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Last updated on September 9, 2025
We use the derivative of x²/2, which is x, 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 x²/2 in detail.
We now understand the derivative of x²/2.
It is commonly represented as d/dx (x²/2) or (x²/2)', and its value is x.
The function x²/2 has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Power Function: (x²/2 is a polynomial function).
Power Rule: Rule for differentiating x²/2.
Constant Multiple Rule: The derivative of a constant times a function.
The derivative of x²/2 can be denoted as d/dx (x²/2) or (x²/2)'. The formula we use to differentiate x²/2 is: d/dx (x²/2) = x (x²/2)' = x The formula applies to all x.
We can derive the derivative of x²/2 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
Using Constant Multiple Rule
We will now demonstrate that the differentiation of x²/2 results in x using the above-mentioned methods:
By First Principle
The derivative of x²/2 can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of x²/2 using the first principle, we will consider f(x) = x²/2.
Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = x²/2, we write f(x + h) = (x + h)²/2.
Substituting these into equation (1), f'(x) = limₕ→₀ [(x + h)²/2 - x²/2] / h = limₕ→₀ [(x² + 2xh + h²)/2 - x²/2] / h = limₕ→₀ [2xh + h²]/2h = limₕ→₀ [x + h/2] As h approaches 0, f'(x) = x.
Hence, proved.
Using Power Rule
To prove the differentiation of x²/2 using the power rule, We use the formula: d/dx (xⁿ) = n*xⁿ⁻¹
For x², n = 2.
So, d/dx (x²) = 2*x¹ = 2x. Since we have x²/2, d/dx (x²/2) = (1/2)*d/dx (x²) = (1/2)*2x = x.
Using Constant Multiple Rule
We will now prove the derivative of x²/2 using the constant multiple rule.
The formula we use is: d/dx (c*f(x)) = c*d/dx (f(x))
Let c = 1/2 and f(x) = x².
So, d/dx (x²/2) = (1/2)*d/dx (x²) = (1/2)*2x = 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 x²/2.
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 x²/2, we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the rate of change (continuing for higher-order derivatives).
When x is 0, the derivative of x²/2 = 0, which means there is no change at this point. When x is negative, the derivative of x²/2 is negative, indicating a downward slope.
Students frequently make mistakes when differentiating x²/2. 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 (x²/2)·x³.
Here, we have f(x) = (x²/2)·x³.
Using the product rule, f'(x) = u′v + uv′
In the given equation, u = x²/2 and v = x³.
Let’s differentiate each term, u′= d/dx (x²/2) = x v′= d/dx (x³) = 3x²
Substituting into the given equation, f'(x) = (x)·(x³) + (x²/2)·(3x²)
Let’s simplify terms to get the final answer, f'(x) = x⁴ + (3/2)x⁴ = (5/2)x⁴
Thus, the derivative of the specified function is (5/2)x⁴.
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 construction company is laying a pipeline represented by the function y = x²/2 where y represents the elevation at a distance x. If x = 1 meter, measure the slope of the pipeline.
We have y = x²/2 (slope of the pipeline)...(1)
Now, we will differentiate the equation (1)
Take the derivative of x²/2: dy/dx = x
Given x = 1 (substitute this into the derivative) dy/dx = 1
Hence, we get the slope of the pipeline at a distance x = 1 as 1.
We find the slope of the pipeline at x = 1 as 1, which means that at a given point, the height of the pipeline would rise at a rate equal to the horizontal distance.
Derive the second derivative of the function y = x²/2.
The first step is to find the first derivative, dy/dx = x...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx (x) d²y/dx² = 1
Therefore, the second derivative of the function y = x²/2 is 1.
We use the step-by-step process, where we start with the first derivative. We then differentiate x to find the second derivative, which is a constant 1.
Prove: d/dx ((3x²)/2) = 3x.
Let’s start using the constant multiple rule:
Consider y = (3x²)/2
To differentiate, we use the constant multiple rule: dy/dx = (3/2)*d/dx (x²)
Since the derivative of x² is 2x, dy/dx = (3/2)*2x dy/dx = 3x
Hence proved.
In this step-by-step process, we used the constant multiple rule to differentiate the equation. Then, we replace x² with its derivative. As a final step, we simplify to derive the equation.
Solve: d/dx ((x²/2)/x)
To differentiate the function, we use the quotient rule: d/dx ((x²/2)/x) = (d/dx (x²/2)·x - (x²/2)·d/dx(x))/x²
We will substitute d/dx (x²/2) = x and d/dx (x) = 1 (x·x - (x²/2)·1) / x² = (x² - x²/2) / x² = (2x²/2 - x²/2) / x² = (x²/2) / x² = 1/2
Therefore, d/dx ((x²/2)/x) = 1/2.
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.
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.
: He loves to play the quiz with kids through algebra to make kids love it.