Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\) using the rules of this section. $$ y=\left(x^{4}+2 x\right)\left(x^{3}+2 x^{2}+1\right) $$

Step-by-Step Solution

Answer

The derivative \( D_{x} y = 7x^6 + 12x^5 + 12x^3 + 12x^2 + 2 \).

Step 1: Identify the Function for Differentiation

The function given is a product of two functions: \[ y = (x^4 + 2x)(x^3 + 2x^2 + 1) \]To differentiate it, we will use the product rule for derivatives, which states: if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product \( y = u(x) \, v(x) \) is given by:\[ D_{x} y = u'(x) \, v(x) + u(x) \, v'(x) \] Here, \( u(x) = x^4 + 2x \) and \( v(x) = x^3 + 2x^2 + 1 \).

Step 2: Differentiate Each Function Separately

First, we find the derivative of \( u(x) = x^4 + 2x \):- The derivative of \( x^4 \) is \( 4x^3 \).- The derivative of \( 2x \) is \( 2 \).Therefore, \( u'(x) = 4x^3 + 2 \).Next, we find the derivative of \( v(x) = x^3 + 2x^2 + 1 \):- The derivative of \( x^3 \) is \( 3x^2 \).- The derivative of \( 2x^2 \) is \( 4x \).- The derivative of \( 1 \) is \( 0 \).Therefore, \( v'(x) = 3x^2 + 4x \).

Step 3: Apply the Product Rule

Substitute the derivatives back into the product rule formula:\[ D_{x} y = u'(x) \, v(x) + u(x) \, v'(x) \]\[ D_{x} y = (4x^3 + 2)(x^3 + 2x^2 + 1) + (x^4 + 2x)(3x^2 + 4x) \]

Step 4: Distribute and Simplify Each Term

Now, expand and simplify each part:1. \( (4x^3 + 2)(x^3 + 2x^2 + 1) \) yields: - \( 4x^3 \cdot x^3 = 4x^6 \) - \( 4x^3 \cdot 2x^2 = 8x^5 \) - \( 4x^3 \cdot 1 = 4x^3 \) - \( 2 \cdot x^3 = 2x^3 \) - \( 2 \cdot 2x^2 = 4x^2 \) - \( 2 \cdot 1 = 2 \) Combine results: \( 4x^6 + 8x^5 + 6x^3 + 4x^2 + 2 \)2. \( (x^4 + 2x)(3x^2 + 4x) \) yields: - \( x^4 \cdot 3x^2 = 3x^6 \) - \( x^4 \cdot 4x = 4x^5 \) - \( 2x \cdot 3x^2 = 6x^3 \) - \( 2x \cdot 4x = 8x^2 \) Combine results: \( 3x^6 + 4x^5 + 6x^3 + 8x^2 \)

Step 5: Combine and Simplify Final Expression

Add the results from Step 4:- Combine the terms: \[ (4x^6 + 3x^6) + (8x^5 + 4x^5) + (6x^3 + 6x^3) + (4x^2 + 8x^2) + 2 \] \[ = 7x^6 + 12x^5 + 12x^3 + 12x^2 + 2 \]So, the derivative \( D_{x} y = 7x^6 + 12x^5 + 12x^3 + 12x^2 + 2 \).

Key Concepts

Derivative

A derivative represents the rate at which a function is changing at any given point. Think of it like a speedometer in a car, measuring how quickly things are moving. When you take the derivative of a function, you're determining how the function's output value changes as you slightly tweak its input value. Mathematically, for a function \(f(x)\), the derivative is denoted as \(f'(x)\). This gives you a function that tells you the slope of the original function at any point along its curve.

• The process of finding a derivative is called differentiation.
• Derivatives are foundational in calculus, giving insights into behavior and trends.
• There are many rules in differentiation, each to tackle different types of functions.

In our exercise, the derivation process included identifying the function parts \(u(x)\) and \(v(x)\), differentiating them separately, and then applying the product rule.

Polynomial Functions

Polynomial functions are types of functions that include terms made up of constants and variables raised to non-negative integer powers. They come in simple or complex forms, but always follow this general template: \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\). In such a function, each term's degree is defined by its exponent on the variable \(x\).

• Polynomials can depict a wide range of behaviors, from simple lines to complex curves.
• The highest power in a polynomial determines its degree, which tells how it grows and behaves at its endpoints.
• They are smooth and continuous, which makes them ideal for calculus operations like differentiation.

In our problem, we are dealing with two polynomial expressions that are multiplied together, requiring careful application of differentiation rules to find the derivative.

Differentiation Techniques

Differentiation techniques form the toolkit used to find derivatives of functions efficiently and accurately. These include various rules that simplify the process, especially when dealing with complex expressions. Let's explore some crucial techniques:

• Product Rule: Used when differentiating products of two functions. It states if \( y = u(x) \, v(x) \), then \[D_{x} y = u'(x) \, v(x) + u(x) \, v'(x)\]
• Sum Rule: If you have a sum of functions, like \(f(x) + g(x)\), you differentiate each part separately and add the results: \((f + g)'(x) = f'(x) + g''(x)\).
• Chain Rule: Helpful when the function composition is involved, this rule is essential for understanding more complex nested functions.

For our example, the key differentiation technique used was the **Product Rule**. We differentiated each polynomial separately, plugged them into the product rule formula, and simplified the result to find the derivative accurately.