Problem 40 - College Algebra and Calculus: An Applied Approach

Question

find \(f^{\prime}(x)\). $$ f(x)=\left(x^{2}+2 x\right)(x+1) $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = (x^{2}+2 x)(x+1)\) is \(f'(x) = 3x^2 + 4x\).

Step 1: Identify the Functions

We need to identify the two functions that make up the overall function \(f(x)\). Let's say that:\(u(x) = x^2 + 2x\) and \(v(x) = x + 1\). We now can apply the product rule.

Step 2: Derive Function \(u(x)\) and \(v(x)\)

Next, we derive the expressions for \(u(x)\) and \(v(x)\). The derivative of \(u(x) = x^2 + 2x\) is \(u'(x) = 2x + 2\) and the derivative of \(v(x) = x+1\) is \(v'(x) = 1\).

Step 3: Apply the Product Rule

After obtaining our derivative expressions, we can apply the product rule. The product rule is represented as \( (uv)' = u'v + uv' \). After filling in our functions and their derivatives, \( (x^2 + 2x)(x +1)'\) becomes \( (2x + 2 )(x+1) + (x^2 + 2x)(1)\).

Step 4: Simplify the Equation

Now, let's simplify the equation. This simplifies to \(2x^2 + 2x + x^2 + 2x = 3x^2 + 4x\).

Key Concepts

Product Rule

When differentiating a product of two functions, we utilize the Product Rule. This rule helps us efficiently find the derivative of the multiplication between two separate functions.
The Product Rule states: If you have two functions, say, \( u(x) \) and \( v(x) \), then the derivative of their product is given by: \[ (uv)' = u'v + uv' \]
Here are some key points to keep in mind:

• First, independently differentiate each function.
• Multiply the derivative of the first function with the second function.
• Add the product of the first function with the derivative of the second function.

The Product Rule can simplify the process significantly when working with expressions that involve products. We saw this in the original solution, where \( u(x) = x^2 + 2x \) and \( v(x) = x + 1 \). Each function was differentiated individually before applying the rule to find the derivative.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. It is used to compute the derivative, which is the primary focus when studying calculus problems such as those involving the Product Rule.
To differentiate a function, you must:

• Recognize the type of function you are dealing with (e.g., polynomial, trigonometric, etc.).
• Use the right differentiation rules for each type of function. For polynomials, like in the exercise, rules are simple and often involve reducing the power of a variable by one and multiplying by its original power.
• Calculate the derivative using these rules, which will tell you the slope or rate of change of the function at any point \( x \).

In our example, the function \( f(x) = (x^2 + 2x)(x+1) \) was differentiated by first breaking it into parts and applying differentiation rules accordingly. We derived \( u(x) = x^2 + 2x \) to get \( u'(x) = 2x + 2 \), showing how the rules of differentiation break down and solve seemingly complex problems.

Polynomial Functions

Polynomial functions are a class of functions that can be represented in the form of \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \), where all exponents are whole numbers, and coefficients are constants. They are pervasive in algebra and calculus exams due to their straightforward differentiability.
Key characteristics of polynomial functions include:

• They are continuous, with no gaps or jumps.
• The degree of the polynomial is the highest exponent in the equation.
• They allow for easy application of basic differentiation rules.

In our initial problem, both \( u(x) = x^2 + 2x \) and \( v(x) = x + 1 \) are polynomial functions. When differentiating these functions, we directly apply the power rule, which simplifies contrast calculations where the highest degree of the polynomial terms are known and easily manageable. This makes polynomial functions particularly friendly when learning the basics, like the Product Rule and differentiation overall.