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106 LearnersLast updated on September 27, 2025
Derivative of xy²
We use the derivative of xy² to understand how the function changes with respect to slight variations in x or y. Derivatives are fundamental in calculating rates of change and can be applied in various real-life scenarios. We will now discuss the derivative of xy² in detail.
What is the Derivative of xy²?
We now understand the derivative of xy². It is commonly represented as d/dx (xy²) or (xy²)', and its value depends on the application of the product and chain rules. The function xy² is differentiable, indicating it has a clearly defined derivative within its domain.
The key concepts are mentioned below:
Product Rule: A rule for differentiating functions that are products of two other functions.
Chain Rule: A rule for differentiating compositions of functions.
Partial Derivative: A derivative taken with respect to one variable while keeping other variables constant.
Derivative of xy² Formula
The derivative of xy² with respect to x can be denoted as d/dx (xy²). The formula we use to differentiate xy² is: d/dx (xy²) = y² + 2xy(dy/dx)
The formula applies when y is a function of x, making it crucial to apply the product rule and chain rule effectively.
Proofs of the Derivative of xy²
We can derive the derivative of xy² using various methods. To demonstrate this, we will use the rules of differentiation.
Several methods are:
- Using Product Rule
- Using Chain Rule
We will now demonstrate the differentiation of xy² using these methods:
Using Product Rule
To differentiate xy² using the product rule, consider u = x and v = y². d/dx (uv) = u'v + uv' Differentiate each term: u' = d/dx (x) = 1 v' = d/dx (y²) = 2y(dy/dx) Apply the product rule: d/dx (xy²) = (1)(y²) + (x)(2y)(dy/dx) Thus, the derivative of xy² is y² + 2xy(dy/dx).
Using Chain Rule
Consider xy² as a composition of functions. Let u = y², then xy² = x(u). d/dx (x(u)) = x(du/dx) + u(dx/dx) Since dx/dx = 1 and du/dx = 2y(dy/dx), d/dx (xy²) = x(2y(dy/dx)) + y² Therefore, the derivative of xy² is y² + 2xy(dy/dx).
Higher-Order Derivatives of xy²
When a function is differentiated multiple times, the resulting derivatives are referred to as higher-order derivatives. Higher-order derivatives can be challenging but provide insights into the behavior of functions. For example, the second derivative can tell us about the curvature or concavity of the function.
For the first derivative of a function, we write f′(x), indicating the rate of change of the function at a certain point. The second derivative, denoted as f′′(x), is derived from the first derivative and can indicate acceleration or deceleration.
For the nth Derivative of xy², we generally denote it as fⁿ(x), representing the change in the rate of change.
Special Cases:
When y is constant, the derivative simplifies to 2xy(dy/dx), showing that changes in x affect the function linearly.
If x is constant, the derivative becomes 0, indicating no change as x remains fixed.
Common Mistakes and How to Avoid Them in Derivatives of xy²
Students frequently make mistakes when differentiating functions like xy². These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Mistake 1
Not applying the Product Rule correctly
Students may overlook the application of the product rule, especially when the function consists of two multiplicative terms.
Always ensure to apply the product rule when differentiating functions like xy².
Mistake 2
Ignoring the Chain Rule
In cases where y is a function of x, students might forget to apply the chain rule, leading to incorrect results.
Always consider dy/dx when differentiating terms involving y.
Mistake 3
Misinterpreting Partial Derivatives
When dealing with multivariable functions, students might confuse partial derivatives with regular derivatives.
Ensure you differentiate with respect to the appropriate variable and recognize when others are constants.
Mistake 4
Forgetting Constants
Students might forget to account for constants that multiply the function, leading to errors.
Always multiply the entire differentiated result by the constant factor.
Mistake 5
Overcomplicating Simple Derivatives
Sometimes, students might overcomplicate simple derivatives by introducing unnecessary steps.
Focus on applying the rules correctly and simplifying the results.
Examples Using the Derivative of xy²
Problem 1
Calculate the derivative of (xy² + x²y).
Here, we have f(x, y) = xy² + x²y. Using the product rule for each term separately: For xy²: u = x, v = y² d/dx (xy²) = y² + 2xy(dy/dx) For x²y: u = x², v = y d/dx (x²y) = 2xy + x²(dy/dx) Combine the derivatives: f'(x, y) = y² + 2xy(dy/dx) + 2xy + x²(dy/dx) Thus, the derivative of the specified function is y² + 2xy(dy/dx) + 2xy + x²(dy/dx).
Explanation
We find the derivative of the given function by applying the product rule to each term separately, then combining the results to get the final derivative.
Problem 2
A cylindrical tank has a height represented by h = xy², where x is the radius of the base. If x = 3 meters and y = 2 meters, find the rate of change of the height h with respect to x.
We have h = xy² (height of the tank)...(1) Differentiate the equation (1) with respect to x: dh/dx = y² + 2xy(dy/dx) Given x = 3 meters, y = 2 meters, and assuming dy/dx = 0 since y is a constant: dh/dx = (2)² + 2(3)(2)(0) = 4 Hence, the rate of change of the height h with respect to x is 4 meters per unit change in x.
Explanation
We find the rate of change of height by differentiating h = xy² with respect to x and substituting the given values, assuming y remains constant.
Problem 3
Derive the second derivative of the function h = xy².
The first step is to find the first derivative: dh/dx = y² + 2xy(dy/dx)...(1) Now differentiate equation (1) again to get the second derivative: d²h/dx² = d/dx [y² + 2xy(dy/dx)] Assuming y = y(x), use the product rule: d²h/dx² = 0 + 2[(dy/dx) + x(d²y/dx²)] Therefore, the second derivative of the function h = xy² is 2[(dy/dx) + x(d²y/dx²)].
Explanation
We use the step-by-step process, starting with the first derivative and applying differentiation rules to find the second derivative of the function.
Problem 4
Prove: d/dx (x²y²) = 2xy² + 2x²y(dy/dx).
Let’s start using the product rule: Consider z = x²y² To differentiate, apply the product rule: d/dx (x²y²) = d/dx (x²)y² + x²d/dx (y²) Differentiate each term: d/dx (x²) = 2x d/dx (y²) = 2y(dy/dx) Substitute these into the equation: d/dx (x²y²) = (2x)y² + x²(2y)(dy/dx) = 2xy² + 2x²y(dy/dx) Hence proved.
Explanation
In this step-by-step process, we applied the product rule to differentiate the equation, then substituted the derivatives of each component to derive the equation.
Problem 5
Solve: d/dx (x/y).
To differentiate the function, use the quotient rule: d/dx (x/y) = [d/dx (x) · y - x · d/dx (y)] / y² Differentiate each term: d/dx (x) = 1 d/dx (y) = dy/dx Substitute these into the equation: d/dx (x/y) = (1 · y - x · dy/dx) / y² = (y - x(dy/dx)) / y² Therefore, d/dx (x/y) = (y - x(dy/dx)) / y².
Explanation
In this process, we differentiate the given function using the quotient rule and simplify the equation to obtain the final result.
FAQs on the Derivative of xy²
1.Find the derivative of xy² with respect to x.
2.Can we use the derivative of xy² in real life?
3.Is it possible to take the derivative of xy² when y is a constant?
4.What rule is used to differentiate x/y?
5.Are the derivatives of xy² and (xy)² the same?
Important Glossaries for the Derivative of xy²
- Derivative: The derivative of a function indicates how the given function changes with respect to changes in variables.
- Product Rule: A rule used for differentiating functions that are products of two or more functions.
- Chain Rule: A rule used for differentiating compositions of functions.
- Partial Derivative: A derivative taken with respect to one variable, keeping other variables constant.
- Quotient Rule: A rule used for differentiating functions that are ratios of two functions.
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Jaskaran Singh Saluja
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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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