Derivative Of X Square Root Of X

Finding the Derivative of x√x: A Comprehensive Guide

The seemingly simple function, x√x, presents a valuable opportunity to explore several differentiation techniques. This comprehensive guide will not only provide the solution but also delve into the underlying principles, offering various approaches and solidifying your understanding of calculus. We'll cover different methods, explain the rationale behind each step, and touch upon related concepts for a complete and thorough understanding.

Understanding the Function: x√x

Before we delve into the derivative, let's first understand the function itself. x√x represents the product of x and the square root of x. Mathematically, we can rewrite this as:

f(x) = x√x = x * x^(1/2)

Using the rules of exponents, we can simplify this to:

f(x) = x^(3/2)

This simpler form will be significantly easier to work with when calculating the derivative.

Method 1: Power Rule of Differentiation

The most straightforward method to find the derivative of x^(3/2) is by applying the power rule. The power rule states that the derivative of xⁿ is nx^n-1, where n is a constant.

Let's apply this rule to our function:

f(x) = x^(3/2)

The derivative, denoted as f'(x) or df/dx, is:

f'(x) = (3/2)x^((3/2)-1)

Simplifying, we get:

f'(x) = (3/2)x^(1/2)

Or, more simply:

f'(x) = (3/2)√x

This is the derivative of x√x. The power rule is efficient and effective, particularly for functions expressed as simple powers of x.

Method 2: Product Rule of Differentiation

While the power rule provided a concise solution, we can also tackle this using the product rule, which is essential for understanding derivatives of more complex functions involving products of functions. The product rule states that the derivative of a product of two functions, u(x) and v(x), is given by:

d(uv)/dx = u(dv/dx) + v(du/dx)

Let's define our function as:

u(x) = x and v(x) = √x = x^(1/2)

Now, we find the derivatives of u(x) and v(x):

du/dx = 1

dv/dx = (1/2)x^(-1/2) = 1/(2√x)

Applying the product rule:

d(uv)/dx = x * (1/(2√x)) + √x * 1

d(uv)/dx = x/(2√x) + √x

To simplify, we can multiply the first term's numerator and denominator by √x:

d(uv)/dx = (x√x)/(2x) + √x

Simplifying further:

d(uv)/dx = (√x)/2 + √x

Combining the terms:

d(uv)/dx = (3/2)√x

Again, we arrive at the same derivative: f'(x) = (3/2)√x This demonstrates the versatility of calculus; multiple methods can often lead to the same correct answer.

Method 3: Chain Rule (Implicit Differentiation)

Although less direct for this specific problem, understanding how the chain rule applies is crucial for more complex scenarios. We can rewrite the function as:

y = x√x

Then, we can take the natural logarithm of both sides:

ln(y) = ln(x√x) = ln(x) + ln(√x) = ln(x) + (1/2)ln(x) = (3/2)ln(x)

Now we can differentiate both sides implicitly with respect to x:

(1/y) * (dy/dx) = (3/2)(1/x)

Solving for dy/dx:

dy/dx = y * (3/(2x))

Substituting y = x√x:

dy/dx = x√x * (3/(2x))

Simplifying:

dy/dx = (3/2)√x

Once more, we obtain the derivative: f'(x) = (3/2)√x

This method, although more involved, showcases the utility of logarithmic differentiation and the chain rule.

Applications and Significance of the Derivative

Understanding the derivative of x√x has applications across various fields. The derivative represents the instantaneous rate of change of the function. In practical terms:

• Economics: If x represents the quantity of a product and x√x represents the revenue, the derivative would give the marginal revenue – the additional revenue generated by producing one more unit.
• Physics: If x represents time and x√x represents distance, the derivative would represent the instantaneous velocity.
• Engineering: The derivative could represent the rate of change of a physical quantity like temperature or pressure in a system described by the function x√x.

The derivative helps us analyze how the function behaves and provides critical information about its slope at any given point.

Exploring Higher-Order Derivatives

We can further analyze the function by finding its higher-order derivatives. The second derivative, denoted as f''(x) or d²f/dx², represents the rate of change of the first derivative.

For our function, f'(x) = (3/2)√x = (3/2)x^(1/2), the second derivative is:

f''(x) = (3/2) * (1/2)x^(-1/2) = (3/4)x^(-1/2) = 3/(4√x)

The second derivative provides information about the concavity of the function – whether it's curving upwards or downwards. Higher-order derivatives provide increasingly nuanced information about the function's behavior.

Dealing with Potential Errors

When working with derivatives, especially involving roots, it's crucial to be mindful of potential errors:

• Domain Restrictions: The original function x√x and its derivative (3/2)√x are only defined for non-negative values of x (x ≥ 0). This needs to be considered when applying the derivative in real-world applications.
• Simplification Errors: Carefully simplify algebraic expressions to avoid mistakes. Double-check your work at each step, ensuring that all operations are performed correctly.
• Incorrect Application of Rules: Remember the specific rules of differentiation (power rule, product rule, chain rule) and apply them precisely to avoid errors.

Conclusion

Finding the derivative of x√x, although seemingly simple, offers a valuable opportunity to reinforce fundamental calculus concepts. Whether you use the power rule, the product rule, or implicit differentiation, the result remains the same: (3/2)√x. Understanding the various methods and their applications enhances your problem-solving skills and prepares you to tackle more complex derivative problems. Remember to always consider the context of the problem and be mindful of potential errors in your calculations. The careful application of these techniques is crucial to successfully navigating the world of calculus and its many practical applications. The deep understanding gained from this detailed exploration will serve as a solid foundation for tackling more challenging derivatives in the future.