Problem 35 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\frac{1+\ln t}{1-\ln t}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differentiation Rule

The function given, \( y = \frac{1 + \ln t}{1 - \ln t} \), is a rational function of \( t \). To differentiate a quotient, we need to use the Quotient Rule, which states: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] where \( u = 1 + \ln t \) and \( v = 1 - \ln t \).

Step 2: Find the Derivative of the Numerator

Differentiate the numerator \( u = 1 + \ln t \) with respect to \( t \). The derivative of 1 is 0 and the derivative of \( \ln t \) is \( \frac{1}{t} \). Hence, \( u' = \frac{1}{t} \).

Step 3: Find the Derivative of the Denominator

Now, differentiate the denominator \( v = 1 - \ln t \) with respect to \( t \). The derivative of 1 is also 0, and the derivative of \( -\ln t \) is \( -\frac{1}{t} \). So, \( v' = -\frac{1}{t} \).

Step 4: Apply the Quotient Rule

Plug the derivatives from Steps 2 and 3 into the quotient rule formula: \[ \left( \frac{1 + \ln t}{1 - \ln t} \right)' = \frac{\left(\frac{1}{t}\right)(1 - \ln t) - (1 + \ln t)\left(-\frac{1}{t}\right)}{(1 - \ln t)^2} \].

Step 5: Simplify the Expression

Expand the numerator: \( \frac{1}{t} \cdot (1 - \ln t) = \frac{1}{t} - \frac{\ln t}{t} \) and \( -(1 + \ln t)\cdot\left(-\frac{1}{t}\right) = \frac{1}{t} + \frac{\ln t}{t} \). Combine these to get: \[ \frac{1}{t} - \frac{\ln t}{t} + \frac{1}{t} + \frac{\ln t}{t} = \frac{2}{t} \]. So, the derivative simplifies to \( \frac{2}{t(1 - \ln t)^2} \).

Key Concepts

Quotient Rule

In calculus, the Quotient Rule is a method used to find the derivative of a quotient of two functions. If you have a function that is the division of two other functions, the Quotient Rule provides a formula to differentiate them. It states that for two functions \( u \) and \( v \), where \( y = \frac{u}{v} \), the derivative \( y' \) is given by \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]. Here's how it works step by step:

• **Differentiate the numerator**: Find the derivative of the function \( u \). This is written as \( u' \).
• **Differentiate the denominator**: Similarly, find the derivative of the function \( v \), which is \( v' \).
• **Plug into the formula**: Substitute \( u' \), \( v \), \( u \), and \( v' \) into the Quotient Rule formula to find the derivative of the quotient.

Carefully applying each step will allow you to correctly differentiate rational functions using the Quotient Rule.

Derivative of logarithmic function

Understanding how to differentiate logarithmic functions is essential in calculus, especially when dealing with rational functions. The logarithmic function \( \ln t \) has a straightforward derivative: For \( y = \ln t \), the derivative is \( \frac{1}{t} \). It is important to remember the basic rules:

• **The constant rule**: The derivative of a constant is 0. For example, the derivative of the constant 1 is 0.
• **The natural logarithm rule**: The derivative of \( \ln t \) is \( \frac{1}{t} \).

In rational functions like \( y = \frac{1 + \ln t}{1 - \ln t} \), you'll need these derivatives to apply the Quotient Rule effectively. When differentiating such a function, always differentiate the terms involving \( \ln t \) separately, then combine them to apply in the quotient or product rules as needed.

Rational functions

Rational functions are fractions that involve polynomials in both the numerator and the denominator. In simpler terms, they are expressed as \( \frac{u}{v} \) where \( u \) and \( v \) are polynomials or expressions involving variables, like the example \( y = \frac{1 + \ln t}{1 - \ln t} \). Key aspects to consider when working with rational functions include:

• **Identify the components**: Recognize which part of the function is the numerator and which is the denominator.
• **Apply rules appropriately**: Utilize differentiation rules such as the Quotient Rule specifically for rational functions.
• **Find derivatives of individual parts**: Separate the function into its components to differentiate them separately, such as \( 1 + \ln t \) and \( 1 - \ln t \).

These functions often appear in calculus problems, making solid understanding and the ability to apply derivative rules crucial when solving them. As with the exercise problem, apply the Quotient Rule once you've determined the derivatives of each part of the function.