Problem 87 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=2\), assuming that \(f(2)=\) \(-1\) and \(f^{\prime}(2)=1\). $$ y=\frac{f(x)}{f(x)+x} $$

Step-by-Step Solution

Answer

The derivative of \( y \) at \( x = 2 \) is 3.

Step 1: Understand the Problem

We're tasked with finding the derivative of the function \( y = \frac{f(x)}{f(x)+x} \) at \( x = 2 \). Given that \( f(2) = -1 \) and \( f'(2) = 1 \).

Step 2: Apply the Quotient Rule

To differentiate \( y = \frac{u(x)}{v(x)} \), where \( u(x) = f(x) \) and \( v(x) = f(x) + x \), we use the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).

Step 3: Find the Derivatives of u(x) and v(x)

Determine \( u'(x) = f'(x) \) and \( v'(x) = f'(x) + 1 \) since \( v(x) = f(x) + x \).

Step 4: Substitute into the Quotient Rule Formula

Recognize the derivative of \( y \) is \( y' = \frac{f'(x)(f(x)+x) - f(x)(f'(x)+1)}{(f(x)+x)^2} \). This is by plugging \( u'(x) \) and \( v'(x) \) into the quotient rule.

Step 5: Simplify the Expression

After expanding and simplifying, \( y' = \frac{f'(x) \cdot x - f(x)}{(f(x)+x)^2} \).

Step 6: Evaluate at x = 2

Substitute \( x = 2 \), \( f(2) = -1 \), and \( f'(2) = 1 \) into the formula: \[ y'(2) = \frac{1 \cdot 2 - (-1)}{(-1+2)^2} = \frac{2 + 1}{1^2} = 3. \]

Key Concepts

Quotient Rule

When you're dealing with the differentiation of a function that is expressed as a quotient of two other functions, the Quotient Rule becomes your go-to technique. This rule is essential for solving derivatives where one function is divided by another. It's expressed as follows: If you have a function in the form of \( y = \frac{u(x)}{v(x)} \), the derivative \( y' \) or \( \frac{d}{dx}\left(\frac{u}{v}\right) \) is computed as:

• \( y' = \frac{u'v - uv'}{v^2} \)

In this formula:

• \( u(x) \) is the numerator function, and \( v(x) \) is the denominator function.
• \( u'(x) \) is the derivative of the numerator function.
• \( v'(x) \) is the derivative of the denominator function.
• \( v(x)^2 \) appears in the denominator to account for both parts of the quotient as a squared term.

This concise rule helps you maneuver through calculus problems where functions are interrelated in ratios or divisions.

Derivative Evaluation

Evaluating a derivative at a specific point involves a few careful steps to ensure precision. Once you have differentiated a function using the appropriate rules, such as the Quotient Rule, you can plug in the specific value of \( x \) to find the slope of the tangent to the curve at that point. Here's how you can do it:

• First, ensure you have the correct derivative function form. For instance, from the exercise, the derivative is: \( y' = \frac{f'(x)(f(x)+x) - f(x)(f'(x)+1)}{(f(x)+x)^2} \).
• Next, substitute the given values of \( x \), \( f(x) \), and \( f'(x) \) directly into this formula. For \( x = 2 \), with \( f(2) = -1 \) and \( f'(2) = 1 \), perform the substitution.
• Finally, simplify to find the specific value. For our example, substituting these values yields \( y'(2) = 3 \).

By carefully conducting these steps, you apply derivative evaluation to determine exact values crucial for understanding function behaviors at particular points.

Calculus Problem Solving

Solving calculus problems effectively hinges on a clear approach and understanding of necessary rules. Initially, identifying which differentiation rule to apply sets the foundation for the solution. In this given problem, recognizing it requires the Quotient Rule was the first step. Here's a structured approach to tackle calculus problems efficiently:

• Understand the Problem: Break down what you are tasked to find and identify the given information.
• Choose the Right Rule: Different problems will require different differentiation rules, such as the Quotient, Product, or Chain Rule.
• Perform the Differentiation: Apply the chosen rule meticulously, being cautious with algebraic manipulations.
• Simplify the Expression: Algebraic simplification is necessary to facilitate easy evaluation in the next step.
• Evaluate and Review: Plug in any specific values like here with \( x = 2 \), review your steps, and report the results clearly.

This systematic method not only helps in finding derivatives but also enhances overall problem-solving skills in calculus, ensuring you understand the concepts and their applications thoroughly.