Last updated on July 21st, 2025
We use the derivative of 3/x, which is -3/x², as a tool to measure 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 talk about the derivative of 3/x in detail.
We now understand the derivative of 3/x. It is commonly represented as d/dx (3/x) or (3/x)', and its value is -3/x².
The function 3/x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Reciprocal Function: (f(x) = 3/x).
Power Rule: A rule for differentiating functions of the form x^n.
Negative Exponent: 3/x can be written as 3x^(-1).
The derivative of 3/x can be denoted as d/dx (3/x) or (3/x)'.
The formula we use to differentiate 3/x is: d/dx (3/x) = -3/x² The formula applies to all x where x ≠ 0.
We can derive the derivative of 3/x using proofs. To show this, we will use the rules of differentiation.
There are several methods we use to prove this, such as: By First Principle Using Power Rule Using Product Rule
We will now demonstrate that the differentiation of 3/x results in -3/x² using the above-mentioned methods: By First Principle The derivative of 3/x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 3/x using the first principle, we will consider f(x) = 3/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) = 3/x, we write f(x + h) = 3/(x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [3/(x + h) - 3/x] / h = limₕ→₀ [3x - 3(x + h)] / [x(x + h)h] = limₕ→₀ [-3h] / [x(x + h)h] = limₕ→₀ -3 / [x(x + h)] = -3 / x² Hence, the derivative of 3/x is -3/x².
Using Power Rule To prove the differentiation of 3/x using the power rule, We express 3/x as 3x⁻¹.
Using the power rule, d/dx [x^n] = n*x^(n-1), d/dx [3x⁻¹] = -1*3x^(-1-1) = -3x⁻² = -3/x²
Thus, the derivative of 3/x is -3/x².
Using Product Rule We will now prove the derivative of 3/x using the product rule.
The step-by-step process is demonstrated below: Here, we use the formula, 3/x = 3 * x⁻¹ Let u = 3 and v = x⁻¹
Using the product rule formula: d/dx [u*v] = u'v + uv' u' = d/dx (3) = 0 v' = d/dx (x⁻¹) = -1*x⁻² d/dx (3/x) = 0 * x⁻¹ + 3 * (-1)x⁻² = -3x⁻² = -3/x²
Therefore, the derivative of 3/x is -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 3/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 3/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 = 0, the derivative is undefined because 3/x has a vertical asymptote there. When x = 1, the derivative of 3/x = -3/1², which is -3.
Students frequently make mistakes when differentiating 3/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 (3/x * x³)
Here, we have f(x) = (3/x) * x³. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 3/x and v = x³. Let’s differentiate each term, u′ = d/dx (3/x) = -3/x² v′ = d/dx (x³) = 3x²
Substituting into the given equation, f'(x) = (-3/x²) * x³ + (3/x) * 3x²
Let’s simplify terms to get the final answer, f'(x) = -3x + 9 Thus, the derivative of the specified function is -3x + 9.
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 water tank has a drain that empties at a rate proportional to 3/x, where x is the time in minutes. If x = 2 minutes, calculate the change in the drainage rate.
We have y = 3/x (drainage rate)...(1) Now, we will differentiate the equation (1).
Take the derivative 3/x: dy/dx = -3/x² Given x = 2 (substitute this into the derivative) dy/dx = -3/2² = -3/4
Hence, the change in the drainage rate at x = 2 minutes is -3/4.
We find the change in the drainage rate at x = 2 minutes as -3/4, which means that at this point, the rate of drainage decreases by 3/4 units per minute.
Derive the second derivative of the function y = 3/x.
The first step is to find the first derivative, dy/dx = -3/x²...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-3/x²]
Here we use the power rule, d²y/dx² = 6/x³ Therefore, the second derivative of the function y = 3/x is 6/x³.
We use the step-by-step process, where we start with the first derivative.
Using the power rule, we differentiate -3/x². We then simplify the terms to find the final answer.
Prove: d/dx (3/x²) = -6/x³.
Let’s start using the power rule: Consider y = 3/x² y = 3x⁻² To differentiate, we use the power rule: dy/dx = -2 * 3 * x⁻³ = -6x⁻³ = -6/x³ Hence proved.
In this step-by-step process, we used the power rule to differentiate the equation 3x⁻². Then, we simplify the expression to derive the equation.
Solve: d/dx (3/x + x)
To differentiate the function, we use the sum rule: d/dx (3/x + x) = d/dx (3/x) + d/dx (x) We will substitute d/dx (3/x) = -3/x² and d/dx (x) = 1 = -3/x² + 1 Therefore, d/dx (3/x + x) = -3/x² + 1
In this process, we differentiate the given function using the sum 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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