Problem 20 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\log \left(1+x^{2}\right)\) at \(a=0\)

Step-by-Step Solution

Answer

The linear approximation is 0.

Step 1: Identify Given Function and Point

We are given the function \( f(x) = \log(1 + x^2) \) and need to calculate its linear approximation around the point \( a = 0 \).

Step 2: Calculate the Function Value at the Point

Substitute \( a = 0 \) into the function: \( f(0) = \log(1 + 0^2) = \log(1) = 0 \).

Step 3: Find the Derivative of the Function

Differentiate \( f(x) = \log(1 + x^2) \) using the chain rule: \( f'(x) = \frac{d}{dx} \log(1 + x^2) = \frac{2x}{1 + x^2} \).

Step 4: Evaluate the Derivative at the Point

Substitute \( a = 0 \) into the derivative: \( f'(0) = \frac{2(0)}{1 + 0^2} = 0 \).

Step 5: Construct the Linear Approximation Formula

Substitute \( f(0) = 0 \) and \( f'(0) = 0 \) into the linear approximation formula: \[ f(x) \approx f(0) + f'(0)(x - 0) = 0 + 0 \cdot (x - 0) = 0 \].

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. Here's how you can think about it. When you have a function inside another function, like our example with the logarithmic function and its inner component, you can differentiate the outer and inner components separately, combining them using multiplication. The formula is essentially the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In the exercise, we have the function \( f(x) = \log(1 + x^2) \). Here, the outer function is \( \log(u) \) and the inner function \( u(x) = 1 + x^2 \). By applying the chain rule:

• First, differentiate the outer function: the derivative of \( \log(u) \) is \( \frac{1}{u} \)
• Next, differentiate the inner function: the derivative of \( 1 + x^2 \) is \( 2x \)
• Multiply both derivatives: \( \frac{1}{1 + x^2} \times 2x \)

This gives us the derivative \( f'(x) = \frac{2x}{1 + x^2} \), as shown in the solution steps.

Derivative Evaluation

Once we have differentiated the function using the chain rule, the next important step is derivative evaluation. This involves calculating the derivative at a specific point, which helps in linear approximation.

In our exercise, after finding the derivative \( f'(x) = \frac{2x}{1 + x^2} \), we evaluate it at the point \( a = 0 \). Substituting \( x = 0 \) into the derivative gives:

• \( f'(0) = \frac{2 \times 0}{1 + 0^2} = 0 \)

This value, \( f'(0) = 0 \), is crucial for constructing the linear approximation. At this point, the slope of the tangent line to the curve is zero, meaning the tangent is horizontal at \( a = 0 \). This indicates that the change around \( a = 0 \) follows a flat line in this linear approximation.

Logarithmic Function

The function \( f(x) = \log(1 + x^2) \) represents a specific type of logarithmic function, which is quite common in calculus problems. Understanding how logarithms behave can make evaluating and differentiating much easier.

Logarithmic functions are the inverses of exponential functions. The basic function \( \log(x) \) grows slowly, especially for values greater than one. When combined with other functions, like \( 1 + x^2 \), it creates a composition that shows more complex behavior. This is where linear approximation becomes useful as it simplifies complex behavior near a certain point.

• At \( x = 0 \), we'll see the logarithmic of \( 1 + x^2 \) simplifies. Since \( \log(1) = 0 \), it makes calculations straightforward.
• The logarithmic function smoothly passes through the origin, which means the value and slope are zero at \( x = 0 \) in this example.

Linear approximation with logarithms can be a quick method to estimate function values close to the point of interest by reducing the function to a linear equation, which is easier to handle and understand.