Apply these identities to simplify complex functions before finding their derivatives. In our example, the expression inside the logarithm was simplified using this rule, reducing the derivative calculation to a more palatable form. After simplification, what follows is a much clearer and more systematic path through differentiation. This pre-step streamlines the entire process, ironing out any potential blocks that could complicate finding the derivative of the function.