Derivative Of Square Root Of 2x

Understanding the Derivative of the Square Root of 2x

The derivative of a function describes its instantaneous rate of change at any given point. Understanding how to find the derivative, especially for functions involving roots, is fundamental in calculus and has wide-ranging applications in various fields like physics, engineering, and economics. This comprehensive guide will meticulously walk you through finding the derivative of √(2x), exploring different approaches and solidifying your understanding of the underlying concepts.

Understanding the Problem: d/dx [√(2x)]

Our goal is to find the derivative of the function f(x) = √(2x) with respect to x. This is denoted as d/dx [√(2x)] or f'(x). Before diving into the solution, let's refresh some crucial calculus concepts.

Key Calculus Concepts:

• The Power Rule: This is a cornerstone of differentiation. The power rule states that the derivative of xⁿ is nxⁿ⁻¹. We'll use this extensively.
• The Chain Rule: When dealing with composite functions (functions within functions), the chain rule is indispensable. It states that the derivative of f(g(x)) is f'(g(x)) * g'(x). We will apply this rule to the problem at hand.
• Constant Multiple Rule: The derivative of a constant multiplied by a function is simply the constant multiplied by the derivative of the function. This is expressed as d/dx[cf(x)] = c * d/dx[f(x)], where 'c' is a constant.

Method 1: Using the Power Rule and Chain Rule

This method is the most straightforward approach. We'll rewrite the function to make it easier to apply the power rule.

1. Rewrite the function: We can rewrite √(2x) as (2x)^(1/2). This form allows us to directly apply the power rule.

2. Apply the Chain Rule: Since we have a composite function ((2x) raised to the power of 1/2), we must use the chain rule. Let's break it down:

   • Outer function: u^(1/2), where u = 2x
   • Inner function: u = 2x

3. Differentiate the outer function: The derivative of u^(1/2) with respect to u is (1/2)u^(-1/2).

4. Differentiate the inner function: The derivative of 2x with respect to x is 2.

5. Apply the Chain Rule: Multiply the derivative of the outer function by the derivative of the inner function:

   (1/2)u^(-1/2) * 2 = u^(-1/2)

6. Substitute back: Remember that u = 2x. Substituting this back into our expression, we get:

   (2x)^(-1/2) = 1 / √(2x)

Therefore, the derivative of √(2x) is 1 / √(2x).

Method 2: Implicit Differentiation

This method provides an alternative approach, particularly useful when dealing with more complex functions.

1. Let y = √(2x). This sets up our implicit equation.

2. Square both sides: This eliminates the square root, giving us y² = 2x.

3. Differentiate both sides with respect to x: This leads to:

   2y(dy/dx) = 2

4. Solve for dy/dx: Divide both sides by 2y:

   dy/dx = 1/y

5. Substitute back: Recall that y = √(2x). Substituting this back, we get:

   dy/dx = 1 / √(2x)

Again, we arrive at the same result: the derivative of √(2x) is 1 / √(2x).

Method 3: Using the Definition of the Derivative

This method reinforces the fundamental definition of the derivative using limits. While more computationally intensive, it provides a deeper understanding.

The definition of the derivative is:

f'(x) = lim (h→0) [(f(x + h) - f(x)) / h]

Let's apply this to our function f(x) = √(2x):

1. Substitute f(x) and f(x + h):

   f'(x) = lim (h→0) [√(2(x + h)) - √(2x)] / h

2. Rationalize the numerator: Multiply the numerator and denominator by the conjugate of the numerator:

   f'(x) = lim (h→0) [√(2(x + h)) - √(2x)] / h * [√(2(x + h)) + √(2x)] / [√(2(x + h)) + √(2x)]

3. Simplify: After simplifying the numerator, we get:

   f'(x) = lim (h→0) [2(x + h) - 2x] / [h(√(2(x + h)) + √(2x))]

4. Cancel out terms and simplify further:

   f'(x) = lim (h→0) 2h / [h(√(2(x + h)) + √(2x))]

5. Cancel h: We can cancel out the 'h' terms:

   f'(x) = lim (h→0) 2 / [√(2(x + h)) + √(2x)]

6. Evaluate the limit: As h approaches 0, the expression becomes:

   f'(x) = 2 / [√(2x) + √(2x)] = 2 / [2√(2x)] = 1 / √(2x)

Once more, we obtain the derivative as 1 / √(2x).

Applications and Significance

The derivative of √(2x) has several practical applications across various fields:

• Physics: Calculating instantaneous velocity or acceleration from a position function involving square roots.
• Economics: Determining marginal cost or marginal revenue in optimization problems.
• Engineering: Analyzing rates of change in systems described by square root functions.
• Geometry: Finding the slope of a tangent line to a curve defined by a square root function at a specific point.

Understanding its derivative allows us to analyze the rate of change of this function at any given point, enabling a deeper comprehension of the underlying process or phenomenon it represents.

Understanding the Domain and Range

The original function, f(x) = √(2x), has a domain of x ≥ 0, since you cannot take the square root of a negative number. The range of this function is y ≥ 0.

The derivative, f'(x) = 1/√(2x), shares a similar domain restriction: x > 0. Note that x cannot equal 0 because division by zero is undefined. The range of the derivative is y > 0. The derivative is always positive, indicating that the original function is always increasing for x > 0.

Conclusion

We have successfully explored three distinct methods for deriving the derivative of √(2x), each offering a valuable perspective on differential calculus. The result consistently proves to be 1/√(2x). Understanding these approaches, along with their applications, is crucial for anyone studying calculus or applying calculus to real-world problems. Remember to always consider the domain and range of both the original function and its derivative to fully grasp the implications of your findings. The techniques demonstrated here form a solid foundation for tackling more complex differentiation problems in the future. By mastering these fundamental concepts, you'll be well-equipped to confidently navigate the world of calculus and its diverse applications.