Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\left(3 x^{2}+2 x\right)\left(x^{4}-3 x+1\right) $$

Step-by-Step Solution

Answer

18x^5 + 10x^4 - 27x^2 - 6x + 2

Step 1: Identify the Product Rule

The provided function, \(y = (3x^2 + 2x)(x^4 - 3x + 1)\), is the product of two functions. We need to use the product rule for differentiation, which states \(D_x(u imes v) = u'v + uv'\), where \(u = 3x^2 + 2x\) and \(v = x^4 - 3x + 1\).

Step 2: Differentiate the First Function

Calculate the derivative of the first function \(u = 3x^2 + 2x\). Differentiate term by term to get \(u' = 6x + 2\).

Step 3: Differentiate the Second Function

Calculate the derivative of the second function \(v = x^4 - 3x + 1\). Differentiate term by term to get \(v' = 4x^3 - 3\) since the derivative of \(x^4\) is \(4x^3\), \(-3x\) is \(-3\), and the constant \(1\) becomes \(0\).

Step 4: Apply the Product Rule

Substitute \(u, u', v,\) and \(v'\) into the product rule: \(D_x y = (6x + 2)(x^4 - 3x + 1) + (3x^2 + 2x)(4x^3 - 3)\).

Step 5: Simplify the Expression

Expand each product by distributing: - First term: \((6x + 2)(x^4 - 3x + 1) = 6x^5 - 18x^2 + 6x + 2x^4 - 6x + 2 = 6x^5 + 2x^4 - 18x^2 + 2\)- Second term: \((3x^2 + 2x)(4x^3 - 3) = 12x^5 + 8x^4 - 9x^2 - 6x\) Now combine to get: \(6x^5 + 2x^4 - 18x^2 + 2 + 12x^5 + 8x^4 - 9x^2 - 6x\).

Step 6: Combine Like Terms

Combine the like terms: - For \(x^5\): \(6x^5 + 12x^5 = 18x^5\)- For \(x^4\): \(2x^4 + 8x^4 = 10x^4\)- For \(x^2\): \(-18x^2 - 9x^2 = -27x^2\)- For \(x^1\): \(-6x\)- Constant term: \(+2\).Thus, the derivative is: \(D_x y = 18x^5 + 10x^4 - 27x^2 - 6x + 2\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that refers to the process of finding the derivative of a function. The derivative indicates how a function changes as its input changes. Essentially, it measures the rate at which a function's output changes concerning its input.
Understanding differentiation is crucial as it forms the basis for discovering how mathematical models can predict future patterns or behaviors.

• It’s used to calculate speeds, optimize business profits, and even in physics to find forces and energy.
• When differentiating, you're often looking for an equation or expression that shows how a function behaves as its variables change slightly.

In this exercise, we differentiate a product of two polynomial functions. By using rules such as the product rule and chain rule, we simplify the process of finding the derivative of more complex functions.

Derivative of Polynomial Functions

Finding the derivative of polynomial functions is a straightforward process. Polynomials are mathematical expressions consisting of variables and coefficients combined using only subtraction, addition, multiplication, and non-negative integer exponents.
To differentiate a polynomial function, apply the rule: bring down the exponent as a coefficient and reduce the exponent by one.

• For example, the derivative of \(x^n\) is \(nx^{n-1}\).
• The derivative of a constant is zero because constants do not change.

In this exercise, the function \(y\) is broken down into simpler polynomials: \(3x^2 + 2x\) and \(x^4 - 3x + 1\). By differentiating each part separately, you get \(u' = 6x + 2\) and \(v' = 4x^3 - 3\). Each of these is obtained by applying the basic differentiation rule term by term.

Chain Rule

Though we primarily focus on the product rule in this exercise, understanding the chain rule is essential in calculus. The chain rule is used to differentiate composite functions, which are functions applied within other functions.Think of it like peeling an onion, where you need to find the derivative of the outer layers step-by-step. The chain rule helps you tackle these layers efficiently. The formula for the chain rule is:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]When you deal with functions nested within others, the chain rule guides you to find the derivative effectively.

• It becomes especially useful when dealing with trigonometric, exponential, or logarithmic functions incorporated into polynomials.
• It ensures you don’t miss any inner functions that affect the outer one.

Calculus Problem Solving

Calculus problem solving involves a systematic approach to tackling differential equations, limits, and integrals. Understanding the rules and knowing when to apply them is key. By breaking down complicated problems into simpler parts, calculus makes it easier to manage mathematical challenges.
A significant part of problem-solving is recognizing when to use specific rules like the product rule. The product rule, for example, is essential when dealing with functions that are multiplied together, such as in this exercise. To solve any calculus problem efficiently:

• Identify the type of function or functions involved.
• Select appropriate rules for differentiation or integration.
• Simplify and solve step by step, ensuring all terms and solutions are accounted for.

This methodical process highlights the importance of understanding calculus principles and their applications in real-world contexts, from engineering to economics.