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Last updated on July 21st, 2025

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Derivative of Root 2x

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We use the derivative of √(2x), which is 1/√(2x), as a tool for understanding how the square root function changes in response to a slight change in x. Derivatives are essential in calculating changes in various real-life situations. We will now discuss the derivative of √(2x) in detail.

Derivative of Root 2x for Australian Students
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What is the Derivative of Root 2x?

To find the derivative of √(2x), we represent it as d/dx (√(2x)) or (√(2x))'. The derivative is 1/√(2x), indicating that the function is differentiable within its domain.

 

The key concepts involved include: -

 

Square Root Function: √(2x). 

 

Chain Rule: Used for differentiating composite functions like √(2x). 

 

Derivative of x^n: The general rule for differentiating functions with exponents.

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Derivative of Root 2x Formula

The derivative of √(2x) can be denoted as d/dx (√(2x)) or (√(2x))'. The formula we use to differentiate √(2x) is: d/dx (√(2x)) = 1/√(2x) The formula is applicable for all x > 0, where the square root function is defined.

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Proofs of the Derivative of Root 2x

We can derive the derivative of √(2x) using proofs. To show this, we utilize the rules of differentiation. Here are some methods used to prove this:

 

By First Principle

 

The derivative of √(2x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

Consider f(x) = √(2x). Its derivative can be expressed as the following limit: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given f(x) = √(2x), then f(x + h) = √(2(x + h)).

Substituting these into the equation: f'(x) = limₕ→₀ [√(2(x + h)) - √(2x)] / h Using the identity a - b = (a^2 - b^2)/(a + b)

for simplification: = limₕ→₀ (2(x + h) - 2x) / [h(√(2(x + h)) + √(2x))] = limₕ→₀ [2h] / [h(√(2(x + h)) + √(2x))] = limₕ→₀ 2 / (√(2(x + h)) + √(2x))

As h approaches 0, we get: f'(x) = 2 / (2√(2x)) = 1/√(2x)

Hence, proved.

 

Using Chain Rule

 

To prove the differentiation of √(2x) using the chain rule,

 

let: y = √(2x) = (2x)^(1/2)

 

By the chain rule: d/dx (2x)^(1/2) = (1/2)(2x)^(-1/2) * (d/dx (2x)) = (1/2)(2x)^(-1/2) * 2 = 1/√(2x)

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Higher-Order Derivatives of Root 2x

When a function is differentiated multiple times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a bit complex.

 

To understand them better, consider a scenario where the speed changes (first derivative), and the rate at which the speed changes (second derivative) also varies. Higher-order derivatives make it easier to understand functions like √(2x).

 

For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a specific point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), results from the second derivative, and this pattern continues.

 

For the nth Derivative of √(2x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.

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Special Cases:

When x is 0, the derivative is undefined because √(2x) has an undefined point at x = 0. For any positive x, the derivative of √(2x) = 1/√(2x), which varies with different values of x.

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Common Mistakes and How to Avoid Them in Derivatives of Root 2x

Students often make mistakes when differentiating √(2x). These mistakes can be avoided by understanding the correct solutions. Here are a few common mistakes and ways to solve them:

Mistake 1

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Not simplifying the equation

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Students may forget to simplify the equation, leading to incomplete or incorrect results. They often skip steps and directly arrive at the result, especially when solving using the chain or product rule. Ensure that each step is written in order. It may seem tedious, but it's crucial to avoid errors in the process.

Mistake 2

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Forgetting the Domain of Root 2x

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Students might not remember that √(2x) is undefined at x = 0 or for negative values of x. Consider the domain of the function when differentiating to understand that the function is not continuous at certain points.

Mistake 3

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Incorrect use of Chain Rule

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While differentiating functions like √(2x^2), students may misapply the chain rule. Incorrect differentiation: d/dx (√(2x^2)) = 1/√(2x^2) without considering the inner derivative.

 

Correct approach: d/dx (√(2x^2)) = (1/2)(2x^2)^(-1/2) * d/dx (2x^2)

 

Always ensure the chain rule is applied correctly.

Mistake 4

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Not paying attention to constants

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Students sometimes forget to multiply constants outside the square root. For example, incorrectly writing d/dx (5√(2x)) = 1/√(2x) without considering the constant. Check the constants and ensure they are included in the differentiation. Correct: d/dx (5√(2x)) = 5/√(2x).

Mistake 5

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Neglecting to use Simplification Techniques

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Students often forget to use algebraic simplification techniques, such as rationalizing the denominator, which can lead to more complex expressions. Always simplify expressions where possible to make calculations easier.

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Examples Using the Derivative of Root 2x

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Problem 1

Calculate the derivative of (√(2x)·x^3)

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Here, we have f(x) = √(2x)·x^3.

 

Using the product rule, f'(x) = u′v + uv′ In the given equation, u = √(2x) and v = x^3.

 

Differentiate each term: u′= d/dx (√(2x)) = 1/√(2x) v′= d/dx (x^3) = 3x^2

 

Substituting into the equation: f'(x) = (1/√(2x)).(x^3) + (√(2x)).(3x^2) = x^3/√(2x) + 3x^2√(2x)

 

Thus, the derivative of the specified function is x^3/√(2x) + 3x^2√(2x).

Explanation

We find the derivative of the given function by dividing it into two parts. First, find the derivatives and then combine them using the product rule to get the final result.

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Problem 2

A company uses a cable to support a bridge, modeled by y = √(2x), where y represents the height of the cable above the ground at a distance x. Calculate the rate at which the cable height changes when x = 4 meters.

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We have y = √(2x) (height of the cable)...(1)

 

Now, we will differentiate equation (1) Take the derivative √(2x): dy/dx = 1/√(2x)

 

Given x = 4, substitute this into the derivative: dy/dx = 1/√(2*4) = 1/√8 = 1/(2√2)

 

Hence, the rate at which the cable height changes at x = 4 meters is 1/(2√2).

Explanation

We find the rate of change of the cable height at x = 4 meters, indicating how quickly the height increases per unit of horizontal distance.

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Problem 3

Derive the second derivative of the function y = √(2x).

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First, find the first derivative: dy/dx = 1/√(2x)...(1)

 

Now differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/√(2x)] = -1/(2(2x)^(3/2)) * 2 = -1/(2x√(2x))

 

Therefore, the second derivative of the function y = √(2x) is -1/(2x√(2x)).

Explanation

We use a step-by-step process, starting with the first derivative. By applying the chain rule and simplifying, we find the second derivative of √(2x).

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Problem 4

Prove: d/dx (2√(2x)) = 2/√(2x).

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Using the chain rule: Consider y = 2√(2x)

 

To differentiate, we use the chain rule: dy/dx = 2 * d/dx [√(2x)] = 2 * (1/√(2x)) = 2/√(2x)

 

Hence proved.

Explanation

In this step-by-step process, we use the chain rule to differentiate the equation and then substitute √(2x) with its derivative, resulting in the final expression.

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Problem 5

Solve: d/dx (√(2x)/x)

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To differentiate the function, use the quotient rule: d/dx (√(2x)/x) = (d/dx (√(2x))·x - √(2x)·d/dx(x))/x²

 

Substitute d/dx (√(2x)) = 1/√(2x) and d/dx (x) = 1: = (1/√(2x)·x - √(2x)·1)/x² = (x/√(2x) - √(2x))/x² = (x - 2x)/x²√(2x) = -x/x²√(2x) = -1/(x√(2x))

 

Therefore, d/dx (√(2x)/x) = -1/(x√(2x)).

Explanation

In this process, we differentiate the given function using the quotient rule. After simplifying, we obtain the final result for the derivative.

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FAQs on the Derivative of Root 2x

1.Find the derivative of √(2x).

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2.Can we use the derivative of √(2x) in real life?

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3.Is it possible to take the derivative of √(2x) at the point where x = 0?

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4.What rule is used to differentiate √(2x)/x?

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5.Are the derivatives of √(2x) and √x the same?

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Important Glossaries for the Derivative of Root 2x

  • Derivative: The derivative of a function indicates how the function changes in response to a slight change in x.

 

  • Square Root Function: A function that involves the square root of a variable, such as √(2x).

 

  • Chain Rule: A rule for differentiating composite functions.

 

  • Quotient Rule: A rule used to differentiate ratios of two functions.

 

  • Undefined Point: A value of x for which a function does not produce a valid result.
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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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