Problem 7 - Calculus Single Variable

Question

In Exercises 7-18, use the Product Rule to compute the derivative of the given expression with respect to \(x\). (In each of Exercises 7,8,14,16, and 18, do not avoid using the Product Rule by first expanding the product.) $$ x^{2}(x+3) $$

Step-by-Step Solution

Answer

The derivative of the expression is \( 3x^2 + 6x \).

Step 1: Identify the functions

Recognize that the expression can be seen as a product of two functions. Let \( u = x^2 \) and \( v = x + 3 \).

Step 2: Find the derivatives of the functions

Compute the derivative of \( u \) with respect to \( x \), which is \( u' = 2x \). Also, compute the derivative of \( v \) with respect to \( x \), which is \( v' = 1 \).

Step 3: Apply the Product Rule

Recall the Product Rule, which states \((uv)' = u'v + uv'\). Apply this rule: \((x^2 (x+3))' = (2x)(x+3) + (x^2)(1)\).

Step 4: Simplify the expression

Distribute and combine like terms: \( (2x)(x+3) + x^2 = 2x^2 + 6x + x^2 = 3x^2 + 6x \). This is the derivative of the given expression.

Key Concepts

Understanding Derivatives

Derivatives represent the rate at which a function is changing at any given point. They are a fundamental concept in calculus, showing how a change in one quantity can affect another. To find a derivative, we examine how a function behaves as its input approaches a particular value. This involves understanding one-sided limits and slope interpretations. If we consider a function like \( f(x) \), its derivative, \( f'(x) \), tells us the slope of the tangent to the curve described by the function at any given point \( x \).
Derivatives are essential because they provide critical insights such as:

• The velocity of an object at a specific point in time (if the function represents position).
• The rate of change of different quantities in relation to each other (common in physics and engineering).
• Optimizing business processes or economic models by determining maximum or minimum values.

The Process of Differentiation

Differentiation is the action or process of computing a derivative. It involves applying specific rules and formulas to determine the derivative of a given function. One common technique is the Product Rule, which is particularly useful when differentiating a product of two functions. Different rules might apply depending on the structure of the function you're dealing with.
To effectively differentiate a function using the Product Rule, you need to:

• Identify the distinct functions within the product. For example, in \( x^2(x+3) \), we can view \( x^2 \) and \( x + 3 \) as separate entities.
• Find the derivatives of these components individually.
• Apply the Product Rule formula:\((uv)' = u'v + uv'\), where \( u \) and \( v \) are functions of \( x \).
• Finally, simplify the result for the desired derivative.

Differentiation by using rules like the Product Rule makes solving complex calculus problems more manageable, especially when dealing with multiple components in a function.

Diving into Calculus

Calculus is a branch of mathematics focusing on perpetual motion and change. It's split into two main disciplines: differential calculus and integral calculus. Differential calculus deals explicitly with derivatives, giving tools to calculate rates of change.

• The necessity of calculus grows with its ability to handle various real-world phenomena from physics to biology.
• Using derivatives, you can better predict patterns and optimize results in sectors like economics, physics, and statistics.
• In the world of academia, calculus provides the methodologies for solving advanced problems involving continuous change.

By grappling with derivatives and the processes of differentiation, students begin to harness the power of calculus to unlock deeper insights into both theoretical and practical applications. Whether analyzing the growth rate of populations or understanding the trajectory of moving objects, calculus provides a robust framework for discovery and innovation.