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Last updated on September 12, 2025
We use the derivative of 10/x, which is -10/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 discuss the derivative of 10/x in detail.
We now understand the derivative of 10/x. It is commonly represented as d/dx (10/x) or (10/x)', and its value is -10/x².
The function 10/x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Reciprocal Function: (10/x).
Power Rule: Rule for differentiating x raised to a power.
Negative Exponents: Understanding how to differentiate functions with negative exponents.
The derivative of 10/x can be denoted as d/dx (10/x) or (10/x)'.
The formula we use to differentiate 10/x is: d/dx (10/x) = -10/x² (or) (10/x)' = -10/x²
The formula applies to all x where x ≠ 0.
We can derive the derivative of 10/x using proofs. To show this, we will use basic differentiation rules. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 10/x results in -10/x² using the above-mentioned methods:
The derivative of 10/x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 10/x using the first principle, we will consider f(x) = 10/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) = 10/x, we write f(x + h) = 10/(x + h).
Substituting these into equation (1),
f'(x) = limₕ→₀ [10/(x + h) - 10/x] / h = limₕ→₀ [10x - 10(x + h)] / [hx(x + h)] = limₕ→₀ [-10h] / [hx(x + h)] = limₕ→₀ [-10] / [x(x + h)] = -10/x²
Hence, proved.
To prove the differentiation of 10/x using the power rule, Rewrite 10/x as 10x⁻¹.
Using the power rule: d/dx (xⁿ) = nxⁿ⁻¹ d/dx (10x⁻¹) = 10(-1)x⁻² = -10x⁻²
This simplifies to -10/x². Using Quotient Rule We will now prove the derivative of 10/x using the quotient rule. The step-by-step process is demonstrated below: Let u = 10 and v = x.
By quotient rule: d/dx [u/v] = [v u' - u v'] / [v²]
Let’s substitute u = 10 and v = x, d/dx (10/x) = [x(0) - 10(1)] / x² = -10/x² Thus, d/dx (10/x) = -10/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 10/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 10/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 10/x has a discontinuity there. When x is 1, the derivative of 10/x = -10/1², which is -10.
Students frequently make mistakes when differentiating 10/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 (10/x)·x²
Here, we have f(x) = (10/x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 10/x and v = x².
Let’s differentiate each term, u′= d/dx (10/x) = -10/x² v′= d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (-10/x²)·x² + (10/x)·2x
Let’s simplify terms to get the final answer, f'(x) = -10 + 20/x
Thus, the derivative of the specified function is -10 + 20/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.
A company monitors their production efficiency using the function y = 10/x, where y represents the efficiency at production level x. If x = 5 units, measure the rate of change of efficiency.
We have y = 10/x (efficiency function)...(1)
Now, we will differentiate the equation (1)
Take the derivative of 10/x: dy/dx = -10/x²
Given x = 5 (substitute this into the derivative)
dy/dx = -10/5² = -10/25 = -2/5
Hence, we get the rate of change of efficiency at a production level of x = 5 as -2/5.
We find the rate of change of efficiency at x = 5 as -2/5, which means that at this production level, the efficiency decreases as the production level increases.
Derive the second derivative of the function y = 10/x.
The first step is to find the first derivative, dy/dx = -10/x²...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-10/x²] = 20/x³
Therefore, the second derivative of the function y = 10/x is 20/x³.
We use the step-by-step process, starting with the first derivative. We then apply the power rule again to find the second derivative, which results in 20/x³.
Prove: d/dx [(10/x)²] = -20/x³
Let’s start using the chain rule: Consider y = (10/x)² = [10x⁻¹]²
To differentiate, we use the chain rule: dy/dx = 2[10x⁻¹]·d/dx [10x⁻¹]
Since the derivative of 10x⁻¹ is -10x⁻², dy/dx = 2[10x⁻¹]·[-10x⁻²] = -20/x³
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace 10x⁻¹ with its derivative. As a final step, we substitute back to derive the equation.
Solve: d/dx (10x/x)
To differentiate the function, we use the quotient rule: d/dx (10x/x) = (d/dx (10x)·x - 10x·d/dx(x))/x²
We will substitute d/dx (10x) = 10 and d/dx (x) = 1 = (10·x - 10x·1)/x² = (10x - 10x)/x² = 0/x²
Therefore, d/dx (10x/x) = 0
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result of 0.
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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