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Last updated on October 11, 2025
We use the derivative of 9/x, which is -9/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 9/x in detail.
We now understand the derivative of 9/x. It is commonly represented as d/dx (9/x) or (9/x)', and its value is -9/x². The function 9/x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Reciprocal Function: (9/x = 9 * x⁻¹).
Power Rule: Rule for differentiating xⁿ (where n is a constant).
Negative Power: 9/x can be rewritten as 9x⁻¹.
The derivative of 9/x can be denoted as d/dx (9/x) or (9/x)'.
The formula we use to differentiate 9/x is: d/dx (9/x) = -9/x² (or) (9/x)' = -9/x²
The formula applies to all x where x ≠ 0
We can derive the derivative of 9/x using proofs. To show this, we will use the power rule for derivatives.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 9/x results in -9/x² using the above-mentioned methods:
The derivative of 9/x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 9/x using the first principle, we will consider f(x) = 9/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) = 9/x, we write f(x + h) = 9/(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [9/(x + h) - 9/x] / h = limₕ→₀ [9x - 9(x + h)] / [hx(x + h)] = limₕ→₀ [-9h] / [hx(x + h)] = limₕ→₀ -9 / [x(x + h)] As h approaches 0, the expression becomes -9 / x². Hence, proved.
To prove the differentiation of 9/x using the power rule, We rewrite 9/x as 9x⁻¹. The derivative of xⁿ is nxⁿ⁻¹. Applying the power rule, we get: d/dx (9x⁻¹) = 9 * -1 * x⁻² = -9x⁻² Thus, d/dx (9/x) = -9/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 9/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 9/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 9/x is undefined at that point.
When x is 1, the derivative of 9/x = -9/1², which is -9.
Students frequently make mistakes when differentiating 9/x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (9/x * x²)
Here, we have f(x) = 9/x * x². Simplifying, f(x) = 9x. Differentiating f(x), we get: f'(x) = d/dx (9x) = 9. Thus, the derivative of the specified function is 9.
We find the derivative of the given function by simplifying the function first.
The first step is simplifying the expression and then finding its derivative to get the final result.
A water tank is being filled at a rate of 9/x liters per second, where x is the time in seconds. Calculate the rate of change in the flow rate when x = 3 seconds.
We have the flow rate = 9/x liters per second...(1) Now, we will differentiate the equation (1) Take the derivative of 9/x: d/dx (9/x) = -9/x² Given x = 3 (substitute this into the derivative) d/dx (9/3) = -9/3² = -9/9 = -1 Hence, the rate of change in the flow rate at x=3 seconds is -1 liter per second squared.
We find the rate of change in the flow rate at x=3 seconds as -1, which means that the flow rate decreases by 1 liter per second squared at that point.
Derive the second derivative of the function y = 9/x.
The first step is to find the first derivative, dy/dx = -9/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-9/x²] Here we use the power rule, d²y/dx² = 18/x³ Therefore, the second derivative of the function y = 9/x is 18/x³.
We use the step-by-step process, where we start with the first derivative.
Using the power rule, we differentiate -9/x².
We then simplify the terms to find the final answer.
Prove: d/dx (9x⁻²) = -18x⁻³.
Let’s start using the power rule: Consider y = 9x⁻². To differentiate, we use the power rule: dy/dx = 9 * -2 * x⁻³ dy/dx = -18x⁻³ Hence proved.
In this step-by-step process, we used the power rule to differentiate the equation.
We replaced the power of x with its derivative to derive the equation.
Solve: d/dx (9/x²)
To differentiate the function, we first rewrite it using a negative exponent: d/dx (9x⁻²) = 9 * -2 * x⁻³ = -18x⁻³ Therefore, d/dx (9/x²) = -18/x³.
In this process, we differentiate the given function by rewriting it with a negative exponent.
As a final step, we simplify the equation to obtain the final result.
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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