Calculus, a branch of mathematics, is a powerful tool used to study the behavior of functions and their applications in various fields, including physics, engineering, and economics. One of the fundamental concepts in calculus is differentiation, which deals with the study of rates of change and slopes of curves. In this article, we will focus on mastering the differentiation of a specific function, xe^x, a crucial concept for anyone looking to deepen their understanding of calculus.
The function xe^x is a product of two functions, x and e^x, where e^x is the exponential function. The differentiation of such a function requires the application of the product rule, a fundamental rule in calculus. Understanding how to differentiate xe^x not only helps in solving specific problems but also enhances one's ability to tackle more complex functions.
To differentiate xe^x, we apply the product rule, which states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). For xe^x, let u(x) = x and v(x) = e^x. The derivatives of u(x) and v(x) are u'(x) = 1 and v'(x) = e^x, respectively.
Differentiating xe^x Using the Product Rule
Applying the product rule:
\[ \frac{d}{dx}(xe^x) = \frac{d}{dx}(x) \cdot e^x + x \cdot \frac{d}{dx}(e^x) \] \[ = 1 \cdot e^x + x \cdot e^x \] \[ = e^x + xe^x \] \[ = e^x(1 + x) \]
This result, e^x(1 + x), is the derivative of xe^x. It shows that the rate of change of the function xe^x at any point x is given by e^x(1 + x).
Understanding the Derivative
The derivative e^x(1 + x) can be analyzed to understand the behavior of the function xe^x. For instance, to find the critical points, we set the derivative equal to zero:
\[ e^x(1 + x) = 0 \]
Since e^x > 0 for all x, the only solution occurs when 1 + x = 0, or x = -1. This critical point can help in understanding the intervals where the function xe^x is increasing or decreasing.
| Interval | Sign of e^x(1 + x) | Behavior of xe^x |
|---|---|---|
| (-\infty, -1) | Negative | Decreasing |
| (-1, \infty) | Positive | Increasing |
Key Points
- The derivative of xe^x is e^x(1 + x), found using the product rule.
- The critical point of xe^x is x = -1, where the function changes from decreasing to increasing.
- Understanding derivatives is essential for analyzing the behavior of functions and solving optimization problems.
- The function xe^x and its derivative have applications in various fields, including physics and engineering.
- Mastering the differentiation of xe^x enhances one's understanding of calculus and its applications.
Applications and Further Analysis
The function xe^x and its derivative have significant applications in physics, engineering, and other fields. For example, in electrical engineering, e^x can represent a voltage or current in a circuit, and x can be a parameter that modifies this voltage or current. Understanding how xe^x changes with x is crucial for designing and analyzing circuits.
In conclusion, mastering the differentiation of xe^x is a fundamental skill in calculus that has far-reaching implications in various fields. By applying the product rule and analyzing the derivative, we gain insights into the behavior of the function and its applications.
What is the derivative of (xe^x)?
+The derivative of (xe^x) is (e^x(1 + x)).
How is the product rule applied to differentiate (xe^x)?
+The product rule states that if (f(x) = u(x)v(x)), then (fā(x) = uā(x)v(x) + u(x)vā(x)). For (xe^x), let (u(x) = x) and (v(x) = e^x), so (uā(x) = 1) and (vā(x) = e^x). Applying the product rule: (\frac{d}{dx}(xe^x) = 1 \cdot e^x + x \cdot e^x = e^x + xe^x = e^x(1 + x)).
What is the critical point of (xe^x)?
+The critical point of (xe^x) is (x = -1), found by setting the derivative (e^x(1 + x) = 0).