Last updated on July 15th, 2025
We use the derivative of x^(1/3), which represents how the cubic 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 x^(1/3) in detail.
We now understand the derivative of x^(1/3). It is commonly represented as d/dx (x^(1/3)) or (x^(1/3))', and its value is (1/3)x^(-2/3). The function x^(1/3) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Power Rule: Rule for differentiating x^n where n is any real number. Negative Exponents: Understanding how exponents can be negative in derivatives.
The derivative of x^(1/3) can be denoted as d/dx (x^(1/3)) or (x^(1/3))'. The formula we use to differentiate x^(1/3) is: d/dx (x^(1/3)) = (1/3)x^(-2/3) The formula applies to all x where x is not 0.
We can derive the derivative of x^(1/3) using proofs. To show this, we will use the power rule 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^(1/3) results in (1/3)x^(-2/3) using the above-mentioned methods: By First Principle The derivative of x^(1/3) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of x^(1/3) using the first principle, we will consider f(x) = x^(1/3). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = x^(1/3), we write f(x + h) = (x + h)^(1/3). Substituting these into equation (1), f'(x) = limₕ→₀ [(x + h)^(1/3) - x^(1/3)] / h Using binomial expansion or numerical methods to simplify, f'(x) = (1/3)x^(-2/3) Using Power Rule To prove the differentiation of x^(1/3) using the power rule, We use the formula: d/dx (x^n) = nx^(n-1) For x^(1/3), n = 1/3, so we apply the power rule: d/dx (x^(1/3)) = (1/3)x^(-2/3)
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^(1/3). 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 x^(1/3), 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 is 0, the derivative is undefined because x^(1/3) is not differentiable at x = 0. When x is negative, the derivative is still valid since we can take odd roots of negative numbers, but care must be taken with the real domain.
Students frequently make mistakes when differentiating x^(1/3). 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 (x^(1/3)·x²)
Here, we have f(x) = x^(1/3)·x². Using the product rule, f′(x) = u′v + uv′ In the given equation, u = x^(1/3) and v = x². Let's differentiate each term, u′ = d/dx (x^(1/3)) = (1/3)x^(-2/3) v′ = d/dx (x²) = 2x Substituting into the given equation, f′(x) = ((1/3)x^(-2/3))·x² + x^(1/3)·(2x) Let's simplify terms to get the final answer, f′(x) = (1/3)x^(4/3) + 2x^(4/3) Thus, the derivative of the specified function is (7/3)x^(4/3).
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.
XYZ Corporation is planning to build a curved ramp. The elevation of the ramp is represented by the function y = x^(1/3), where y represents the elevation of the ramp at a distance x. If x = 8 meters, measure the slope of the ramp.
We have y = x^(1/3) (slope of the ramp)...(1) Now, we will differentiate the equation (1) Take the derivative x^(1/3): dy/dx = (1/3)x^(-2/3) Given x = 8 (substitute this into the derivative) dy/dx = (1/3)(8)^(-2/3) dy/dx = (1/3)(1/4) = 1/12 Hence, we get the slope of the ramp at a distance x = 8 as 1/12.
We find the slope of the ramp at x = 8 as 1/12, which means that at a given point, the height of the ramp would rise at a rate of 1/12 of the horizontal distance.
Derive the second derivative of the function y = x^(1/3).
The first step is to find the first derivative, dy/dx = (1/3)x^(-2/3)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [(1/3)x^(-2/3)] Here we use the power rule, d²y/dx² = (1/3)(-2/3)x^(-5/3) = -(2/9)x^(-5/3) Therefore, the second derivative of the function y = x^(1/3) is -(2/9)x^(-5/3).
We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate (1/3)x^(-2/3). We then simplify the terms to find the final answer.
Prove: d/dx ((x²)^(1/3)) = (2/3)x^(-1/3).
Let's start using the chain rule: Consider y = (x²)^(1/3) To differentiate, we use the chain rule: dy/dx = (1/3)(x²)^(-2/3)·d/dx(x²) Since the derivative of x² is 2x, dy/dx = (1/3)(x²)^(-2/3)·2x dy/dx = (2/3)x^(1/3) Hence proved.
In this step-by-step process, we used the chain 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 (x^(1/3)/x)
To differentiate the function, we use the quotient rule: d/dx (x^(1/3)/x) = (d/dx (x^(1/3))·x - x^(1/3)·d/dx(x))/x² We will substitute d/dx (x^(1/3)) = (1/3)x^(-2/3) and d/dx (x) = 1 = ((1/3)x^(-2/3)·x - x^(1/3)·1)/x² = ((1/3)x^(1/3) - x^(1/3))/x² = (-2/3)x^(-5/3) Therefore, d/dx (x^(1/3)/x) = (-2/3)x^(-5/3).
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.
Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Power Rule: A basic rule used to find the derivative of a power of x. Negative Exponents: Exponents that indicate the reciprocal of the base raised to a positive exponent. First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function. Undefined Point: A point where the function or its derivative does not exist.
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