Last updated on July 19th, 2025
We use the derivative of x-2, which is 1, as a tool to measure 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 x-2 in detail.
We now understand the derivative of x-2. It is commonly represented as d/dx (x-2) or (x-2)', and its value is 1. The function x-2 has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Linear Function: A function of the form f(x) = x - 2 is a linear function. Constant Rule: The derivative of any constant is zero. Power Rule: Used when differentiating x^n (here n=1).
The derivative of x-2 can be denoted as d/dx (x-2) or (x-2)'. The formula we use to differentiate x-2 is: d/dx (x-2) = 1 The formula applies to all x, as the derivative of a linear function is constant.
We can derive the derivative of x-2 using proofs. To show this, we will use 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 x-2 results in 1 using the above-mentioned methods: By First Principle The derivative of x-2 can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of x-2 using the first principle, we will consider f(x) = x-2. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = x-2, we write f(x + h) = (x + h) - 2. Substituting these into the equation, f'(x) = limₕ→₀ [(x + h - 2) - (x - 2)] / h = limₕ→₀ [h] / h = limₕ→₀ 1 f'(x) = 1. Hence, proved. Using Power Rule To prove the differentiation of x-2 using the power rule, Consider f(x) = x^1 - 2. Using the power rule, d/dx [x^n] = n * x^(n-1), For n = 1, d/dx (x^1) = 1 * x^(1-1) = 1. As the derivative of a constant is zero, the derivative of -2 is 0. Thus, d/dx (x-2) = 1.
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 x-2. 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 x-2, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change. In this case, all higher-order derivatives are 0.
The derivative of x-2 is always 1 as it is a linear function. At x = 0, the derivative is 1, indicating a uniform slope.
Students frequently make mistakes when differentiating x-2. 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 (x-2)(x+3).
Here, we have f(x) = (x-2)(x+3). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x-2 and v = x+3. Let’s differentiate each term, u′= d/dx (x-2) = 1 v′= d/dx (x+3) = 1 Substituting into the given equation, f'(x) = (1)(x+3) + (x-2)(1) Let’s simplify terms to get the final answer, f'(x) = x + 3 + x - 2 f'(x) = 2x + 1. Thus, the derivative of the specified function is 2x + 1.
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.
A company produces widgets, and its cost function is C(x) = x - 2, where C(x) represents the cost in dollars for producing x widgets. If x = 5 widgets, find the rate of change of the cost.
We have C(x) = x - 2 (cost function)...(1) Now, we will differentiate the equation (1) Take the derivative of x-2: dC/dx = 1 The rate of change of the cost is constant and equals 1 dollar per widget. At x = 5, the rate of change remains 1.
We find the rate of change of the cost, which is constant at 1, indicating that each additional widget costs 1 additional dollar to produce, regardless of the number of widgets being produced.
Derive the second derivative of the function y = x-2.
The first step is to find the first derivative, dy/dx = 1...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1] d²y/dx² = 0. Therefore, the second derivative of the function y = x-2 is 0.
We use the step-by-step process, where we start with the first derivative. The second derivative of a constant is 0, as there is no change in the rate of change for a linear function.
Prove: d/dx ((x-2)²) = 2(x-2).
Let’s start using the chain rule: Consider y = (x-2)² To differentiate, we use the chain rule: dy/dx = 2(x-2). d/dx [x-2] Since the derivative of x-2 is 1, dy/dx = 2(x-2).1 Substituting y = (x-2)², d/dx ((x-2)²) = 2(x-2) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace x-2 with its derivative. As a final step, we substitute y = (x-2)² to derive the equation.
Solve: d/dx ((x-2)/x).
To differentiate the function, we use the quotient rule: d/dx ((x-2)/x) = (d/dx (x-2) * x - (x-2) * d/dx(x))/x² We will substitute d/dx (x-2) = 1 and d/dx (x) = 1 = (1 * x - (x-2) * 1) / x² = (x - (x-2)) / x² = 2/x² Therefore, d/dx ((x-2)/x) = 2/x²
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation 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. Linear Function: A function of the form f(x) = ax + b, where the rate of change is constant. Power Rule: A basic rule of differentiation used for finding derivatives of power functions. Constant Rule: The derivative of any constant is zero. Quotient Rule: A rule used to differentiate functions that are the division of two differentiable functions.
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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