Problem 84 - Calculus for Biology and Medicine

Question

In Problems 84-87, assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=2\), assuming that \(f(2)=\) \(-1\) and \(f^{\prime}(2)=1\). $$ y=\frac{f(x)}{x^{2}+1} $$

Step-by-Step Solution

Answer

The derivative of \(y\) at \(x=2\) is \(\frac{9}{25}\).

Step 1: Use the Quotient Rule

To find the derivative of the function \(y = \frac{f(x)}{x^2 + 1}\), we need to use the Quotient Rule for differentiation. The quotient rule states that if you have a function \(y = \frac{u(x)}{v(x)}\), the derivative \(y'\) is given by: \[ y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \] In our case, \(u(x) = f(x)\) and \(v(x) = x^2 + 1\).

Step 2: Differentiate the Numerator and Denominator

Differentiate \(u(x) = f(x)\) to get \(u'(x) = f'(x)\). For the denominator, differentiate \(v(x) = x^2 + 1\) to get \(v'(x) = 2x\).

Step 3: Substitute in the Quotient Rule Formula

Using the Quotient Rule formula, we substitute: \[ y' = \frac{f'(x)(x^2 + 1) - f(x)(2x)}{(x^2 + 1)^2} \] Simplify the expression within the numerator.

Step 4: Substitute Known Values at \(x = 2\)

Given \(f(2) = -1\) and \(f'(2) = 1\), substitute \(x = 2\) into the derivates: \[ y' = \frac{1(2^2 + 1) - (-1)(2 \cdot 2)}{(2^2 + 1)^2} \] Simplify the expression.

Step 5: Simplify and Evaluate at \(x = 2\)

Compute the simplified expression step by step: - First, calculate \( (2^2 + 1) = 5 \) - Then, substitute these into the formula: \[ y' = \frac{1\cdot 5 - (-1)\cdot 4}{5^2} \] \[ y' = \frac{5 + 4}{25} \] \[ y' = \frac{9}{25} \]

Key Concepts

Differentiation

Differentiation is one of the core concepts in calculus. It allows us to find the rate at which something changes. Imagine you are driving a car, and you want to know how fast you're going at an exact moment. Differentiation gives you this speed or rate of change.

Differentiation is the process of obtaining a derivative. The derivative represents the slope or steepness of a function at any given point. This is particularly useful for understanding how a function behaves, such as whether it's increasing or decreasing.

• Notation: If you have a function like \( f(x) \), its derivative is often written as \( f'(x) \) or \( \frac{df}{dx} \).
• Interpretation: The derivative \( f'(x) \) tells you how \( f(x) \) changes when \( x \) changes by a tiny amount.

Understanding differentiation is crucial for solving many types of calculus problems, especially those related to optimization and motion.

Quotient Rule

The Quotient Rule is a specific formula we use when differentiating a function that is the division of two simpler functions. This rule simplifies the process while ensuring accuracy.

When you have a function \( y = \frac{u(x)}{v(x)} \), the Quotient Rule states that the derivative is given by:

• Formula: \[ y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]

In this formula, \( u(x) \) represents the numerator and \( v(x) \) the denominator. Here's how it plays out step-by-step:
1. Differentiate the numerator \( u(x) \) to get \( u'(x) \).2. Differentiate the denominator \( v(x) \) to get \( v'(x) \).3. Substitute these derivatives into the Quotient Rule formula.

This method ensures you correctly account for how both parts of the fraction change. In our exercise, this involved functions \( u(x) = f(x) \) and \( v(x) = x^2 + 1 \), leading to specific derivatives of each.

Function Derivatives

Function derivatives describe how functions change and are fundamental to understanding complex problems. A derivative tells you the function's sensitivity to changes in its input.

Consider a function like \( f(x) \). Its derivative, \( f'(x) \), offers insights into how \( f(x) \) changes as \( x \) shifts slightly. Let's break down some important aspects of function derivatives:

• Smooth Changes: For smooth or continuous functions, the derivative helps sketch the function's behavior over its domain.
• Critical Points: At places where the derivative is zero, the function may have peaks, valleys, or inflection points.
• Real-life Applications: Derivatives model real-world phenomena, from physics (like velocity and acceleration) to economics and other fields.

In our exercise, we calculated the derivative of a fraction involving \( f(x) \) to determine its rate of change at a specific point. Knowing \( f(2) \) and \( f'(2) \), we substituted these into our derivation to compute the exact change rate at \( x = 2 \). This is an essential step in analyzing and predicting function behaviors.