Problem 42 - Calculus of a Single Variable

Question

In Exercises 41–64, find the derivative of the function. $$ f(x)=\ln (x-1) $$

Step-by-Step Solution

Answer

\The derivative of \( f(x) = \ln(x-1) \) is \( f'(x) = \frac{1}{x-1} \).

Step 1: Identify the function

In this problem, the function \( f(x) \) is given as \( f(x) = \ln(x-1) \). \n\nNotice that the function \( f(x) \) is a logarithmic function. The base of the logarithm is the mathematical constant \( e \), which is approximately equal to 2.71828.

Step 2: Differentiate function

The formula to differentiate the natural logarithm of a function \( u \) is \( \frac{1}{u} * \frac{du}{dx} \). In this case, \( u = x-1 \).\n\nSo, to differentiate \( f(x) = \ln(x-1) \), use the chain rule: first differentiate the outer function, leaving the inner function alone, then differentiate the inner function.\n\nThe derivative of \( \ln(u) \) is \( \frac{1}{u} \), which means the derivative of the outer function is \( \frac{1}{x-1} \).\n\nNext, differentiate the inner function. The derivative of \( x-1 \) is simply 1.

Step 3: Apply the chain rule

Apply the chain rule, multiplying the derivative of the outer function by the derivative of the inner function.\n\nThis gives \( \frac{1}{x-1} * 1 \).

Step 4: Simplify the derivative

The expression \( \frac{1}{x-1} * 1 \) simplifies to \( \frac{1}{x-1} \).\n\nThus, the derivative of \( f(x) = \ln(x-1) \) is \( f’(x) = \frac{1}{x-1} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used when differentiating composite functions, meaning functions within functions. In the given exercise, the function is \( f(x) = \ln(x-1) \) , a common form where the chain rule applies. Here, you have an outer function, \( \ln(u) \), and an inner function, \( u = x-1 \).

• Start with the outer function: Take its derivative, ignoring the inner details. For \( \ln(u) \), this derivative is \( \frac{1}{u} \) .
• Move to the inner function: Differentiate \( u = x-1 \) which simply gives 1.

After differentiating both parts separately, multiply these derivatives as per the chain rule: \( \frac{1}{u} \cdot \frac{du}{dx} \). In this case, solving it highlights how the chain rule efficiently breaks down the differentiation process for layered functions. Contrasting simple and more complex functions shows the utility of the chain rule in handling diverse scenarios.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), the constant approximately equal to 2.71828. It appears frequently in calculus due to its unique properties that simplify differentiation and integration. In the exercise, \( f(x) = \ln(x-1) \), the presence of the natural log signifies a logarithmic function where the concept of working with logs is crucial.

• Recognize that \( \ln(x) \) has a straightforward differentiation formula: its derivative is \( \frac{1}{x} \).
• Understand how natural logs help decompose growth-related problems like compounding (exponential growth).

It’s important to grasp how this differentiable feature makes natural logs special when compared to other logarithms, especially when trying to simplify computations. The natural log retains answers in forms easily interpretable in various growth and decay problems. This exercise leverages the simple and elegant differentiation of \( \ln(x) \) but with an adjustment inside the log function.

Differentiation Techniques

Differentiation involves finding the rate at which a function changes at any given point. In calculus, this knowledge is applied using various techniques depending on the type of function. This exercise utilizes a specific differentiation technique tailored for logarithmic functions, particularly key in decomposing functions like \( \ln(x-1) \).

• For \( \ln(x-1) \), utilize the known derivative formula of a logarithm: remain aware that \( f’(x) = \frac{1}{x} \) extends to complex expressions beyond simple variables.
• Apply derivatives carefully — ensuring you adopt accompanying strategies like the chain rule for nested functions.

The mastery of these techniques is fundamental for progressing through different calculus problems effectively. Whether dealing with basic algebraic, exponential, or logarithmic functions, understanding which technique to use streamlines problem-solving, ensuring accuracy and efficiency across varied mathematical inquiries.