Chapter 6: Problem 4
Decide if the function is an antiderivative of \(f(x)=2 e^{2 x}\) $$F(x)=x e^{2 x}$$
Short Answer
Expert verified
No, \( F(x) = xe^{2x} \) is not an antiderivative of \( f(x) = 2e^{2x} \).
Step by step solution
01
Differentiate F(x)
To determine if \( F(x) = x e^{2x} \) is an antiderivative of \( f(x) = 2e^{2x} \), we first need to differentiate \( F(x) \) with respect to \( x \). We'll use the product rule which states \( (uv)' = u'v + uv' \).Let \( u = x \) and \( v = e^{2x} \).Thus, \( u' = 1 \) and \( v' = 2e^{2x} \) (since the derivative of \( e^{2x} \) is \( 2e^{2x} \)).Using the product rule, the derivative of \( F(x) \) is given by:\( F'(x) = u'v + uv' = (1)(e^{2x}) + (x)(2e^{2x}) = e^{2x} + 2xe^{2x} \).
02
Compare F'(x) with f(x)
Now, we compare \( F'(x) \) with \( f(x) \).We have just calculated that \( F'(x) = e^{2x} + 2xe^{2x} \).Given \( f(x) = 2e^{2x} \), we can see that \( F'(x) eq f(x) \) as \( e^{2x} + 2xe^{2x} \) is not equivalent to \( 2e^{2x} \).
03
Conclusion
Since \( F'(x) eq f(x) \), the function \( F(x) = xe^{2x} \) is not an antiderivative of \( f(x) = 2e^{2x} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a central concept in calculus, which involves finding the derivative of a function. The derivative of a function at a given point describes the rate at which the function's value changes as the input changes. In essence, differentiation reveals how a function "behaves."
To differentiate a function, we apply certain rules and techniques, depending on the type of functions involved. For the function \( F(x) = x e^{2x} \), differentiation helps us understand its rate of change. By determining the derivative \( F'(x) \), we uncover how \( F(x) \) evolves as \( x \) shifts.
To differentiate a function, we apply certain rules and techniques, depending on the type of functions involved. For the function \( F(x) = x e^{2x} \), differentiation helps us understand its rate of change. By determining the derivative \( F'(x) \), we uncover how \( F(x) \) evolves as \( x \) shifts.
- Understand the behavior of functions.
- Determine whether a function is an antiderivative.
- Use rules like the product rule for complex functions.
Product Rule
The product rule is a handy tool for differentiation when dealing with functions that are the product of two functions. Presented as \((uv)' = u'v + uv'\), this rule enables us to differentiate products seamlessly. To apply it accurately, we first identify the individual components.
In the example of \( F(x) = x e^{2x} \), we can set \( u = x \) and \( v = e^{2x} \). Differentiating these separately, \( u' = 1 \) and \( v' = 2e^{2x} \), involves applying basic derivative rules.
In the example of \( F(x) = x e^{2x} \), we can set \( u = x \) and \( v = e^{2x} \). Differentiating these separately, \( u' = 1 \) and \( v' = 2e^{2x} \), involves applying basic derivative rules.
- Differentiate each component separately.
- Calculate \( u'v + uv' \) to find the derivative of the product.
- Combine results to get the differentiated product.
Exponential Function
Exponential functions are a key type of function in mathematics, represented by \( e^{kx} \), where \( e \) is the base of the natural logarithm, approximately 2.718, and \( k \) is a constant. These functions have unique properties in calculus, especially in differentiation. When we differentiate an exponential function like \( e^{2x} \), the derivative is \( 2e^{2x} \), the coefficient \( 2 \) being crucial.
Understanding this pattern simplifies differentiating complex functions involving exponentials. In our task with \( F(x) = x e^{2x} \), recognizing exponential differentiation aids in correctly applying the product rule.
Understanding this pattern simplifies differentiating complex functions involving exponentials. In our task with \( F(x) = x e^{2x} \), recognizing exponential differentiation aids in correctly applying the product rule.
- Recognize the structure of exponential functions.
- Identify constant coefficients and their impact on derivatives.
- Leverage exponential properties in calculations.