Last updated on July 21st, 2025
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.
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.
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.
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:
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.
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)
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.
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.
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:
Calculate the derivative of (√(2x)·x^3)
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).
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.
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.
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).
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.
Derive the second derivative of the function y = √(2x).
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)).
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).
Prove: d/dx (2√(2x)) = 2/√(2x).
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.
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.
Solve: d/dx (√(2x)/x)
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)).
In this process, we differentiate the given function using the quotient rule. After simplifying, we obtain the final result for the derivative.
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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