Derivative of sqrt(3x)

We now understand the derivative of sqrt(3x). It is commonly represented as d/dx (sqrt(3x)) or (sqrt(3x))', and its value is 3/(2sqrt(3x)). The function sqrt(3x) has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below:

Square Root Function: (sqrt(x) = x^(1/2)).

Power Rule: Rule for differentiating powers of x.

Chain Rule: Used to differentiate compositions of functions.

The derivative of sqrt(3x) can be denoted as d/dx (sqrt(3x)) or (sqrt(3x))'. The formula we use to differentiate sqrt(3x) is: d/dx (sqrt(3x)) = 3/(2sqrt(3x)) The formula applies to all x where 3x > 0.

We can derive the derivative of sqrt(3x) using proofs. To show this, we will use the rules of differentiation and chain rule.

There are several methods we use to prove this, such as:

1. By First Principle
2. Using Chain Rule

We will now demonstrate that the differentiation of sqrt(3x) results in 3/(2sqrt(3x)) using the above-mentioned methods:

By First Principle

The derivative of sqrt(3x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of sqrt(3x) using the first principle, we will consider f(x) = sqrt(3x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

Given that f(x) = sqrt(3x), we write f(x + h) = sqrt(3(x + h)).

Substituting these into equation (1), f'(x) = limₕ→₀ [sqrt(3(x + h)) - sqrt(3x)] / h

Multiply and divide by the conjugate: = limₕ→₀ [(sqrt(3(x + h)) - sqrt(3x)) * (sqrt(3(x + h)) + sqrt(3x))] / [h * (sqrt(3(x + h)) + sqrt(3x))] = limₕ→₀ [3(x + h) - 3x] / [h * (sqrt(3(x + h)) + sqrt(3x))] = limₕ→₀ [3h] / [h * (sqrt(3(x + h)) + sqrt(3x))]

Cancel h: = limₕ→₀ 3 / [sqrt(3(x + h)) + sqrt(3x)] = 3 / [2sqrt(3x)]

Hence, proved.

Using Chain Rule

To prove the differentiation of sqrt(3x) using the chain rule, We use the formula: Sqrt(3x) = (3x)^(1/2) Let u = 3x

Then, sqrt(3x) = u^(1/2)

By the chain rule: d/dx [u^(n)] = n * u^(n-1) * (du/dx)… (1)

Let’s substitute u = 3x, n = 1/2 into equation (1), d/dx (sqrt(3x)) = (1/2) * (3x)^(-1/2) * d/dx (3x) = (1/2) * (3x)^(-1/2) * 3 = 3/(2sqrt(3x))

Thus: d/dx (sqrt(3x)) = 3/(2sqrt(3x)).

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(3x).

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(3x), 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 is 0, the derivative is undefined because sqrt(3x) is undefined for x ≤ 0.

• 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 that represents the principal square root of a number, typically written as sqrt(x).

• Chain Rule: A rule in calculus for differentiating compositions of functions, written as d/dx [f(g(x))] = f'(g(x)) * g'(x).

• Quotient Rule: A rule for differentiating the division of two functions, written as d/dx [u/v] = (v * u' - u * v') / v².

• Power Rule: A rule for differentiating powers of x, written as d/dx [x^n] = n * x^(n-1).