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2026-01-01
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>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.</p>
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<p>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.</p>
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<h2>What is the Derivative of xy²?</h2>
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<h2>What is the Derivative of xy²?</h2>
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<p>We now understand the derivative<a>of</a>xy². It is commonly represented as d/dx (xy²) or (xy²)', and its value depends on the application of the<a>product</a>and chain rules. The<a>function</a>xy² is differentiable, indicating it has a clearly defined derivative within its domain.</p>
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<p>We now understand the derivative<a>of</a>xy². It is commonly represented as d/dx (xy²) or (xy²)', and its value depends on the application of the<a>product</a>and chain rules. The<a>function</a>xy² is differentiable, indicating it has a clearly defined derivative within its domain.</p>
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<p>The key concepts are mentioned below:</p>
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<p>The key concepts are mentioned below:</p>
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<p><strong>Product Rule:</strong>A rule for differentiating functions that are products of two other functions.</p>
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<p><strong>Product Rule:</strong>A rule for differentiating functions that are products of two other functions.</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</p>
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<p><strong>Partial Derivative:</strong>A derivative taken with respect to one<a>variable</a>while keeping other variables<a>constant</a>.</p>
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<p><strong>Partial Derivative:</strong>A derivative taken with respect to one<a>variable</a>while keeping other variables<a>constant</a>.</p>
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<h2>Derivative of xy² Formula</h2>
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<h2>Derivative of xy² Formula</h2>
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<p>The derivative of xy² with respect to x can be denoted as d/dx (xy²). The<a>formula</a>we use to differentiate xy² is: d/dx (xy²) = y² + 2xy(dy/dx)</p>
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<p>The derivative of xy² with respect to x can be denoted as d/dx (xy²). The<a>formula</a>we use to differentiate xy² is: d/dx (xy²) = y² + 2xy(dy/dx)</p>
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<p>The formula applies when y is a function of x, making it crucial to apply the product rule and chain rule effectively.</p>
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<p>The formula applies when y is a function of x, making it crucial to apply the product rule and chain rule effectively.</p>
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<h2>Proofs of the Derivative of xy²</h2>
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<h2>Proofs of the Derivative of xy²</h2>
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<p>We can derive the derivative of xy² using various methods. To demonstrate this, we will use the rules of differentiation.</p>
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<p>We can derive the derivative of xy² using various methods. To demonstrate this, we will use the rules of differentiation.</p>
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<p>Several methods are:</p>
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<p>Several methods are:</p>
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<ul><li>Using Product Rule </li>
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<ul><li>Using Product Rule </li>
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<li>Using Chain Rule</li>
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<li>Using Chain Rule</li>
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</ul><p>We will now demonstrate the differentiation of xy² using these methods:</p>
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</ul><p>We will now demonstrate the differentiation of xy² using these methods:</p>
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<h2><strong>Using Product Rule</strong></h2>
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<h2><strong>Using Product Rule</strong></h2>
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<p>To differentiate xy² using the product rule, consider u = x and v = y². d/dx (uv) = u'v + uv' Differentiate each<a>term</a>: 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).</p>
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<p>To differentiate xy² using the product rule, consider u = x and v = y². d/dx (uv) = u'v + uv' Differentiate each<a>term</a>: 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).</p>
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<h2><strong>Using Chain Rule</strong></h2>
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<h2><strong>Using Chain Rule</strong></h2>
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<p>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).</p>
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<p>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).</p>
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<h2>Higher-Order Derivatives of xy²</h2>
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<h2>Higher-Order Derivatives of xy²</h2>
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<p>When a function is differentiated<a>multiple</a>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.</p>
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<p>When a function is differentiated<a>multiple</a>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.</p>
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<p>For the first derivative of a function, we write f′(x), indicating the<a>rate</a>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.</p>
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<p>For the first derivative of a function, we write f′(x), indicating the<a>rate</a>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.</p>
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<p>For the nth Derivative of xy², we generally denote it as fⁿ(x), representing the change in the rate of change.</p>
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<p>For the nth Derivative of xy², we generally denote it as fⁿ(x), representing the change in the rate of change.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When y is constant, the derivative simplifies to 2xy(dy/dx), showing that changes in x affect the function linearly.</p>
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<p>When y is constant, the derivative simplifies to 2xy(dy/dx), showing that changes in x affect the function linearly.</p>
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<p>If x is constant, the derivative becomes 0, indicating no change as x remains fixed.</p>
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<p>If x is constant, the derivative becomes 0, indicating no change as x remains fixed.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of xy²</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of xy²</h2>
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<p>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:</p>
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<p>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:</p>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Calculate the derivative of (xy² + x²y).</p>
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<p>Calculate the derivative of (xy² + x²y).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>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).</p>
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<p>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).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>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.</p>
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<p>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.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function h = xy².</p>
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<p>Derive the second derivative of the function h = xy².</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>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)].</p>
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<p>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)].</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use the step-by-step process, starting with the first derivative and applying differentiation rules to find the second derivative of the function.</p>
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<p>We use the step-by-step process, starting with the first derivative and applying differentiation rules to find the second derivative of the function.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: d/dx (x²y²) = 2xy² + 2x²y(dy/dx).</p>
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<p>Prove: d/dx (x²y²) = 2xy² + 2x²y(dy/dx).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>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.</p>
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<p>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.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>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.</p>
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<p>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.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: d/dx (x/y).</p>
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<p>Solve: d/dx (x/y).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>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.</p>
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<p>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.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we differentiate the given function using the quotient rule and simplify the equation to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the quotient rule and simplify the equation to obtain the final result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs on the Derivative of xy²</h2>
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<h2>FAQs on the Derivative of xy²</h2>
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<h3>1.Find the derivative of xy² with respect to x.</h3>
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<h3>1.Find the derivative of xy² with respect to x.</h3>
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<p>Using the product rule and chain rule, d/dx (xy²) = y² + 2xy(dy/dx).</p>
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<p>Using the product rule and chain rule, d/dx (xy²) = y² + 2xy(dy/dx).</p>
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<h3>2.Can we use the derivative of xy² in real life?</h3>
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<h3>2.Can we use the derivative of xy² in real life?</h3>
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<p>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.</p>
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<p>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.</p>
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<h3>3.Is it possible to take the derivative of xy² when y is a constant?</h3>
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<h3>3.Is it possible to take the derivative of xy² when y is a constant?</h3>
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<p>Yes, when y is a constant, the derivative simplifies to 2xy(dy/dx), showing the linear relationship with x.</p>
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<p>Yes, when y is a constant, the derivative simplifies to 2xy(dy/dx), showing the linear relationship with x.</p>
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<h3>4.What rule is used to differentiate x/y?</h3>
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<h3>4.What rule is used to differentiate x/y?</h3>
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<p>We use the<a>quotient</a>rule to differentiate x/y, where d/dx (x/y) = (y - x(dy/dx)) / y².</p>
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<p>We use the<a>quotient</a>rule to differentiate x/y, where d/dx (x/y) = (y - x(dy/dx)) / y².</p>
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<h3>5.Are the derivatives of xy² and (xy)² the same?</h3>
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<h3>5.Are the derivatives of xy² and (xy)² the same?</h3>
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<p>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.</p>
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<p>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.</p>
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<h2>Important Glossaries for the Derivative of xy²</h2>
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<h2>Important Glossaries for the Derivative of xy²</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes with respect to changes in variables.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes with respect to changes in variables.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used for differentiating functions that are products of two or more functions.</li>
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</ul><ul><li><strong>Product Rule:</strong>A rule used for differentiating functions that are products of two or more functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Partial Derivative:</strong>A derivative taken with respect to one variable, keeping other variables constant.</li>
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</ul><ul><li><strong>Partial Derivative:</strong>A derivative taken with respect to one variable, keeping other variables constant.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used for differentiating functions that are ratios of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used for differentiating functions that are ratios of two functions.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<p>▶</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<p>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.</p>
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<p>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.</p>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>