Problem 34 - University Calculus: Early Transcendentals

Question

Find \(d y\). $$y=\ln \left(\frac{x+1}{\sqrt{x-1}}\right)$$

Step-by-Step Solution

Answer

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

Step 1: Apply Logarithmic Identity

Start by using the property of logarithms that states \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\). This allows us to rewrite the given expression for \(y\) as: \[y = \ln(x+1) - \ln(\sqrt{x-1})\]

Step 2: Simplify the Expression Further

Notice that \(\ln(\sqrt{x-1})\) can be further simplified using the logarithm property \(\ln(a^b) = b\ln(a)\). Hence, \[\ln(\sqrt{x-1}) = \frac{1}{2}\ln(x-1)\]So the expression for \(y\) becomes: \[y = \ln(x+1) - \frac{1}{2}\ln(x-1)\]

Step 3: Differentiate Each Term

Differentiate with respect to \(x\), treating each term separately. For \(\ln(x+1)\), differentiate to get \[\frac{d}{dx}[\ln(x+1)] = \frac{1}{x+1}\]For \(\frac{1}{2}\ln(x-1)\), apply the chain rule to get \[\frac{d}{dx}\left[-\frac{1}{2}\ln(x-1)\right] = -\frac{1}{2} \cdot \frac{1}{x-1} = -\frac{1}{2(x-1)}\]

Step 4: Combine the Derivatives

Combine the results from differentiating each term to find the derivative of the entire function. Thus, \[\frac{dy}{dx} = \frac{1}{x+1} - \frac{1}{2(x-1)}\]

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a technique used in calculus to simplify the process of differentiating complex functions, especially those involving products or quotients. It uses the properties of logarithms to turn multiplication into addition and division into subtraction.
This is particularly helpful in situations where straightforward differentiation could become cumbersome.
In our exercise, we started with the function \( y = \ln \left( \frac{x+1}{\sqrt{x-1}} \right) \). Using the logarithmic identity \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \), we transformed the quotient into a difference of two logarithms.

• Apply \( \ln(a^b) = b\ln(a) \) to further simplify terms, such as \( \ln(\sqrt{x-1}) = \frac{1}{2}\ln(x-1) \).
• This breaks down the original function into simpler parts, making it easier to handle.

Logarithmic differentiation empowers us to manage complex expressions with multiple layers, preparing them for further differentiation steps.

Derivative

The derivative represents the rate at which a function changes at any given point, often associated with the slope of the tangent line to its graph at that point. Calculating derivatives is a fundamental skill in calculus for analyzing functions and interpreting their behaviors.
In the exercise, we worked with the expression \( y = \ln(x+1) - \frac{1}{2}\ln(x-1) \) after simplifying it with logarithmic differentiation.

• The derivative of \( \ln(x+1) \) is calculated as \( \frac{1}{x+1} \).
• The derivative of \( -\frac{1}{2}\ln(x-1) \) is calculated using the chain rule, resulting in \( -\frac{1}{2(x-1)} \).

These computations are basic applications, illustrating how different parts of a function contribute to its overall rate of change at any point.

Chain Rule

The chain rule is an essential differentiation technique in calculus, utilized whenever a function is composed of other functions. It allows us to differentiate nested functions systematically by breaking them down.
It can be visualized as a process of "peeling back layers" of functions while keeping track of their derivatives.
In this context, we applied the chain rule to handle \( -\frac{1}{2}\ln(x-1) \), a part of the expression that wasn't as straightforward.

• View the function \( \ln(x-1) \) as a composite of the outer function \( \ln \, u \) and the inner function \( u = x-1 \).
• The chain rule formula is \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).
• By applying it, we find that the derivative of \( \ln(x-1) \) contributes a factor of \( \frac{1}{x-1} \).

In this way, the chain rule equips us to unravel and differentiate complex structures methodically, ensuring nothing is overlooked.