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119 LearnersLast updated on September 6, 2025
Derivative of x/y
We use the derivative of x/y, which helps us understand how the quotient of x and y changes with slight variations in x and y. Derivatives are essential tools in determining rates of change in real-life situations. Let's explore the derivative of x/y in detail.
What is the Derivative of x/y?
The derivative of x/y is commonly represented as d/dx (x/y) or (x/y)'.
The process involves using the quotient rule, which applies to differentiating a function of the form u/v, where u and v are functions of x.
The derivative of x/y is expressed as (y·d/dx(x) - x·d/dx(y))/y² when y ≠ 0.
This indicates that x/y is differentiable wherever y is non-zero.
The key concepts include:
Quotient Rule: A rule for differentiating functions of the form u/v.
Differentiability: The ability of a function to have a derivative at a specific point.
Derivative of x/y Formula
The formula for differentiating x/y is derived using the quotient rule: d/dx (x/y) = (y·d/dx(x) - x·d/dx(y))/y² This formula is applicable for all x where y ≠ 0.
Proofs of the Derivative of x/y
We can derive the derivative of x/y using the quotient rule.
This method utilizes the fundamental rules of differentiation and is outlined as follows:
Using Quotient Rule
For functions u(x) = x and v(x) = y, the quotient rule states: d/dx (u/v) = (v·u' - u·v')/v²
Substituting u(x) = x and v(x) = y, we have: d/dx (x/y) = (y·1 - x·d/dx(y))/y² = (y - x·dy/dx)/y²
This result provides the derivative of x/y when y ≠ 0.
Higher-Order Derivatives of x/y
Higher-order derivatives involve differentiating a function multiple times.
For x/y, the process follows by differentiating the first derivative again to obtain the second derivative, and so forth.
These derivatives provide deeper insights into the curvature and concavity of the function.
The first derivative is expressed as d/dx (x/y).
The second derivative involves differentiating d/dx (x/y) again, which can become increasingly complex.
The notation for higher-order derivatives continues with f''(x), f'''(x), and so on, for the nth derivative fⁿ(x).
Special Cases:
When y = 0, the derivative is undefined because division by zero is not possible. When y is constant, say y = c, the derivative simplifies to zero because the function x/c has a constant rate of change.
Common Mistakes and How to Avoid Them in Derivatives of x/y
Students often make errors when differentiating x/y. These can be mitigated by understanding the correct procedures. Here are some common mistakes and solutions:
Mistake 1
Misapplying the Quotient Rule
Students might incorrectly apply the quotient rule by not correctly identifying u and v.
For example, they may confuse the formula order: d/dx (u/v) = (v·u' - u·v')/v². To avoid this, ensure each component is properly substituted and simplified.
Mistake 2
Ignoring Division by Zero
Failing to recognize points where y = 0 can lead to undefined expressions. Always check the domain of the function before differentiating to avoid division by zero.
Mistake 3
Forgetting to Differentiate Both u and v
When using the quotient rule, students sometimes forget to differentiate both the numerator and the denominator. It's crucial to apply d/dx to both u and v to obtain the correct derivative.
Mistake 4
Not Simplifying the Result
Students may leave the derivative unsimplified, leading to complex expressions. Always simplify the result to its simplest form for clarity.
Mistake 5
Overlooking Constant Multipliers
Constants in expressions like d/dx (k·x/y) may be disregarded. Ensure constants are correctly factored into the derivative, as they influence the final result.
Examples Using the Derivative of x/y
Problem 1
Calculate the derivative of (x/y²).
Here, we have f(x) = x/y².
Using the quotient rule, f'(x) = (y²·d/dx(x) - x·d/dx(y²))/(y²)² = (y²·1 - x·2y·d/dx(y))/(y⁴) = (y² - 2xy·dy/dx)/y⁴
Therefore, the derivative of the given function is (y² - 2xy·dy/dx)/y⁴.
Explanation
We find the derivative of the given function by applying the quotient rule. Each step involves differentiating the numerator and denominator, then simplifying the result.
Problem 2
A company manufactures a product at a rate represented by the function P = x/y, where P is the production rate, x is the number of workers, and y is the time in hours. If x = 10 and y = 5, find the rate of change of production with respect to time.
Given P = x/y, we differentiate with respect to time,
t: dP/dt = (y·dx/dt - x·dy/dt)/y²
For x = 10, y = 5, dx/dt = 0 (constant workers), and dy/dt = 1 (hours increase), dP/dt = (5·0 - 10·1)/5² = -10/25 = -0.4
Hence, the production rate decreases by 0.4 units per hour.
Explanation
We apply the quotient rule to differentiate the production rate function concerning time. Substituting the given values helps determine the rate of change.
Problem 3
Derive the second derivative of the function f(x) = x/y.
First, we find the first derivative: f'(x) = (y·1 - x·dy/dx)/y²
Now, differentiate f'(x) to find the second derivative: f''(x) = d/dx [(y - x·dy/dx)/y²]
We apply the quotient rule again, which involves differentiating the numerator and denominator separately.
This results in a more complex expression, typically involving higher derivatives of y.
Explanation
We start with the first derivative and proceed to find the second derivative by differentiating the result again, which requires careful application of differentiation rules.
Problem 4
Prove: d/dx (x²/y) = (2xy - x²·dy/dx)/y².
Let’s start using the quotient rule: Consider f(x) = x²/y d/dx (x²/y) = (y·d/dx(x²) - x²·d/dx(y))/y² = (y·2x - x²·dy/dx)/y² Hence proved.
Explanation
In this step-by-step process, we use the quotient rule to differentiate the equation. We then simplify the expression to derive the formula.
Problem 5
Solve: d/dx (x/y + y/x).
To differentiate the function, d/dx (x/y + y/x) = d/dx (x/y) + d/dx (y/x)
Using the quotient rule for each term: = [(y - x·dy/dx)/y²] + [(x·dy/dx - y)/x²]
Therefore, the derivative is: (y - x·dy/dx)/y² + (x·dy/dx - y)/x².
Explanation
In this process, we differentiate each term using the quotient rule. The results are then combined and simplified to obtain the final expression.
FAQs on the Derivative of x/y
1.Find the derivative of x/y.
2.Can the derivative of x/y be applied in real life?
3.Is it possible to differentiate x/y when y = 0?
4.What rule is used to differentiate x/y?
5.Is differentiating x/y the same as differentiating y/x?
Important Glossaries for the Derivative of x/y
- Derivative: A measure of how a function changes as its input changes.
- Quotient Rule: A technique for finding the derivative of a quotient of two functions.
- Differentiability: The condition of a function being differentiable at a point or over an interval.
- Higher-Order Derivatives: Derivatives of a function taken multiple times, providing insights into its curvature.
- Undefined: A term used to describe a mathematical expression that lacks meaning, often due to division by zero.
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Jaskaran Singh Saluja
About the Author
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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: He loves to play the quiz with kids through algebra to make kids love it.
