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Last updated on October 3, 2025
We use the derivative of 1/(3x), which is -1/(3x²), as a tool for understanding 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 discuss the derivative of 1/(3x) in detail.
The derivative of 1/(3x) is commonly represented as d/dx (1/(3x)) or (1/(3x))', and its value is -1/(3x²). The function 1/(3x) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: -
Reciprocal Function: A function in the form of 1/f(x).
Power Rule: A basic rule of differentiation applicable to functions of the form x^n.
Constant Multiple Rule: Allows us to multiply the derivative of a function by a constant.
The derivative of 1/(3x) can be denoted as d/dx (1/(3x)) or (1/(3x))'.
The formula we use to differentiate 1/(3x) is: d/dx (1/(3x)) = -1/(3x²)
The formula applies to all x where x ≠ 0.
We can derive the derivative of 1/(3x) using proofs. To show this, we will use algebraic manipulation 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 1/(3x) results in -1/(3x²) using the above-mentioned methods:
The derivative of 1/(3x) can be proved using the First Principles, which express the derivative as the limit of the difference quotient. To find the derivative of 1/(3x) using the first principle, we will consider f(x) = 1/(3x).
Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = 1/(3x), we write f(x + h) = 1/(3(x + h)).
Substituting these into equation (1), f'(x) = limₕ→₀ [1/(3(x + h)) - 1/(3x)] / h = limₕ→₀ [(3x - 3(x + h)) / (3x(3(x + h)))] / h = limₕ→₀ [-3h / (9x(x + h))]/ h = limₕ→₀ -3 / (9x(x + h))
As h approaches 0, f'(x) = -3 / (9x²) = -1 / (3x²)
Hence, proved.
The function 1/(3x) can be rewritten as\( (1/3)x^-1\).
Differentiating using the power rule, \(d/dx [(1/3)x^-1] = (1/3) * (-1)x^(-1-1) = -1/(3x²)\)
We use the constant multiple rule to differentiate 1/(3x).
Consider g(x) = 1/x, which has a derivative of -1/x².
Then, f(x) = 1/(3x) = (1/3)g(x).
Using the constant multiple rule, f'(x) = (1/3) * g'(x) = (1/3) * (-1/x²) = -1/(3x²).
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 help us understand functions like 1/(3x).
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 1/(3x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.
1. When x is 0, the derivative is undefined because 1/(3x) is undefined at x = 0. 2.
When x is 1, the derivative of 1/(3x) = -1/(3 * 1²), which is -1/3.
Students frequently make mistakes when differentiating 1/(3x). 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 (1/(3x) * ln(x))
Here, we have f(x) = 1/(3x) * ln(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 1/(3x) and v = ln(x).
Let’s differentiate each term, u′ = d/dx (1/(3x)) = -1/(3x²) v′ = d/dx (ln(x)) = 1/x
Substituting into the given equation, f'(x) = (-1/(3x²)) * ln(x) + (1/(3x)) * (1/x)
Let’s simplify terms to get the final answer, f'(x) = -ln(x)/(3x²) + 1/(3x²)
Thus, the derivative of the specified function is -ln(x)/(3x²) + 1/(3x²).
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 uses the function y = 1/(3x) to model the cost per unit of production, where x is the number of units produced. If x = 2 units, calculate the rate of change of cost per unit.
We have y = 1/(3x) (cost per unit)...(1)
Now, we will differentiate the equation (1) Take the derivative of 1/(3x): dy/dx = -1/(3x²) Given x = 2 (substitute this into the derivative) dy/dx = -1/(3 * 2²) = -1/12
Hence, the rate of change of cost per unit at x = 2 is -1/12.
We find the rate of change of cost per unit at x = 2 as -1/12, which means that the cost per unit will decrease at this rate as production increases.
Derive the second derivative of the function y = 1/(3x).
The first step is to find the first derivative, dy/dx = -1/(3x²)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/(3x²)] = -1 * [d/dx (x-2)] * (1/3) = -1 * (-2x-3) * (1/3) = 2/(3x³)
Therefore, the second derivative of the function y = 1/(3x) is 2/(3x³).
We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate
x-2. We then simplify the terms to find the final answer.
Prove: d/dx (1/(3x²)) = -2/(3x³).
Let’s start using the power rule: Consider y = 1/(3x²) = (1/3)x^-2 To differentiate, we use the power rule:
dy/dx = (1/3) * (-2)x(-2-1) = -2/(3x³)
Hence proved.
In this step-by-step process, we used the power rule to differentiate the equation. Then, we replaced the expression with its derivative. As a final step, we simplified the terms to derive the equation.
Solve: d/dx ((1/(3x)) * e^x)
To differentiate the function, we use the product rule:\( d/dx ((1/(3x)) * e^x) = (d/dx (1/(3x)) * e^x + 1/(3x) * d/dx(e^x))\)
We will substitute \(d/dx (1/(3x)) = -1/(3x²) and d/dx (e^x) = e^x = (-1/(3x²)) * e^x + (1/(3x)) * e^x = e^x [-1/(3x²) + 1/(3x)] = e^x [(-1 + x)/(3x²)]\)
Therefore, d/dx \(((1/(3x)) * e^x) = e^x [(-1 + x)/(3x²)]\)
In this process, we differentiate the given function using the product 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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