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Last updated on October 8, 2025
We use the derivative of 3x³, which is 9x², as a tool for understanding how the cubic 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 3x³ in detail.
We now understand the derivative of 3x³. It is commonly represented as d/dx (3x³) or (3x³)', and its value is 9x². The function 3x³ has a clearly defined derivative, indicating it is differentiable for all real numbers.
The key concepts are mentioned below:
Power Rule: A rule for differentiating functions of the form xⁿ.
Constant Multiple Rule: A rule for differentiating a constant times a function.
Polynomial Function: A function consisting of terms that are non-negative integer powers of x.
The derivative of 3x³ can be denoted as d/dx (3x³) or (3x³)'.
The formula we use to differentiate 3x³ is: d/dx (3x³) = 9x²
The formula applies to all real numbers x.
We can derive the derivative of 3x³ using proofs. To show this, we will use the basic rules of differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 3x³ results in 9x² using the above-mentioned methods:
The derivative of 3x³ can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 3x³ using the first principle, we will consider f(x) = 3x³. Its derivative can be expressed as the following limit: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 3x³, we write f(x + h) = 3(x + h)³. Substituting these into equation (1), f'(x) = limₕ→₀ [3(x + h)³ - 3x³] / h = limₕ→₀ [3(x³ + 3x²h + 3xh² + h³) - 3x³] / h = limₕ→₀ [9x²h + 9xh² + 3h³] / h = limₕ→₀ [9x² + 9xh + 3h²] = 9x² (as h approaches 0, the terms with h vanish) Hence, proved.
To prove the differentiation of 3x³ using the power rule, We use the formula: d/dx (xⁿ) = nxⁿ⁻¹ For 3x³, we consider the constant multiple rule: d/dx (3x³) = 3 * d/dx (x³) = 3 * 3x² = 9x² Thus, the derivative is 9x².
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 3x³.
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 3x³, 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 3x³ = 9x², which is 0. There are no undefined points in the domain of 3x³, as it is differentiable for all real numbers.
Students frequently make mistakes when differentiating 3x³. 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 (3x³·4x²)
Here, we have f(x) = 3x³·4x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 3x³ and v = 4x². Let’s differentiate each term, u′= d/dx (3x³) = 9x² v′= d/dx (4x²) = 8x Substituting into the given equation, f'(x) = (9x²)(4x²) + (3x³)(8x) Let’s simplify terms to get the final answer, f'(x) = 36x^4 + 24x^4 Thus, the derivative of the specified function is 60x^4.
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 manufactures boxes, and the volume of each box is represented by the function V = 3x³, where V is the volume in cubic units and x is the side length in units. If x = 2 units, calculate the rate of change of the volume.
We have V = 3x³ (volume of the box)...(1) Now, we will differentiate the equation (1) Take the derivative of 3x³: dV/dx = 9x² Given x = 2 units (substitute this into the derivative) dV/dx = 9(2)² dV/dx = 9(4) = 36 Hence, we get the rate of change of the volume at x = 2 units as 36 cubic units per unit length.
We find the rate of change of the volume at x = 2 units as 36, which means that at this point, the volume of the box increases by 36 cubic units per unit increase in side length.
Derive the second derivative of the function y = 3x³.
The first step is to find the first derivative, dy/dx = 9x²...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [9x²] d²y/dx² = 18x
Therefore, the second derivative of the function y = 3x³ is 18x.
We use a step-by-step process, where we start with the first derivative.
Using the basic rules of differentiation, we find the first derivative and then differentiate again to find the second derivative.
Prove: d/dx ((3x³)²) = 18x²·9x².
Let’s start using the chain rule: Consider y = (3x³)² To differentiate, we use the chain rule: dy/dx = 2(3x³) * d/dx [3x³] Since the derivative of 3x³ is 9x², dy/dx = 2(3x³) * 9x² dy/dx = 18x² * 9x² Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace 3x³ with its derivative.
As a final step, we substitute back to derive the equation.
Solve: d/dx (3x³/x)
To differentiate the function, we use the quotient rule: d/dx (3x³/x) = (d/dx (3x³) * x - 3x³ * d/dx(x)) / x² We will substitute d/dx (3x³) = 9x² and d/dx (x) = 1 = (9x² * x - 3x³ * 1) / x² = (9x³ - 3x³) / x² = 6x² / x² = 6 Therefore, d/dx (3x³/x) = 6.
In this process, we differentiate the given function using the product rule and 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.
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