Derivative of 8/x

We now understand the derivative of 8/x. It is commonly represented as d/dx (8/x) or (8/x)', and its value is -8/x². The function 8/x has a clearly defined derivative, indicating it is differentiable within its domain.

The key concepts are mentioned below:

Reciprocal Function: (8/x is a constant multiplied by 1/x).

Power Rule: Rule for differentiating 1/x (since it can be written as x⁻¹).

Negative Exponent: x⁻¹ = 1/x.

The derivative of 8/x can be denoted as d/dx (8/x) or (8/x)'.

The formula we use to differentiate 8/x is: d/dx (8/x) = -8/x² (or) (8/x)' = -8/x² The formula applies to all x where x ≠ 0.

We can derive the derivative of 8/x using proofs. To show this, we will use the rules of differentiation along with algebraic manipulation. There are several methods we use to prove this, such as:

• By First Principle
   
• Using Power Rule
   
• Using Product Rule

We will now demonstrate that the differentiation of 8/x results in -8/x² using the above-mentioned methods:

By First Principle

The derivative of 8/x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 8/x using the first principle, we will consider f(x) = 8/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) = 8/x, we write f(x + h) = 8/(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [8/(x + h) - 8/x] / h = limₕ→₀ [8x - 8(x + h)] / [h(x + h)x] = limₕ→₀ [-8h] / [h(x + h)x] = limₕ→₀ [-8] / [(x + h)x] = -8/x² Hence, proved.

Using Power Rule

To prove the differentiation of 8/x using the power rule, We can write 8/x as 8·x⁻¹. The derivative of x⁻¹ is -x⁻². Therefore, d/dx (8·x⁻¹) = 8·(-x⁻²) = -8/x². Thus, the derivative of 8/x is -8/x².

Using Product Rule

We will now prove the derivative of 8/x using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, 8/x = 8·x⁻¹ Let u = 8 and v = x⁻¹. Using the product rule formula: d/dx [u·v] = u'·v + u·v' u' = 0 (since 8 is a constant) v' = d/dx (x⁻¹) = -x⁻² d/dx (8/x) = u'·v + u·v' = 0·x⁻¹ + 8·(-x⁻²) = -8/x² Thus, the derivative of 8/x is -8/x².

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 8/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 8/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 is 0, the derivative is undefined because 8/x is undefined at x = 0. When x is 1, the derivative of 8/x = -8/1², which is -8.

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

• Reciprocal Function: A function of the form 1/x or any constant multiplied by 1/x.

• Power Rule: A basic rule in calculus used to find the derivative of a power function.

• Undefined Point: A point where the function does not have a defined value.

• Higher-Order Derivative: Derivatives that are obtained after the first derivative, indicating further rates of change.