Last updated on July 16th, 2025
We use the derivative of 2/x, which is -2/x², to understand 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 discuss the derivative of 2/x in detail.
We now understand the derivative of 2/x. It is commonly represented as d/dx (2/x) or (2/x)', and its value is -2/x². The function 2/x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: - Reciprocal Function: A function of the form 1/x. - Power Rule: A rule for differentiating functions of the form xⁿ. - Negative Exponent: Represents the reciprocal of a number.
The derivative of 2/x can be denoted as d/dx (2/x) or (2/x)'. The formula we use to differentiate 2/x is: d/dx (2/x) = -2/x² The formula applies to all x ≠ 0.
We can derive the derivative of 2/x using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation. There are several methods we use to prove this, such as: - By First Principle - Using Power Rule We will now demonstrate that the differentiation of 2/x results in -2/x² using the above-mentioned methods: By First Principle The derivative of 2/x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 2/x using the first principle, consider f(x) = 2/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = 2/x, we write f(x + h) = 2/(x + h). Substituting into the equation, f'(x) = limₕ→₀ [2/(x + h) - 2/x] / h = limₕ→₀ [2x - 2(x + h)] / [hx(x + h)] = limₕ→₀ [-2h] / [hx(x + h)] = limₕ→₀ [-2] / [x(x + h)] = -2/x² Thus, f'(x) = -2/x², hence proved. Using Power Rule To prove the differentiation of 2/x using the power rule, Rewrite 2/x as 2x⁻¹. Differentiate using the power rule: d/dx (2x⁻¹) = 2(-1)x⁻² = -2x⁻² = -2/x² Thus, the derivative of 2/x is -2/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 2/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 2/x, we generally use fⁿ(x) for the nth derivative, which tells us the change in the rate of change (continuing for higher-order derivatives).
When x = 0, the derivative is undefined because 2/x has a vertical asymptote there. For x > 0, the derivative of 2/x = -2/x², which is negative, indicating a decreasing function.
Students frequently make mistakes when differentiating 2/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 (2/x · x³)
Here, we have f(x) = (2/x) * x³. Simplify the expression: f(x) = 2x². Differentiate using the power rule: f'(x) = d/dx (2x²) = 4x. Thus, the derivative of the specified function is 4x.
We find the derivative of the given function by simplifying the expression and then applying the power rule to obtain the final result.
A water tank is being drained such that the rate of water flow is given by V = 2/x liters per second, where x is the time in seconds. Calculate the rate of change of flow when x = 5 seconds.
We have V = 2/x (rate of water flow)...(1) Now, we will differentiate the equation (1): dV/dx = -2/x². Given x = 5, substitute this into the derivative: dV/dx = -2/(5²) = -2/25. Hence, the rate of change of flow at x = 5 seconds is -2/25 liters per second.
We find the rate of change of flow by differentiating the function and substituting x = 5 to obtain the rate at that specific time.
Derive the second derivative of the function V = 2/x.
The first step is to find the first derivative: dV/dx = -2/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²V/dx² = d/dx [-2/x²] = 4/x³. Therefore, the second derivative of the function V = 2/x is 4/x³.
We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate -2/x² and simplify to find the second derivative.
Prove: d/dx (4/x²) = -8/x³.
Let’s start using the power rule: Consider y = 4/x² Rewrite as y = 4x⁻². Differentiate using the power rule: dy/dx = 4(-2)x⁻³. dy/dx = -8x⁻³. Thus, d/dx (4/x²) = -8/x³, hence proved.
In this step-by-step process, we use the power rule to differentiate the equation. We rewrite the function with a negative exponent and apply the power rule to derive the equation.
Solve: d/dx (2/x + x²).
To differentiate the function, we differentiate each term separately: d/dx (2/x) = -2/x² d/dx (x²) = 2x Combine the results: d/dx (2/x + x²) = -2/x² + 2x. Therefore, the derivative is -2/x² + 2x.
In this process, we differentiate each term separately using the power rule and then combine the derivatives to obtain the final result.
Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Reciprocal Function: A function that is the reciprocal of another function, such as 1/x. Power Rule: A basic differentiation rule used for finding the derivative of power functions. Negative Exponent: Represents the reciprocal of a number, used in rewriting functions for differentiation. Undefined Points: Points where a function does not exist, such as division by zero.
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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