Last updated on July 15th, 2025
We use the derivative of 1/y, which is -1/y², as a measuring tool for how the function 1/y changes in response to a slight change in y. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 1/y in detail.
We now understand the derivative of 1/y. It is commonly represented as d/dy (1/y) or (1/y)', and its value is -1/y². The function 1/y has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Reciprocal Function: 1/y is the reciprocal of y. Power Rule: Rule for differentiating y^(-1). Negative Power: Differentiating negative powers of y.
The derivative of 1/y can be denoted as d/dy (1/y) or (1/y)'. The formula we use to differentiate 1/y is: d/dy (1/y) = -1/y² (or) (1/y)' = -1/y² The formula applies to all y where y ≠ 0.
We can derive the derivative of 1/y using proofs. To show this, we will use the power rule along with basic differentiation concepts. 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 1/y results in -1/y² using the above-mentioned methods: By First Principle The derivative of 1/y can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 1/y using the first principle, we will consider f(y) = 1/y. Its derivative can be expressed as the following limit. f'(y) = limₕ→₀ [f(y + h) - f(y)] / h Given that f(y) = 1/y, we write f(y + h) = 1/(y + h). Substituting these into the equation, f'(y) = limₕ→₀ [1/(y + h) - 1/y] / h = limₕ→₀ [y - (y + h)] / [h(y + h)y] = limₕ→₀ [-h] / [h(y + h)y] = limₕ→₀ [-1] / [(y + h)y] As h approaches zero, f'(y) = -1/y² Hence, proved. Using Power Rule To prove the differentiation of 1/y using the power rule, We write 1/y as y^(-1). Differentiating using the power rule: d/dy [y^(-1)] = -1 * y^(-1 - 1) = -1/y². Hence, the derivative of 1/y is -1/y².
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 1/y. For the first derivative of a function, we write f′(y), 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′′(y). Similarly, the third derivative, f′′′(y), is the result of the second derivative, and this pattern continues. For the nth derivative of 1/y, we generally use fⁿ(y) for the nth derivative of a function f(y), which tells us the change in the rate of change (continuing for higher-order derivatives).
When y is 0, the derivative is undefined because 1/y has a vertical asymptote there. When y is 1, the derivative of 1/y = -1/1², which is -1.
Students frequently make mistakes when differentiating 1/y. 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/y² · y³)
Here, we have f(y) = 1/y² · y³. Using the product rule, f'(y) = u′v + uv′ In the given equation, u = 1/y² and v = y³. Let’s differentiate each term, u′= d/dy (1/y²) = -2/y³ v′= d/dy (y³) = 3y² Substituting into the given equation, f'(y) = (-2/y³) · (y³) + (1/y²) · (3y²) Let’s simplify terms to get the final answer, f'(y) = -2 + 3 f'(y) = 1 Thus, the derivative of the specified function is 1.
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 elevation of a hill is represented by the function y = 1/z where y represents the height at a distance z. If z = 2 meters, measure the rate of change of elevation.
We have y = 1/z (elevation of the hill)...(1) Now, we will differentiate the equation (1) Take the derivative of 1/z: dy/dz = -1/z² Given z = 2 (substitute this into the derivative) dy/dz = -1/2² dy/dz = -1/4 Hence, we get the rate of change of the elevation at a distance z = 2 as -1/4.
We find the rate of change of the elevation at z = 2 as -1/4, which means that at a given point, the height of the hill decreases as the horizontal distance increases.
Derive the second derivative of the function y = 1/y.
The first step is to find the first derivative, dy/dy = -1/y²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dy² = d/dy [-1/y²] d²y/dy² = 2/y³ Therefore, the second derivative of the function y = 1/y is 2/y³.
We use the step-by-step process, where we start with the first derivative. We then differentiate -1/y² to find the final answer using the power rule.
Prove: d/dy ((1/y)²) = -2/y³.
Let’s start using the chain rule: Consider y = (1/y)² = [1/y]² To differentiate, we use the chain rule: dy/dy = 2(1/y) * d/dy [1/y] Since the derivative of 1/y is -1/y², dy/dy = 2(1/y) * (-1/y²) = -2/y³ Substituting y = (1/y)², d/dy ((1/y)²) = -2/y³ Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace 1/y with its derivative. As a final step, we substitute y = (1/y)² to derive the equation.
Solve: d/dy (1/y²)
To differentiate the function, we use the power rule: d/dy (1/y²) = d/dy (y^(-2)) = -2y^(-3) = -2/y³ Therefore, d/dy (1/y²) = -2/y³.
In this process, we differentiate the given function using the power 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 y. Reciprocal Function: A function that represents the reciprocal of a variable, commonly written as 1/y. Power Rule: A basic rule of differentiation used to find the derivative of functions of the form y^n. First Derivative: The initial result of a function, which gives us the rate of change of a specific function. Asymptote: A line that a function approaches but never touches, often occurring in reciprocal functions like 1/y.
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