Problem 10 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\). \(f(x)=\left(x^{2}+x\right)\left(x^{2}-x\right)\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 4x^3 - 2x \).

Step 1: Identify the Function Type

The function \( f(x) = (x^2 + x)(x^2 - x) \) is a product of two functions: \( u(x) = x^2 + x \) and \( v(x) = x^2 - x \). We'll use the product rule to find \( f'(x) \).

Step 2: Recall the Product Rule Formula

The product rule states that if \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \). We need to find the derivatives \( u'(x) \) and \( v'(x) \).

Step 3: Differentiate u(x)

The function \( u(x) = x^2 + x \). Differentiating term by term, we get: \[ u'(x) = 2x + 1 \]

Step 4: Differentiate v(x)

The function \( v(x) = x^2 - x \). Differentiating term by term, we get: \[ v'(x) = 2x - 1 \]

Step 5: Apply the Product Rule

Plug the derivatives \( u'(x) \) and \( v'(x) \) into the product rule formula: \[ f'(x) = (2x + 1)(x^2 - x) + (x^2 + x)(2x - 1) \] We'll expand and simplify this expression.

Step 6: Expand the Terms

First expand \( (2x + 1)(x^2 - x) \): \[ 2x(x^2) - 2x(x) + 1(x^2) - 1(x) = 2x^3 - 2x^2 + x^2 - x = 2x^3 - x^2 - x \] Now expand \( (x^2 + x)(2x - 1) \): \[ x^2(2x) - x^2(1) + x(2x) - x(1) = 2x^3 - x^2 + 2x^2 - x = 2x^3 + x^2 - x \]

Step 7: Simplify the Expression

Add the expanded terms from Step 6: \[ (2x^3 - x^2 - x) + (2x^3 + x^2 - x) = 4x^3 - 2x \]

Step 8: Verify the Simplified Derivative

The derivative is correct after combining like terms, so: \[ f'(x) = 4x^3 - 2x \]

Key Concepts

Product Rule

The product rule is an essential technique in calculus used to find the derivative of a product of two functions. When you have a function that is the multiplication of two separate functions, such as \( f(x) = u(x)v(x) \), the product rule helps to determine how the derivative, \( f'(x) \), behaves.

According to the product rule, the derivative of the product \( u(x)v(x) \) is given by:

• \( f'(x) = u'(x)v(x) + u(x)v'(x) \)

This means you take the derivative of the first function \( u(x) \), multiply it by the second function \( v(x) \), then add the product of the first function \( u(x) \) and the derivative of the second function \( v(x) \).

This rule is particularly useful when dealing with functions that cannot be simplified into a single polynomial or when their interaction as a product is crucial to the function's behavior.

Polynomial Functions

Polynomial functions are mathematical expressions involving sums of powers of variables with coefficients. In simple terms, they are sums of terms like \( ax^n \), where \( a \) is the coefficient, \( x \) is the variable, and \( n \) is a non-negative integer. These functions can be found everywhere in mathematics and are easy to manipulate while offering rich mathematical behaviors.

A polynomial, such as \( f(x) = x^2 + x \), consists of multiple terms with varying powers of \( x \). Each term added together makes the complete polynomial.

Polynomial functions have critical properties:

• They are continuous and smooth, with no abrupt changes or breaks.
• The degree of the polynomial is determined by the highest power, like \( x^2 \) in the function above.
• They include operations such as addition, subtraction, multiplication, and non-negative exponentiation of variables.

Understanding polynomials is key when applying the product rule, especially if the function can be broken into simpler polynomial components before differentiation.

Differentiation Rules

Differentiation is the process of finding the derivative, which represents the rate of change of a function. This is a foundational concept in calculus, allowing us to understand and express how functions behave, change, and relate to one another.

There are several fundamental differentiation rules you will frequently use:

• **Constant Rule**: The derivative of a constant is zero.
• **Power Rule**: For a function \( x^n \), the derivative is \( nx^{n-1} \).
• **Sum Rule**: The derivative of the sum of functions is the sum of their derivatives.
• **Difference Rule**: The derivative of the difference of functions is the difference of their derivatives.

When differentiating any complex function, these rules enable us to break down the function into manageable parts. By applying rules like the product rule or the power rule, you can find derivatives step-by-step efficiently.

In the context of the exercise, these rules allow us to compute derivatives of polynomial expressions and apply the product rule seamlessly to obtain the derivative \( f'(x) \). High proficiency in these rules is essential for handling a wide range of calculus problems.