Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{x^{2}-x+1}{x^{2}+1} $$

Step-by-Step Solution

Answer

\( D_{x} y = \frac{-x^2 + 1}{(x^2 + 1)^2} \)

Step 1: Identify the Rule to Use

This problem involves finding the derivative of a quotient of two functions. Therefore, we need to use the Quotient Rule for differentiation. The Quotient Rule is given by:\[ D_{x} \frac{u}{v} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^{2}} \]where \( u \) is the numerator and \( v \) is the denominator of the function.

Step 2: Define Functions and Their Derivatives

Let's define:\[ u = x^{2} - x + 1 \]\[ v = x^{2} + 1 \]Now, we compute the derivatives of these functions:\[ \frac{du}{dx} = 2x - 1 \]\[ \frac{dv}{dx} = 2x \]

Step 3: Apply the Quotient Rule

Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the Quotient Rule formula:\[ D_{x} \frac{x^{2}-x+1}{x^{2}+1} = \frac{(x^{2} + 1)(2x - 1) - (x^{2} - x + 1)(2x)}{(x^{2} + 1)^{2}} \]

Step 4: Simplify the Numerator

Expand and simplify the numerator:- First term: \((x^{2} + 1)(2x - 1) = 2x^{3} - x^{2} + 2x - 1 \)- Second term: \((x^{2} - x + 1)(2x) = 2x^{3} - 2x^{2} + 2x \)Combine the terms:\[ 2x^{3} - x^{2} + 2x - 1 - (2x^{3} - 2x^{2} + 2x) \]\[ = -x^{2} + 1 \]

Step 5: Final Expression for the Derivative

Substitute the simplified numerator back into the quotient and simplify:\[ D_{x} \frac{x^{2} - x + 1}{x^{2} + 1} = \frac{-x^{2} + 1}{(x^{2} + 1)^{2}} \]

Key Concepts

Understanding the Derivative

The concept of a derivative is fundamental in calculus. It provides insight into how functions change. At its core, a derivative measures the rate at which a function's output changes with respect to changes in the input. Imagine you are driving a car and looking at the speedometer; the number you see is the derivative of your position with respect to time. In mathematical terms, if you have a function, like our example function \( y = \frac{x^2-x+1}{x^2+1} \), the derivative tells us how \( y \) changes as \( x \) changes. To find the derivative of more complex functions, like the one in our problem, we often need specialized rules. The Quotient Rule is a tool that deals specifically with quotients - that is, one function divided by another.

Understanding Functions in Calculus

In calculus, functions are building blocks. A function is simply a relationship between two things, where each input has exactly one output. For example, the function \( y = \frac{x^2-x+1}{x^2+1} \) tells us exactly how the value of \( y \) depends on \( x \). Functions can be simple, like \( f(x) = x^2 \), or complex, like the ones in fractions or involving exponentials. When we deal with derivatives, we often break down complex functions into simpler parts to apply rules like the Power Rule, Product Rule, or Quotient Rule.

The Process of Simplification

Simplification is key in calculus and makes equations manageable and easier to understand. When you apply the Quotient Rule, like in our example, the goal is not just to find any expression for the derivative, but the simplest form possible. This allows us to see patterns, predict behavior, or even solve other problems more efficiently.Start by expanding products and combining like terms. In our exercise, the numerator ends up being surprisingly straightforward: from an expression involving multiple terms, it simplifies to \(-x^2 + 1\). Simplifying isn't always this straightforward, but it's an essential skill and often reveals deeper insights into the function's behavior.

Connecting with Calculus Techniques

Calculus is a mathematical language that allows us to study change, and derivatives are a central part of that. Beyond procedures, understanding the why behind steps like applying the Quotient Rule is crucial. This rule helps determine the derivative of a fraction of two functions, which is a common scenario in calculus problems. Breaking it down, we took two functions, \(u\) and \(v\), found their derivatives, and inserted them into a specific formula. This illustrates how rules and structured processes in calculus simplify our understanding of change — from curves on graphs to real-world phenomena, like motion and growth patterns. Understanding these connections plays a pivotal role in mastering calculus.