Problem 28 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=(1-x)^{-n}\) at \(a=0 .\) (Assume that \(n\) is a positive integer.)

Step-by-Step Solution

Answer

The linear approximation is \( f(x) \approx 1 + nx \).

Step 1: Identify the Function Components

First, notice that the function we are working with is \( f(x) = (1-x)^{-n} \). We need to identify the function and its derivative at the point \( a = 0 \) for the linear approximation formula.

Step 2: Evaluate the Function at a

Next, calculate \( f(a) \), where \( a = 0 \). This is simply \( f(0) = (1-0)^{-n} = 1 \).

Step 3: Find the Derivative of the Function

The derivative of \( f(x) = (1-x)^{-n} \) is found using the chain rule. The derivative is \( f'(x) = n(1-x)^{-n-1} \).

Step 4: Evaluate the Derivative at a

Calculate \( f'(a) \) where \( a = 0 \). Substituting in, we find \( f'(0) = n(1-0)^{-n-1} = n \).

Step 5: Substitute into the Linear Approximation Formula

Replace \( f(a) \) and \( f'(a) \) into the linear approximation formula: \[ f(x) \approx f(0) + f'(0)(x-0) = 1 + n \cdot (x - 0) = 1 + nx \].

Step 6: Write the Final Linear Approximation

The linear approximation of \( f(x) = (1-x)^{-n} \) at \( a = 0 \) is \( f(x) \approx 1 + nx \).

Key Concepts

Understanding Calculus

Calculus is a branch of mathematics that deals with change. It provides tools to find how things change, accumulate, grow, or shrink. Central to calculus are two fundamental ideas: differentiation and integration. These concepts allow us to:

• Understand rates of change, such as velocity and acceleration.
• Calculate areas under curves and volumes of objects.

In the case of linear approximation, calculus helps us find a linear function that closely resembles a more complex function near a specific point. This is particularly useful because linear functions are simple, making calculations easier. In the given exercise, the task is to use calculus to approximate a complex function, using its behavior understood through derivatives, to provide a linear representation around a particular point.

Delving into Derivatives

Derivatives represent how a function changes as the input changes, calculated as the "slope" of the function at any given point. Imagine a car driving on a curvy road. At any instant, the speedometer tells you how fast the car is going, which is akin to finding the derivative. For the function \( f(x) = (1-x)^{-n} \), the derivative helps us understand how rapidly it changes as \( x \) shifts.

• Finding the derivative of a function involves applying rules and formulas of differentiation.
• A critical utility of derivatives is in linear approximation — determining how a function's change can be approximated linearly at a point.

In the exercise, the derivative \( f'(x) = n(1-x)^{-n-1} \) was calculated to find the slope of the tangent at \( a = 0 \). This slope is then used to construct a linear approximation.

Exploring the Chain Rule in Differentiation

The chain rule is an essential tool in calculus for finding derivatives of composed functions. It states that to differentiate a composite function, you multiply the derivative of the outer function by the derivative of the inner function. For a function like \( (1-x)^{-n} \), which is a composition of a basic power function and a linear function, the chain rule becomes quite handy.

• First, identify the outer function \( y = (1-x)^{-n} \) and the inner function \( u = 1-x \).
• The derivative of the outer function with respect to the inner is \( -n(1-x)^{-n-1} \), while the derivative of the inner function \( 1-x \) is \(-1\).

By applying the chain rule, you can derive that \( f'(x) = n(1-x)^{-n-1} \). The chain rule's systematic approach simplifies finding the derivative of functions involving compositions, which is a cornerstone practice in calculus, particularly when linear approximations are involved.