Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The derivative is \( D_x y = 5x^4 + 6x^2 + 2x \).

Step 1: Identify the Differentiation Rule

Observe the expression for \( y \): \( y = (x^2 + 2)(x^3 + 1) \). This expression involves a product of two functions. Therefore, we should apply the Product Rule for differentiation.

Step 2: Apply the Product Rule

The Product Rule states that if \( y = u imes v \), then the derivative \( D_x y \) is \( u'v + uv' \), where \( u = x^2 + 2 \) and \( v = x^3 + 1 \). Compute \( u' \) and \( v' \) first.

Step 3: Differentiate the First Function

For \( u = x^2 + 2 \), differentiate with respect to \( x \) to get \( u' = 2x \). The derivative of a constant is zero, so only \( x^2 \) is differentiated.

Step 4: Differentiate the Second Function

For \( v = x^3 + 1 \), differentiate with respect to \( x \) to get \( v' = 3x^2 \). As with the previous step, the derivative of a constant is zero.

Step 5: Substitute into the Product Rule Formula

Substitute the derivatives into the Product Rule formula: \( D_x y = (x^2 + 2)(3x^2) + (2x)(x^3 + 1) \).

Step 6: Simplify the Expression

Expand the expression: \((x^2 + 2)(3x^2) = 3x^4 + 6x^2 \) and \((2x)(x^3 + 1) = 2x^4 + 2x \). Combine to get \( D_x y = 3x^4 + 6x^2 + 2x^4 + 2x \).

Step 7: Combine Like Terms

Combine like terms: \( 3x^4 + 2x^4 = 5x^4 \). The full expression for the derivative is \( D_x y = 5x^4 + 6x^2 + 2x \).

Key Concepts

Product Rule in Differentiation

The product rule is an essential tool in calculus for finding the derivative of a product of two functions. When you have two functions multiplied together, such as

• \( y = u \cdot v \)

this rule is invaluable. The product rule states that the derivative of this expression, with respect to a variable like \( x \), is given by:

• \( D_x y = u'v + uv' \)

Here, \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \), respectively.

This rule is particularly useful when each part of the product is a function of \( x \), and you want to find how \( y \) changes as \( x \) changes. Essentially, it allows you to consider how each component of the product contributes to the overall rate of change through differentiation.

• "\( u'v \)" involves differentiating \( u \) while keeping \( v \) unchanged.
• "\( uv' \)" involves differentiating \( v \) while keeping \( u \) unchanged.

Understanding Derivatives

A derivative is a fundamental concept in mathematics, capturing how a function changes as its input changes. It measures the rate of change or the slope of the function.

For a polynomial function, finding the derivative involves using simple rules of differentiation. For example, if you have \( f(x) = ax^n \), the derivative is found using the rule:

• \( \frac{d}{dx}[ax^n] = nax^{n-1} \)

This means you multiply the power \( n \) by the coefficient \( a \), and then reduce the power by 1.

Derivatives tell us about the behavior of a graph at any point. They show where the function is increasing or decreasing and are crucial in finding maxima and minima points, which are key in many real-world applications, helping to find the most efficient or cost-effective solutions.

Working with Polynomial Expressions

Polynomial expressions consist of variables raised to powers multiplied by coefficients. An example is:

• \( P(x) = ax^n + bx^{n-1} + \, ... \, + k \)

They are a cornerstone of calculus and algebra. When differentiating such expressions, each term is handled individually; this is thanks to the sum rule of differentiation.

• The derivative of \( f(x)=ax^n \) is \( \frac{d}{dx}[ax^n] = nax^{n-1} \),
• and for a constant \( c \), \( \frac{d}{dx}[c] = 0 \).

This makes polynomial expressions straightforward to differentiate. You just have to apply the rule to each term separately.

In practice, this means you will see how each individual term in the polynomial contributes to the total change of the expression by its own rate of change. Overall, handling polynomials means carefully applying these rules to obtain smooth, continuous functions that help describe a wide range of phenomena in practical scenarios.