Problem 17
Question
Find the derivative of each function by using the Product Rule. Simplify your answers. $$ f(x)=\left(2 x^{2}+1\right)(1-x) $$
Step-by-Step Solution
Verified Answer
The derivative of \( f(x) = (2x^2 + 1)(1 - x) \) is \( f'(x) = -6x^2 + 4x - 1 \).
1Step 1: Identify the functions
The function is given as \( f(x) = (2x^2 + 1)(1 - x) \). Identify the two separate functions that comprise the product. Here, the two functions are \( u(x) = 2x^2 + 1 \) and \( v(x) = 1 - x \).
2Step 2: Find the derivatives of the functions
Differentiate each of the functions separately. \( u(x) = 2x^2 + 1 \) and its derivative is \( u'(x) = 4x \). Similarly, \( v(x) = 1 - x \) and its derivative is \( v'(x) = -1 \).
3Step 3: Apply the Product Rule
The Product Rule states that the derivative of a product of two functions is \( (uv)' = u'v + uv' \). Substitute the derivatives and functions: \[ f'(x) = (4x)(1 - x) + (2x^2 + 1)(-1) \].
4Step 4: Expand and simplify the expression
Distribute in each of the terms and combine like terms: First term: \( (4x)(1) - (4x)(x) = 4x - 4x^2 \). Second term: \( -(2x^2 + 1) = -2x^2 - 1 \). Combine: \( f'(x) = 4x - 4x^2 - 2x^2 - 1 \).
5Step 5: Combine like terms
Combine the terms obtained in the previous step: \( f'(x) = -6x^2 + 4x - 1 \). Reorder to obtain the simplified derivative, \( f'(x) = -6x^2 + 4x - 1 \).
Key Concepts
Product RuleFunction DifferentiationSimplifying Algebraic Expressions
Product Rule
The Product Rule is an essential tool in calculus used to find the derivative of two multiplied functions. This rule becomes especially handy when dealing with products of functions that are not easy to simplify. The rule states that if you have a function that is the product of two other functions, say \( u(x) \) and \( v(x) \), then the derivative \((uv)'\) is given by:
- \( (uv)' = u'v + uv' \)
Function Differentiation
Differentiation is the process of finding the derivative of a function, which represents the rate of change or the slope of that function. When looking at our example with the function \( f(x)=(2x^2 + 1)(1-x) \), function differentiation involves separating \( f(x) \) into its components, differentiating each part, and applying rules like the Product Rule.
When you differentiate \( u(x) = 2x^2 + 1 \), the derivative is \( u'(x) = 4x \). This is calculated by applying basic differentiation rules for polynomials. Similarly, the derivative of \( v(x) = 1 - x \) is \( v'(x) = -1 \). Differentiation helps us understand how functions behave and change, which is crucial for solving problems in calculus, physics, engineering, and many other fields.
Often, functions are more complex than just simple polynomials, and techniques like the Product Rule are necessary to find their derivatives accurately. Practicing function differentiation will help you become more comfortable with identifying and applying the correct methods for various types of functions.
When you differentiate \( u(x) = 2x^2 + 1 \), the derivative is \( u'(x) = 4x \). This is calculated by applying basic differentiation rules for polynomials. Similarly, the derivative of \( v(x) = 1 - x \) is \( v'(x) = -1 \). Differentiation helps us understand how functions behave and change, which is crucial for solving problems in calculus, physics, engineering, and many other fields.
Often, functions are more complex than just simple polynomials, and techniques like the Product Rule are necessary to find their derivatives accurately. Practicing function differentiation will help you become more comfortable with identifying and applying the correct methods for various types of functions.
Simplifying Algebraic Expressions
Simplifying algebraic expressions is about making an expression easier to work with, often after applying calculus operations like the Product Rule. Once you apply the Product Rule to the given functions, you end up with an expanded form: \( f'(x) = 4x - 4x^2 - 2x^2 - 1 \).
The goal is to combine like terms to get a concise and manageable expression. In this case, the \( x^2 \) terms \(-4x^2 \) and \(-2x^2 \) are combined to give \(-6x^2 \). The final step is arranging the terms in a conventional order, usually from the highest power to the lowest. Therefore, we get: \( f'(x) = -6x^2 + 4x - 1 \).
Simplifying expressions in this way is crucial for clarity and precision, allowing for easier interpretation and further calculations. This practice also prevents errors in subsequent analysis or applications of the expression, making it a necessary skill for anyone studying calculus. The process highlights the importance of attention to detail and accurate arithmetic, as combining and rearranging terms incorrectly can lead to mistakes in final answers.
The goal is to combine like terms to get a concise and manageable expression. In this case, the \( x^2 \) terms \(-4x^2 \) and \(-2x^2 \) are combined to give \(-6x^2 \). The final step is arranging the terms in a conventional order, usually from the highest power to the lowest. Therefore, we get: \( f'(x) = -6x^2 + 4x - 1 \).
Simplifying expressions in this way is crucial for clarity and precision, allowing for easier interpretation and further calculations. This practice also prevents errors in subsequent analysis or applications of the expression, making it a necessary skill for anyone studying calculus. The process highlights the importance of attention to detail and accurate arithmetic, as combining and rearranging terms incorrectly can lead to mistakes in final answers.
Other exercises in this chapter
Problem 17
Find the following limits without using a graphing calculator or making tables. $$ \lim _{x \rightarrow 3} \sqrt{2} $$
View solution Problem 17
Find the derivative of each function. $$ f(x)=\frac{6}{\sqrt[3]{x}} $$
View solution Problem 18
Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=\sqrt{x^{6}+3 x-1} $$
View solution Problem 18
Find the second derivative of each function. $$ f(x)=\frac{x}{x-2} $$
View solution