Problem 17 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\ln (1+x)\) at \(a=0\)

Step-by-Step Solution

Answer

The linear approximation of \( f(x) = \ln(1+x) \) at \( x = 0 \) is \( f(x) \approx x \).

Step 1: Identify the Function and the Point

The function given is \( f(x) = \ln(1+x) \), and we need to calculate the linear approximation at \( a = 0 \). In other words, we want to approximate \( f(x) \) near \( x = 0 \).

Step 2: Calculate the Function Value at the Point

Compute \( f(a) \) where \( a = 0 \):\[ f(0) = \ln(1+0) = \ln(1) = 0 \].

Step 3: Find the Derivative of the Function

First, we need to find the derivative of \( f(x) = \ln(1+x) \). Using the chain rule, the derivative is:\[ f'(x) = \frac{d}{dx}[\ln(1+x)] = \frac{1}{1+x} \].

Step 4: Evaluate the Derivative at the Point

Evaluate the derivative \( f'(x) \) at \( a = 0 \):\[ f'(0) = \frac{1}{1+0} = 1 \].

Step 5: Form the Linear Approximation Formula

Substituting \( f(a) \), \( f'(a) \), and \( a \) into the linear approximation formula \( f(x) \approx f(a) + f'(a)(x-a) \):\[ f(x) \approx 0 + 1 \cdot (x-0) \].

Step 6: Simplify the Linear Approximation

Simplify the approximation:\[ f(x) \approx x \].

Key Concepts

Linear Approximation

Linear approximation is a method used to approximate the value of a function near a given point using the concept of a tangent line. The formula for linear approximation is given by:

• \( f(x) \approx f(a) + f'(a)(x-a) \)

This formula represents a first-order Taylor expansion around the point \( a \). It provides a way to estimate the function's value by using the derivative or the slope of the function at that point. In simpler terms, it uses the information we have about the function at \( a \) to make educated guesses about its values nearby. This technique is particularly useful when the function itself is complicated or difficult to calculate directly.
In the context of our exercise, we approximate \( f(x) = \ln(1+x) \) around \( x = 0 \), using this linear model to simplify calculations.

Derivative

The derivative of a function is an essential concept in calculus. It tells us the rate of change of the function at any given point. To find a derivative, we apply differentiation rules to expressions.
For the function \( f(x) = \ln(1+x) \), the derivative is calculated as follows:

• According to the rules of differentiation, if \( f(x) = \ln(g(x)) \), then \( f'(x) = \frac{g'(x)}{g(x)} \).
• For \( f(x) = \ln(1+x) \), simplify to get \( f'(x) = \frac{1}{1+x} \).

This derivative tells us how the natural logarithmic function behaves as \( x \) changes slightly. By knowing the derivative, we understand how to adjust the linear approximation, predicting how \( f(x) \) grows or shrinks as \( x \) varies near zero.

Natural Logarithm

A natural logarithm is a logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to \( 2.71828 \). The natural logarithm is denoted as \( \ln(x) \). It is frequently used in calculus to simplify the process of differentiation and integration involving exponential functions.
The function \( \ln(1+x) \) is particularly interesting because it provides a simple form for calculating growth rates without requiring exponential scaling.

• At \( x = 0 \), \( \ln(1+x) = 0 \), emphasizing that small changes in \( x \) translate linearly into changes in \( \ln(1+x) \).
• This feature makes it extremely useful for approximating values within narrow bounds and for analyzing the behavior close to points like \( x = 0 \).

Understanding the nature of the natural logarithm helps in harnessing its properties for practical applications like financial modeling and biological growth analysis.

Chain Rule

The chain rule is a fundamental technique in calculus used when differentiating compositions of functions. It allows us to evaluate the derivative of a complex expression by breaking it down into simpler parts. The chain rule is expressed as:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In our exercise, when differentiating \( \ln(1+x) \), the chain rule helps in managing the internal function \( 1+x \). It makes the differentiation straightforward, enabling us to derive:

• \( f'(x) = \frac{d}{dx}[\ln(1+x)] = \frac{1}{1+x} \cdot \frac{d}{dx}(1+x) \).
• Since \( \frac{d}{dx}(1+x) = 1 \), the derivative simplifies to \( \frac{1}{1+x} \).

The chain rule simplifies the differentiation process, especially for functions nested within other functions. Mastering this rule is crucial for tackling more complex calculus problems efficiently.