Derivative of 4 Square Root of x

We now understand the derivative of 4√x. It is commonly represented as d/dx (4√x) or (4√x)', and its value is 2/x^(1/2). The function 4√x has a well-defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below: -

Square Root Function: 4√x or 4x^(1/2). 

Power Rule: Rule for differentiating functions of the form x^n. 

Chain Rule: Differentiating composite functions.

The derivative of 4√x can be denoted as d/dx (4√x) or (4√x)'.

The formula we use to differentiate 4√x is: d/dx (4√x) = 2/x^(1/2) or (4√x)' = 2/x^(1/2).

This formula applies to all x where x > 0.

We can derive the derivative of 4√x using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation. There are several methods we use to prove this, such as: - By First Principle - Using Power Rule - Using Chain Rule We will now demonstrate that the differentiation of 4√x results in 2/x^(1/2) using the above-mentioned methods: By First Principle The derivative of 4√x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 4√x using the first principle, we will consider f(x) = 4√x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 4√x, we write f(x + h) = 4√(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [4√(x + h) - 4√x] / h We can rationalize the numerator by multiplying by the conjugate: = limₕ→₀ [ (4√(x + h) - 4√x)(4√(x + h) + 4√x) ] / [ h(4√(x + h) + 4√x) ] = limₕ→₀ [ (x + h) - x ] / [ h(4√(x + h) + 4√x) ] = limₕ→₀ h / [ h(4√(x + h) + 4√x) ] = limₕ→₀ 1 / [ 4√(x + h) + 4√x ] = 1 / [ 2√x ] As h approaches 0, 4√(x + h) approaches 4√x, so: f'(x) = 2/x^(1/2). Hence, proved. Using Power Rule To prove the differentiation of 4√x using the power rule, We rewrite the function as 4x^(1/2). By the power rule, d/dx [x^n] = n*x^(n-1), d/dx [4x^(1/2)] = 4 * (1/2) * x^(-1/2) = 2/x^(1/2). Hence, proved. Using Chain Rule We will now prove the derivative of 4√x using the chain rule. The process is as follows: Express the function as a composition: Let u = x^(1/2), then 4√x = 4u. The derivative of u with respect to x is: du/dx = (1/2)x^(-1/2) The derivative of 4u with respect to u is: d/du (4u) = 4 Using the chain rule, d/dx (4u) = d/du (4u) * du/dx = 4 * (1/2)x^(-1/2) = 2/x^(1/2). Thus, proved.

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 4√x.

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 4√x, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change. (continuing for higher-order derivatives).

When x=0, the derivative is undefined because the function involves division by zero. When x=1, the derivative of 4√x = 2/√1 = 2.

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

• Square Root Function: A function that involves the square root of a variable, typically written as √x or x^(1/2). 

• Power Rule: A basic differentiation rule used to find the derivative of functions of the form x^n. 

• Chain Rule: A rule for differentiating composite functions, allowing us to separate inner and outer functions. 

• Undefined Points: Points where the function does not exist, often due to division by zero or other issues.