Derivative of 8/x
2026-02-28 09:06 Diff
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Last updated on September 26, 2025

We use the derivative of 8/x, which is -8/x², as a measuring tool for how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 8/x in detail.

What is the 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.

Derivative of 8/x Formula

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.

Proofs of the Derivative of 8/x

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.

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Higher-Order Derivatives of 8/x

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).

Special Cases:

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.

Common Mistakes and How to Avoid Them in Derivatives of 8/x

Students frequently make mistakes when differentiating 8/x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Problem 1

Calculate the derivative of (8/x)·ln(x)

Okay, lets begin

Here, we have f(x) = (8/x)·ln(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 8/x and v = ln(x). Let’s differentiate each term, u′ = d/dx (8/x) = -8/x² v′ = d/dx (ln(x)) = 1/x Substituting into the given equation, f'(x) = (-8/x²)·ln(x) + (8/x)·(1/x) Let’s simplify terms to get the final answer, f'(x) = -8ln(x)/x² + 8/x² Thus, the derivative of the specified function is -8ln(x)/x² + 8/x².

Explanation

We find the derivative of the given function by dividing the function into two parts.

The first step is finding its derivative and then combining them using the product rule to get the final result.

Well explained 👍

Problem 2

Jupiter Company is tracking their inventory with a function y = 8/x, where y represents the inventory level at time x. If x = 2 hours, determine the rate of change of inventory.

Okay, lets begin

We have y = 8/x (inventory level)...(1) Now, we will differentiate the equation (1) Take the derivative of 8/x: dy/dx = -8/x² Given x = 2 (substitute this into the derivative) dy/dx = -8/2² = -8/4 = -2 Hence, the rate of change of inventory at x = 2 hours is -2.

Explanation

We find the rate of change of inventory at x = 2 hours as -2, which means that at that time, the inventory level is decreasing by 2 units per hour.

Well explained 👍

Problem 3

Derive the second derivative of the function y = 8/x.

Okay, lets begin

The first step is to find the first derivative, dy/dx = -8/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-8/x²] Using the power rule, d²y/dx² = 16/x³ Therefore, the second derivative of the function y = 8/x is 16/x³.

Explanation

We use the step-by-step process, where we start with the first derivative.

Using the power rule, we differentiate -8/x².

We then simplify the terms to find the final answer.

Well explained 👍

Problem 4

Prove: d/dx (1/x²) = -2/x³.

Okay, lets begin

Let’s start using the power rule: Consider y = 1/x² = x⁻² To differentiate, we use the power rule: dy/dx = -2x⁻³ Substituting y = 1/x², d/dx (1/x²) = -2/x³ Hence proved.

Explanation

In this step-by-step process, we used the power rule to differentiate the equation.

Then, we replace x⁻² with its derivative.

As a final step, we substitute y = 1/x² to derive the equation.

Well explained 👍

Problem 5

Solve: d/dx (8x/x²)

Okay, lets begin

To differentiate the function, we simplify first: 8x/x² = 8/x Then differentiate: d/dx (8/x) = -8/x² Therefore, d/dx (8x/x²) = -8/x²

Explanation

In this process, we simplify the given function before differentiating.

We then apply the power rule to obtain the final result.

Well explained 👍

FAQs on the Derivative of 8/x

1.Find the derivative of 8/x.

Using the power rule to differentiate 8/x gives 8·x⁻¹. d/dx (8/x) = -8/x² (simplified).

2.Can we use the derivative of 8/x in real life?

Yes, we can use the derivative of 8/x in real life to calculate the rate of change of any quantity that follows an inverse relationship, especially in fields such as mathematics, physics, and economics.

3.Is it possible to take the derivative of 8/x at the point where x = 0?

No, x = 0 is a point where 8/x is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).

4.What rule is used to differentiate 8/x²?

We use the power rule to differentiate 8/x², d/dx (8/x²) = -16/x³.

5.Are the derivatives of 8/x and x⁻² the same?

No, they are different. The derivative of 8/x is -8/x², while the derivative of x⁻² is -2x⁻³.

6.Can we find the derivative of the 8/x formula?

To find, consider y = 8/x. We use the power rule: y' = 8·d/dx(x⁻¹) = -8/x².

Important Glossaries for the Derivative of 8/x

  • 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.

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Jaskaran Singh Saluja

About the Author

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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: He loves to play the quiz with kids through algebra to make kids love it.