Derivative of sqrt(2x)

We now understand the derivative of sqrt(2x). It is commonly represented as d/dx (sqrt(2x)) or (sqrt(2x))', and its value is 1/sqrt(2x).

The function sqrt(2x) has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below: Square Root Function: sqrt(2x) = (2x)^(1/2).

Chain Rule: Rule for differentiating sqrt(2x) (since it consists of an inner function 2x and an outer function sqrt).

The derivative of sqrt(2x) can be denoted as d/dx (sqrt(2x)) or (sqrt(2x))'.

The formula we use to differentiate sqrt(2x) is: d/dx (sqrt(2x)) = 1/sqrt(2x)

The formula applies to all x where x > 0.

We can derive the derivative of sqrt(2x) using proofs.

To show this, we will use the chain rule along with the rules of differentiation.

There are several methods we use to prove this, such as: By First Principle Using Chain Rule By First Principle The derivative of sqrt(2x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of sqrt(2x) using the first principle, we will consider f(x) = sqrt(2x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h

Given that f(x) = sqrt(2x), we write f(x + h) = sqrt(2(x + h)).

Substituting these into equation (1), f'(x) = limₕ→₀ [sqrt(2(x + h)) - sqrt(2x)] / h = limₕ→₀ [sqrt(2(x + h)) * sqrt(2x) - sqrt(2x) * sqrt(2(x + h))] / [h * sqrt(2x) * sqrt(2(x + h))]

Multiplying and dividing by the conjugate, f'(x) = limₕ→₀ [2(x + h) - 2x] / [h * sqrt(2x) * sqrt(2(x + h)) * (sqrt(2(x + h)) + sqrt(2x))] = limₕ→₀ [2h] / [h * sqrt(2x) * sqrt(2(x + h)) * (sqrt(2(x + h)) + sqrt(2x))] Canceling h, f'(x) = limₕ→₀ 2 / [sqrt(2x) * sqrt(2(x + h)) * (sqrt(2(x + h)) + sqrt(2x))]

As h approaches 0, f'(x) = 1 / sqrt(2x).

Using Chain Rule To prove the differentiation of sqrt(2x) using the chain rule, We use the formula: sqrt(2x) = (2x)^(1/2) Consider f(x) = 2x

So we get, sqrt(2x) = f(x)^(1/2) By chain rule: d/dx [f(x)^(1/2)] = (1/2) * f(x)^(-1/2) * f'(x)

Let’s substitute f(x) = 2x d/dx (sqrt(2x)) = (1/2) * (2x)^(-1/2) * 2 = 1/sqrt(2x)

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky.

To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes.

Higher-order derivatives make it easier to understand functions like sqrt(2x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point.

The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.

For the nth Derivative of sqrt(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 sqrt(2x) is not defined for x < 0. When x is 1, the derivative of sqrt(2x) = 1/sqrt(2)

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

• Square Root Function: A function represented as sqrt(x) or (x)^(1/2).

• Chain Rule: A rule for differentiating compositions of functions.

• First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function.

• Quotient Rule: A method for finding the derivative of a quotient of two functions.