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Last updated on September 26, 2025
We use the derivative of x³/3, which is x², as a tool to understand how the cubic function changes with respect to a change in x. Derivatives help us calculate changes, such as profit or loss, in various real-life contexts. We will now discuss the derivative of x³/3 in detail.
We now understand the derivative of x³/3. It is commonly represented as d/dx (x³/3) or (x³/3)', and its value is x². The function x³/3 has a clearly defined derivative, indicating it is differentiable across its domain.
The key concepts are mentioned below:
Cubic Function: (x³/3 is a cubic polynomial).
Power Rule: Rule for differentiating x³/3.
Simplification: Simplifying the expression before differentiating can make the process straightforward.
The derivative of x³/3 can be denoted as d/dx (x³/3) or (x³/3)'.
The formula we use to differentiate x³/3 is: d/dx (x³/3) = x² The formula applies to all real numbers x.
We can derive the derivative of x³/3 using proofs. To show this, we will use basic calculus principles along with differentiation rules.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of x³/3 results in x² using the above-mentioned methods:
The derivative of x³/3 can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of x³/3 using the first principle, we will consider f(x) = x³/3. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x+h) - f(x)] / h … (1) Given that f(x) = x³/3, we write f(x+h) = (x+h)³/3. Substituting these into equation (1), f'(x) = limₕ→₀ [(x+h)³/3 - x³/3] / h = limₕ→₀ [((x³ + 3x²h + 3xh² + h³)/3) - x³/3] / h = limₕ→₀ [3x²h + 3xh² + h³]/(3h) = limₕ→₀ [x² + xh + h²/3] = x² + limₕ→₀ [xh + h²/3] = x² + 0 Thus, f'(x) = x². Hence, proved.
To prove the differentiation of x³/3 using the power rule, Consider f(x) = x³/3 We know that d/dx (xⁿ) = n*xⁿ⁻¹ Therefore, d/dx (x³) = 3x² Now, consider x³/3 = (1/3)x³ Applying the constant multiple rule, d/dx [(1/3)x³] = (1/3) * d/dx (x³) = (1/3) * 3x² = x² Thus, d/dx (x³/3) = 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 x³/3.
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 x³/3, 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 = 0, the derivative is 0 because x² = 0. For any positive or negative value of x, the derivative x² will always be positive, indicating that the function is always increasing for x > 0 and decreasing for x < 0.
Students frequently make mistakes when differentiating x³/3. 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 (x³/3 + x²).
Here, we have f(x) = x³/3 + x². Using the power rule, f'(x) = d/dx (x³/3) + d/dx (x²) = x² + 2x Thus, the derivative of the specified function is x² + 2x.
We find the derivative of the given function by differentiating each term separately using the power rule and then combining them to get the final result.
A company is modeling its profit function, which is represented by P(x) = x³/3, where P(x) is the profit when x units are produced. If the company produces 5 units, calculate the rate of change of profit.
We have P(x) = x³/3 (profit function)...(1) Now, we will differentiate the equation (1) Take the derivative of x³/3: dP/dx = x² Given x = 5 (substitute this into the derivative) dP/dx = 5² = 25 Hence, the rate of change of profit when producing 5 units is 25.
We find the rate of change of profit at x = 5 as 25, which means that at this level of production, the profit increases at a rate of 25 units.
Derive the second derivative of the function f(x) = x³/3.
The first step is to find the first derivative, f'(x) = x²...(1) Now we will differentiate equation (1) to get the second derivative: f''(x) = d/dx [x²] f''(x) = 2x Therefore, the second derivative of the function f(x) = x³/3 is 2x.
We use the step-by-step process, starting with the first derivative and then applying the power rule again to find the second derivative.
Prove: d/dx (x³) = 3x².
To prove this, we use the power rule: Consider y = x³ We know that d/dx (xⁿ) = n*xⁿ⁻¹ Therefore, d/dx (x³) = 3x² Hence, proved.
In this step-by-step process, we used the power rule to differentiate the equation, confirming that the derivative of x³ is 3x².
Solve: d/dx (x³/3 + 1).
To differentiate the function, f(x) = x³/3 + 1 Using the power rule, f'(x) = d/dx (x³/3) + d/dx (1) = x² + 0 = x² Therefore, d/dx (x³/3 + 1) = x².
In this process, we differentiate the given function by using the power rule and simplifying 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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