Problem 12
Question
Find the derivative of each function by using the Product Rule. Simplify your answers. $$ f(x)=\left(x^{3}-1\right)\left(x^{3}+1\right) $$
Step-by-Step Solution
Verified Answer
The derivative is \( f'(x) = 6x^5 \).
1Step 1: Identify the function components
The given function is expressed as the product of two polynomials: \( u(x) = x^3 - 1 \) and \( v(x) = x^3 + 1 \). We will need these expressions in simplified form to apply the Product Rule.
2Step 2: State the Product Rule formula
The derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] where \( u'(x) \) and \( v'(x) \) are the derivatives of \( u(x) \) and \( v(x) \), respectively.
3Step 3: Calculate the derivatives of \( u(x) \) and \( v(x) \)
Determine the derivatives:- \( u'(x) = \frac{d}{dx}(x^3 - 1) = 3x^2 \) since the derivative of \( x^3 \) is \( 3x^2 \) and the derivative of a constant is \( 0 \).- \( v'(x) = \frac{d}{dx}(x^3 + 1) = 3x^2 \) using the same logic as for \( u'(x) \).
4Step 4: Apply the Product Rule
Substitute the function components and their derivatives into the Product Rule formula:\[ f'(x) = (3x^2)(x^3 + 1) + (x^3 - 1)(3x^2) \]This uses the values calculated in the previous step.
5Step 5: Simplify the expression
Expand and simplify the derivative expression:First, expand both terms:- \( 3x^2(x^3 + 1) = 3x^5 + 3x^2 \)- \( 3x^2(x^3 - 1) = 3x^5 - 3x^2 \) Now, combine like terms:- \( f'(x) = 3x^5 + 3x^2 + 3x^5 - 3x^2 \)- The simplified result is \( f'(x) = 6x^5 \).
Key Concepts
Derivatives of PolynomialsSimplifying ExpressionsCalculus Problem Solving
Derivatives of Polynomials
Understanding derivatives is a core aspect of calculus, especially when dealing with polynomials. To find the derivative of a polynomial, we apply the power rule. This rule states that if you have a term in the form of \( ax^n \), the derivative is \( n \cdot ax^{n-1} \).
For example, in the expression \( x^3 \), the derivative is \( 3x^2 \) because we multiply the exponent 3 by the coefficient 1 and then decrease the exponent by 1. Constants, like -1 or +1 in our expression \( x^3 - 1 \) or \( x^3 + 1 \), have a derivative of 0 since they don’t change as \( x \) changes.
The derivation of such expressions enables you to analyze how quantities change in relation to each other, a fundamental aspect of calculus.
For example, in the expression \( x^3 \), the derivative is \( 3x^2 \) because we multiply the exponent 3 by the coefficient 1 and then decrease the exponent by 1. Constants, like -1 or +1 in our expression \( x^3 - 1 \) or \( x^3 + 1 \), have a derivative of 0 since they don’t change as \( x \) changes.
The derivation of such expressions enables you to analyze how quantities change in relation to each other, a fundamental aspect of calculus.
- Applying the derivative to each part of a polynomial is straightforward with practice.
- Always ensure to apply the power rule correctly to avoid errors.
Simplifying Expressions
Simplifying expressions is essential in calculus to keep your results neat and understandable. Many expressions, once expanded, have like terms that can be combined.
In our example, we expanded \( 3x^2(x^3 + 1) \) and \( 3x^2(x^3 - 1) \) during the application of the Product Rule which resulted in the terms \( 3x^5 + 3x^2 \) and \( 3x^5 - 3x^2 \). By combining like terms, namely \( 3x^2 + 3x^2 \) and \( 3x^5 + 3x^5 \), we simplified the final expression to \( 6x^5 \).
This step ensures that your calculations remain consistent and clear, showcasing the changes concisely.
In our example, we expanded \( 3x^2(x^3 + 1) \) and \( 3x^2(x^3 - 1) \) during the application of the Product Rule which resulted in the terms \( 3x^5 + 3x^2 \) and \( 3x^5 - 3x^2 \). By combining like terms, namely \( 3x^2 + 3x^2 \) and \( 3x^5 + 3x^5 \), we simplified the final expression to \( 6x^5 \).
This step ensures that your calculations remain consistent and clear, showcasing the changes concisely.
- After expanding expressions, look for like terms that can be combined.
- Simplification often reveals the most elegant form of your derivative.
Calculus Problem Solving
Solving calculus problems often involves several key steps, from understanding the problem to finding useful methods for solutions. The Product Rule, used in this case, is a critical tool for differentiating products of functions.
When faced with derivatives involving function products, always start by identifying individual functions. As we did with \( u(x) = x^3 - 1 \) and \( v(x) = x^3 + 1 \). Once you've done this, calculate the derivatives of these components separately and then apply the Product Rule formula.
This systematic approach ensures accurate, efficient problem-solving:
When faced with derivatives involving function products, always start by identifying individual functions. As we did with \( u(x) = x^3 - 1 \) and \( v(x) = x^3 + 1 \). Once you've done this, calculate the derivatives of these components separately and then apply the Product Rule formula.
This systematic approach ensures accurate, efficient problem-solving:
- Carefully identify and separate function components in the expression.
- Apply known differentiation rules accurately, like the Product Rule in this problem.
- Simplify your results for clarity, reducing the chance of errors in interpretation.
Other exercises in this chapter
Problem 12
a. Show that the definition of the derivative applied to the function \(f(x)=\sqrt[3]{x}\) at \(x=0\) gives \(f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\sqrt[
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Find the derivative of each function. $$ h(x)=\frac{4}{x^{3}} $$
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Use the Generalized Power Rule to find the derivative of each function. $$ g(x)=\left(2 x^{2}-7 x+3\right)^{4} $$
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Find the average rate of change of the given function between the following pairs of \(x\) -values. [Hint: See pages 95-96.] a. \(x=3\) and \(x=5\) b. \(x=3\) a
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