Derivative of xy²
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Last updated on September 27, 2025

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).

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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:

Problem 1

Calculate the derivative of (xy² + x²y).

Okay, lets begin

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.

Well explained 👍

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.

Okay, lets begin

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.

Well explained 👍

Problem 3

Derive the second derivative of the function h = xy².

Okay, lets begin

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.

Well explained 👍

Problem 4

Prove: d/dx (x²y²) = 2xy² + 2x²y(dy/dx).

Okay, lets begin

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.

Well explained 👍

Problem 5

Solve: d/dx (x/y).

Okay, lets begin

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.

Well explained 👍

FAQs on the Derivative of xy²

1.Find the derivative of xy² with respect to x.

Using the product rule and chain rule, d/dx (xy²) = y² + 2xy(dy/dx).

2.Can we use the derivative of xy² in real life?

Yes, derivatives like xy² can be applied in real-life situations involving rates of change, such as calculating the growth rate of an area or volume.

3.Is it possible to take the derivative of xy² when y is a constant?

Yes, when y is a constant, the derivative simplifies to 2xy(dy/dx), showing the linear relationship with x.

4.What rule is used to differentiate x/y?

We use the quotient rule to differentiate x/y, where d/dx (x/y) = (y - x(dy/dx)) / y².

5.Are the derivatives of xy² and (xy)² the same?

No, they are different. The derivative of xy² involves the product rule and is y² + 2xy(dy/dx), while the derivative of (xy)² involves a different application of the chain and product rules.

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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