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Last updated on 20 September 2025
We use the derivative of 1/x, which is -1/x², as a tool for understanding how the function changes with respect to x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 1/x in detail.
We now understand the derivative of 1/x. It is commonly represented as d/dx (1/x) or (1/x)', and its value is -1/x². The function 1/x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Reciprocal Function: (1/x is the reciprocal of x).
Quotient Rule: Rule for differentiating 1/x (since it is a quotient of constants and x).
Exponential Functions: We can rewrite 1/x as x⁻¹.
The derivative of 1/x can be denoted as d/dx (1/x) or (1/x)'.
The formula we use to differentiate 1/x is: d/dx (1/x) = -1/x² (or) (1/x)' = -1/x²
The formula applies to all x where x ≠ 0.
We can derive the derivative of 1/x using proofs. To show this, we will use the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 1/x results in -1/x² using the above-mentioned methods:
The derivative of 1/x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 1/x using the first principle, we will consider f(x) = 1/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) = 1/x, we write f(x + h) = 1/(x + h).
Substituting these into equation (1),
f'(x) = limₕ→₀ [1/(x + h) - 1/x] / h = limₕ→₀ [(x - (x + h)) / (x(x + h))] / h = limₕ→₀ [-h / (x² + xh)] / h = limₕ→₀ -1 / (x² + xh) = -1/x²
Hence, proved.
To prove the differentiation of 1/x using the power rule, We rewrite 1/x as x⁻¹.
By the power rule: d/dx [xⁿ] = nxⁿ⁻¹ Let n = -1, d/dx (x⁻¹) = -1 * x⁻² = -1/x²
We will now prove the derivative of 1/x using the quotient rule. The step-by-step process is demonstrated below:
Here, 1/x can be viewed as (1)/(x)
Using the quotient rule formula:
d/dx [u/v] = (v * u' - u * v') / v²
Let u = 1 (constant), v = x u' = 0, v' = 1
d/dx (1/x) = (x * 0 - 1 * 1) / x² = -1/x²
Thus, d/dx (1/x) = -1/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 1/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 1/x, we generally use f⁽ⁿ⁾(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 1/x has a vertical asymptote there. When x is 1, the derivative of 1/x = -1/1², which is -1.
Students frequently make mistakes when differentiating 1/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 (1/x * x²)
Here, we have f(x) = (1/x) * x². Using simplification, f(x) = x.
The derivative of x is, f'(x) = d/dx (x) = 1.
Thus, the derivative of the specified function is 1.
We simplify the given function first by cancelling x in the numerator and denominator.
Then, we find the derivative of the simplified function to get the final result.
AXB Engineering Company is studying the rate of change in electrical resistance represented by the function R = 1/x, where R is the resistance in ohms and x is the distance in meters. If x = 2 meters, calculate the rate of change in resistance.
We have R = 1/x (resistance function)...(1)
Now, we will differentiate the equation (1) Take the derivative 1/x: dR/dx = -1/x²
Given x = 2 (substitute this into the derivative)
dR/dx = -1/2² dR/dx = -1/4
Hence, the rate of change in resistance at a distance x = 2 meters is -1/4 ohms per meter.
We find the rate of change in resistance at x = 2 as -1/4, which means that for every meter increase in distance, the resistance decreases by 1/4 ohms.
Derive the second derivative of the function y = 1/x.
The first step is to find the first derivative, dy/dx = -1/x²...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/x²]
Rewrite -1/x² as -x⁻² and use the power rule: d²y/dx² = 2/x³.
Therefore, the second derivative of the function y = 1/x is 2/x³.
We use the step-by-step process, where we start with the first derivative.
Using the power rule, we differentiate -x⁻². We then substitute the result to find the final answer.
Prove: d/dx (1/x²) = -2/x³.
Let’s start using the power rule: Consider y = 1/x² Rewrite as y = x⁻² T
o differentiate, we use the power rule: dy/dx = -2x⁻³
Substituting y = 1/x², d/dx (1/x²) = -2/x³
Hence proved.
In this step-by-step process, we used the power rule to differentiate the equation. Then, we replace 1/x² with its derivative. As a final step, we substitute y = 1/x² to derive the equation.
Solve: d/dx (1/x + x)
To differentiate the function, we differentiate each term separately: d/dx (1/x + x) = d/dx (1/x) + d/dx (x)
Use the power rule for each term: = -1/x² + 1
Therefore, d/dx (1/x + x) = -1/x² + 1.
In this process, we differentiate each term in the given function separately. We use the power rule for both terms. As a final step, we combine the results to obtain the final answer.
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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