How To Differentiate X Ln X

How to Differentiate x ln x: A Comprehensive Guide

The derivative of a function describes its instantaneous rate of change. Understanding how to differentiate functions is fundamental in calculus and has broad applications in various fields, from physics and engineering to economics and finance. This comprehensive guide will delve into the differentiation of the function x ln x, exploring various methods, providing detailed explanations, and highlighting common pitfalls to avoid. We’ll also discuss the broader context of logarithmic differentiation and its usefulness in tackling more complex problems.

Understanding the Product Rule

The function x ln x is a product of two functions: x and ln x. Therefore, we cannot simply differentiate each term individually. Instead, we must utilize the product rule of differentiation. The product rule states:

d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

Where:

• f(x) and g(x) are differentiable functions.
• f'(x) and g'(x) are their respective derivatives.

In our case:

• f(x) = x
• g(x) = ln x

Finding the Derivatives of f(x) and g(x)

Before applying the product rule, we need to find the derivatives of f(x) and g(x):

1. Derivative of f(x) = x:

The derivative of x with respect to x is simply 1. This is a basic power rule derivative:

f'(x) = d/dx (x) = 1

2. Derivative of g(x) = ln x:

The derivative of the natural logarithm function ln x is 1/x:

g'(x) = d/dx (ln x) = 1/x

Applying the Product Rule to x ln x

Now, we can apply the product rule using the derivatives we just calculated:

d/dx (x ln x) = f'(x)g(x) + f(x)g'(x)

Substituting our values for f'(x), g(x), f(x), and g'(x):

d/dx (x ln x) = (1)(ln x) + (x)(1/x)

This simplifies to:

d/dx (x ln x) = ln x + 1

Therefore, the derivative of x ln x is ln x + 1.

Illustrative Examples and Applications

Let's explore some examples to solidify our understanding:

Example 1: Finding the slope of the tangent line.

Find the slope of the tangent line to the curve y = x ln x at x = e.

Solution: The slope of the tangent line is given by the derivative evaluated at x = e. We know that d/dx (x ln x) = ln x + 1. Substituting x = e:

Slope = ln(e) + 1 = 1 + 1 = 2

The slope of the tangent line at x = e is 2.

Example 2: Optimization problems.

Suppose the cost function of a company is given by C(x) = x ln x, where x is the number of units produced. Find the marginal cost when x = 5.

Solution: The marginal cost is the derivative of the cost function. Using our derived formula:

Marginal Cost = dC/dx = ln x + 1

Substituting x = 5:

Marginal Cost = ln(5) + 1 ≈ 1.609 + 1 = 2.609

The marginal cost when producing 5 units is approximately 2.609.

Example 3: Related Rates Problems.

If the area of a rectangle is given by A(x) = x ln x, where x is the length of one side, and x is increasing at a rate of 2 units per second, find the rate of change of the area when x = 3.

Solution: We need to find dA/dt. Using the chain rule: dA/dt = (dA/dx)(dx/dt).

We know dA/dx = ln x + 1 and dx/dt = 2.

When x = 3:

dA/dt = (ln 3 + 1)(2) ≈ (1.099 + 1)(2) ≈ 4.198

The area is increasing at a rate of approximately 4.198 square units per second.

Beyond the Basics: Logarithmic Differentiation

The product rule is sufficient for differentiating x ln x. However, for more complex functions involving products, quotients, and powers of x, logarithmic differentiation offers a powerful alternative. Logarithmic differentiation simplifies the differentiation process by utilizing the properties of logarithms.

Consider a function of the form y = f(x)^g(x). Directly differentiating this using the product or chain rule can be cumbersome. Logarithmic differentiation streamlines the process:

1. Take the natural logarithm of both sides: ln y = g(x) ln[f(x)]

2. Differentiate implicitly with respect to x: (1/y) (dy/dx) = g'(x) ln[f(x)] + g(x) [f'(x)/f(x)]

3. Solve for dy/dx: dy/dx = y [g'(x) ln[f(x)] + g(x) f'(x)/f(x)]

4. Substitute the original function for y: Replace y with f(x)^g(x) in the expression for dy/dx.

Common Mistakes to Avoid

Several common mistakes can hinder the accurate differentiation of x ln x and similar functions:

• Forgetting the product rule: Attempting to differentiate x and ln x separately and then adding them is incorrect.
• Incorrect derivative of ln x: Remember that the derivative of ln x is 1/x, not x.
• Algebraic errors: Carefully simplify the expression after applying the product rule to avoid errors.
• Misunderstanding logarithmic differentiation: Ensure you correctly apply the rules of logarithms and implicit differentiation when using this technique.

Conclusion

Differentiating x ln x, while seemingly straightforward, serves as a valuable introduction to fundamental differentiation techniques. Mastering the product rule and understanding the application of these principles in various problem scenarios, such as optimization and related rates, is crucial for developing proficiency in calculus. Furthermore, exploring logarithmic differentiation expands our ability to handle more intricate functions, strengthening our understanding of calculus’ versatility and practical significance across diverse fields. Through practice and a thorough understanding of these concepts, you'll confidently navigate the world of differentiation.