Derivative Of X 2e X

Unveiling the Mystery: Finding the Derivative of x²eˣ

Finding the derivative of a function is a cornerstone of calculus. This article delves into the process of differentiating x²eˣ, a function combining polynomial and exponential terms, providing a comprehensive understanding suitable for students and enthusiasts alike. We'll explore the necessary rules, demonstrate the step-by-step solution, and address frequently asked questions. By the end, you'll not only know the derivative but also grasp the underlying principles and techniques. This detailed explanation will equip you to tackle similar problems with confidence.

Introduction: Understanding the Fundamentals

Before diving into the specifics of finding the derivative of x²eˣ, let's refresh our understanding of some fundamental calculus concepts. This will lay a solid foundation for the problem at hand.

• Derivatives: The derivative of a function measures its instantaneous rate of change at a given point. Geometrically, it represents the slope of the tangent line to the function's graph at that point.

• The Product Rule: This rule is crucial when dealing with functions that are the product of two or more simpler functions. The product rule states: If f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). In simpler terms, the derivative of a product is the derivative of the first function multiplied by the second, plus the first function multiplied by the derivative of the second.

• Derivatives of Elementary Functions: We need to know the derivatives of basic functions such as polynomials and exponential functions. The derivative of xⁿ is nxⁿ⁻¹, and the derivative of eˣ is eˣ.

Now, armed with these basics, we're ready to tackle the derivative of x²eˣ.

Step-by-Step Differentiation of x²eˣ

The function x²eˣ is a product of two functions: u(x) = x² and v(x) = eˣ. We will apply the product rule to find its derivative.

1. Identify u(x) and v(x):

• u(x) = x²
• v(x) = eˣ

2. Find the derivatives of u(x) and v(x):

• u'(x) = d(x²)/dx = 2x (using the power rule)
• v'(x) = d(eˣ)/dx = eˣ (the derivative of eˣ is itself)

3. Apply the Product Rule:

Recall the product rule: f'(x) = u'(x)v(x) + u(x)v'(x)

Substituting our values:

f'(x) = (2x)(eˣ) + (x²)(eˣ)

4. Simplify the Result:

We can factor out eˣ from both terms:

f'(x) = eˣ(2x + x²)

Therefore, the derivative of x²eˣ is eˣ(2x + x²). This can also be written as x eˣ (x + 2). Both forms are correct and equivalent.

A Deeper Dive: Understanding the Implications

The derivative, eˣ(2x + x²), provides valuable information about the original function, x²eˣ. Let's explore some of these implications:

• Rate of Change: The derivative tells us the instantaneous rate of change of x²eˣ at any given point x. For example, at x = 1, the rate of change is e¹(2(1) + 1²) = 3e. This means the function is increasing rapidly at x = 1.

• Critical Points: Critical points occur where the derivative is zero or undefined. Setting f'(x) = 0, we get eˣ(2x + x²) = 0. Since eˣ is always positive, this simplifies to 2x + x² = 0, which factors to x(x + 2) = 0. Thus, the critical points are x = 0 and x = -2. These points are potential locations for local maxima or minima.

• Concavity: The second derivative, f''(x), provides information about the concavity of the function. By finding the second derivative and analyzing its sign, we can determine where the graph of x²eˣ is concave up or concave down. This helps us understand the shape of the curve more completely.

Let's calculate the second derivative:

f'(x) = eˣ(2x + x²)

To find the second derivative, we'll use the product rule again:

u(x) = eˣ, u'(x) = eˣ v(x) = 2x + x², v'(x) = 2 + 2x

f''(x) = u'(x)v(x) + u(x)v'(x) = eˣ(2x + x²) + eˣ(2 + 2x) = eˣ(x² + 4x + 2)

Analyzing the second derivative, we find that f''(x) = 0 when x² + 4x + 2 = 0. Solving this quadratic equation (using the quadratic formula) gives us two points of inflection. The sign of f''(x) in the intervals determined by these points of inflection will indicate whether the function is concave up or concave down in those regions.

This in-depth analysis demonstrates the power of derivatives in understanding the behavior and characteristics of functions.

Extending the Concept: Higher-Order Derivatives

The process of differentiation can be repeated to find higher-order derivatives. For example, the third derivative, f'''(x), would be found by differentiating f''(x). These higher-order derivatives offer even more nuanced insights into the function's behavior, particularly in fields like physics and engineering.

Finding the third derivative of f(x) = x²eˣ:

We've already found: f''(x) = eˣ(x² + 4x + 2)

Applying the product rule again:

u(x) = eˣ, u'(x) = eˣ v(x) = x² + 4x + 2, v'(x) = 2x + 4

f'''(x) = eˣ(x² + 4x + 2) + eˣ(2x + 4) = eˣ(x² + 6x + 6)

Frequently Asked Questions (FAQ)

Q1: Why is the product rule necessary here?

A1: The product rule is essential because x²eˣ is a product of two distinct functions, x² and eˣ. We cannot simply differentiate each term separately; the product rule accounts for the interaction between these two functions.

Q2: Can I use other differentiation techniques?

A2: While the product rule is the most straightforward approach for this specific problem, more advanced techniques like logarithmic differentiation could be employed, but they are unnecessarily complex for this case. The product rule offers the most efficient and easily understood solution.

Q3: What if the exponent was different? For example, x²e^(2x)?

A3: The process remains similar. You would still use the product rule. The derivative of e^(2x) is 2e^(2x) (using the chain rule). The solution would involve a slightly more complex simplification but the core principle remains the same.

Q4: What are the practical applications of this derivative?

A4: Derivatives, and this specific example, find applications in various fields. In physics, they can describe the rate of change of velocity (acceleration). In economics, they help model marginal cost or revenue. In computer science, derivatives are used in optimization algorithms. The specific application depends on the context in which the function x²eˣ is being used.

Q5: How can I check my answer?

A5: You can use online calculus tools or software to verify your derivative. Many calculators and computer algebra systems can compute derivatives. Comparing your result with the output from a reliable source will help confirm your understanding and identify any potential errors.

Conclusion: Mastering the Derivative of x²eˣ and Beyond

This comprehensive guide has walked you through the process of finding the derivative of x²eˣ, highlighting the application of the product rule and delving into the implications of the result. We've explored how the derivative provides critical information about the function's behavior, including its rate of change, critical points, and concavity. Furthermore, we've expanded on the concept to include higher-order derivatives and addressed common questions. Remember, practice is key to mastering calculus. By working through similar problems and understanding the underlying principles, you'll build a strong foundation in differential calculus. The ability to differentiate functions like x²eˣ is a valuable skill with wide-ranging applications in various fields.