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Last updated on September 22, 2025
We use the derivative of the cube root function, which is 1/(3∛(x²)), as a measuring tool for how the cube root 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 the cube root function in detail.
We now understand the derivative of the cube root function. It is commonly represented as d/dx (∛x) or (∛x)', and its value is 1/(3∛(x²)). The function ∛x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Cube Root Function: The cube root of x is represented as ∛x.
Power Rule: Rule for differentiating functions of the form xⁿ.
Chain Rule: Used when differentiating functions that are composite.
The derivative of ∛x can be denoted as d/dx (∛x) or (∛x)'. The formula we use to differentiate ∛x is: d/dx (∛x) = 1/(3∛(x²)) The formula applies to all x except where x=0.
We can derive the derivative of ∛x using proofs. To show this, we will use the power rule along with the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of ∛x results in 1/(3∛(x²)) using the above-mentioned methods:
The derivative of ∛x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of ∛x using the first principle, we will consider f(x) = ∛x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = ∛x, we write f(x + h) = ∛(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [∛(x + h) - ∛x] / h
We simplify using the binomial expansion for roots: = limₕ→₀ [ (x + h)^(1/3) - x^(1/3) ] / h
Using the limit definition, we find: f'(x) = 1/(3x^(2/3))
Hence, proved.
To prove the differentiation of ∛x using the chain rule, We express the cube root as a power: ∛x = x^(1/3). Using the chain rule, the derivative of x^(1/3) is: d/dx (x^(1/3)) = (1/3)x^(-2/3) Which is equivalent to: 1/(3∛(x²))
The power rule states if y = x^n, then dy/dx = nx^(n-1). For ∛x, rewrite it as x^(1/3).
Applying the power rule, d/dx (x^(1/3)) = (1/3)x^(-2/3) This simplifies to: 1/(3∛(x²))
Thus, it is proved.
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.
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, 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 is undefined because ∛x is not differentiable at x=0. When x is 1, the derivative of ∛x = 1/(3∛(1²)), which is 1/3.
Students frequently make mistakes when differentiating ∛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 (∛x·x²)
Here, we have f(x) = ∛x·x².
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ∛x and v = x².
Let’s differentiate each term, u′ = d/dx (∛x) = 1/(3∛(x²)) v′ = d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (1/(3∛(x²))). (x²) + (∛x). (2x)
Let’s simplify terms to get the final answer, f'(x) = x²/(3∛(x²)) + 2x∛x
Thus, the derivative of the specified function is x²/(3∛(x²)) + 2x∛x.
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.
The local bakery's dough rising is modeled by the function y = ∛x where y represents the volume of dough after x hours. If x = 8 hours, measure the rate at which the dough's volume is increasing.
We have y = ∛x (rate of dough rising)...(1)
Now, we will differentiate the equation (1) Take the derivative ∛x: dy/dx = 1/(3∛(x²))
Given x = 8 (substitute this into the derivative) dy/dx = 1/(3∛(8²)) ∛(8²) = ∛64 = 4
Therefore, dy/dx = 1/(3*4) = 1/12
Hence, we get the rate of dough rising at x=8 as 1/12.
We find the rate at which the dough's volume increases at x=8 as 1/12, which means that at this given point, the volume of the dough rises at a rate of 1/12 cubic units per hour.
Derive the second derivative of the function y = ∛x.
The first step is to find the first derivative, dy/dx = 1/(3∛(x²))...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/(3∛(x²))]
Here we use the chain rule, d²y/dx² = -1/(9x∛(x⁴))
Therefore, the second derivative of the function y = ∛x is -1/(9x∛(x⁴)).
We use the step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate 1/(3∛(x²)). We then substitute the identity and simplify the terms to find the final answer.
Prove: d/dx ((∛x)²) = 2∛x/(3∛(x²)).
Let’s start using the chain rule: Consider y = (∛x)² = [x^(1/3)]² = x^(2/3)
To differentiate, we use the power rule: dy/dx = (2/3)x^(-1/3) Since ∛x = x^(1/3), dy/dx = 2∛x/(3∛(x²))
Hence proved.
In this step-by-step process, we used the power rule to differentiate the equation. Then, we replace x with its power form. As a final step, we substitute ∛x = x^(1/3) to derive the equation.
Solve: d/dx (∛x/x)
To differentiate the function, we use the quotient rule: d/dx (∛x/x) = (d/dx (∛x). x - ∛x. d/dx(x))/x²
We will substitute d/dx (∛x) = 1/(3∛(x²)) and d/dx (x) = 1 (1/(3∛(x²)). x - ∛x. 1) / x² = (x/(3∛(x²)) - ∛x) / x² = (x - 3∛(x³))/ (3x²∛(x²))
Therefore, d/dx (∛x/x) = (x - 3∛(x³))/ (3x²∛(x²))
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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