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2026-01-01
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>We use the derivative of 2xy, which helps us understand how the function changes in response to a slight change in x or y. Derivatives are useful in calculating rates of change in real-life situations. We will now discuss the derivative of 2xy in detail.</p>
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<p>We use the derivative of 2xy, which helps us understand how the function changes in response to a slight change in x or y. Derivatives are useful in calculating rates of change in real-life situations. We will now discuss the derivative of 2xy in detail.</p>
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<h2>What is the Derivative of 2xy?</h2>
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<h2>What is the Derivative of 2xy?</h2>
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<p>The derivative of 2xy is commonly represented as d/dx (2xy) or (2xy)'.</p>
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<p>The derivative of 2xy is commonly represented as d/dx (2xy) or (2xy)'.</p>
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<p>The<a>function</a>2xy is differentiable, and its derivative within its domain can be expressed using the<a>product</a>rule.</p>
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<p>The<a>function</a>2xy is differentiable, and its derivative within its domain can be expressed using the<a>product</a>rule.</p>
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<p>The key concepts are mentioned below: Product Rule: Rule for differentiating products<a>of functions</a>.</p>
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<p>The key concepts are mentioned below: Product Rule: Rule for differentiating products<a>of functions</a>.</p>
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<p>Partial Derivatives: Differentiation concerning one<a>variable</a>while keeping others<a>constant</a>.</p>
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<p>Partial Derivatives: Differentiation concerning one<a>variable</a>while keeping others<a>constant</a>.</p>
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<h2>Derivative of 2xy Formula</h2>
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<h2>Derivative of 2xy Formula</h2>
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<p>The derivative of 2xy can be denoted as d/dx (2xy) or (2xy)'. When differentiating with respect to x, using the product rule, we get: d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>The derivative of 2xy can be denoted as d/dx (2xy) or (2xy)'. When differentiating with respect to x, using the product rule, we get: d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>The<a>formula</a>applies in scenarios where both x and y can independently vary.</p>
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<p>The<a>formula</a>applies in scenarios where both x and y can independently vary.</p>
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<h2>Proofs of the Derivative of 2xy</h2>
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<h2>Proofs of the Derivative of 2xy</h2>
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<p>To derive the derivative of 2xy, we will use the product rule and partial derivatives.</p>
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<p>To derive the derivative of 2xy, we will use the product rule and partial derivatives.</p>
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<p>Here's how it's done: Using Product Rule To prove the differentiation of 2xy using the product rule, consider u = 2x and v = y.</p>
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<p>Here's how it's done: Using Product Rule To prove the differentiation of 2xy using the product rule, consider u = 2x and v = y.</p>
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<p>Using the product rule: d/dx [u.v] = u'.v + u.v' u' = d/dx (2x) = 2 v' = dy/dx Therefore, d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>Using the product rule: d/dx [u.v] = u'.v + u.v' u' = d/dx (2x) = 2 v' = dy/dx Therefore, d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>Using Partial Derivatives The derivative of 2xy with respect to x also involves considering partial derivatives.</p>
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<p>Using Partial Derivatives The derivative of 2xy with respect to x also involves considering partial derivatives.</p>
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<p>Treat y as a constant: ∂/∂x (2xy) = 2y For differentiation with respect to y, treat x as a constant: ∂/∂y (2xy) = 2x</p>
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<p>Treat y as a constant: ∂/∂x (2xy) = 2y For differentiation with respect to y, treat x as a constant: ∂/∂y (2xy) = 2x</p>
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<p>Thus, the derivative can be expressed as a<a>combination</a>of these partial derivatives.</p>
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<p>Thus, the derivative can be expressed as a<a>combination</a>of these partial derivatives.</p>
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<h2>Higher-Order Derivatives of 2xy</h2>
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<h2>Higher-Order Derivatives of 2xy</h2>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times.</p>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times.</p>
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<p>For example, differentiating 2xy with respect to x and then again yields: The first derivative with respect to x is: d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>For example, differentiating 2xy with respect to x and then again yields: The first derivative with respect to x is: d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>The second derivative involves differentiating the first derivative: d²/dx² (2xy) = 2(dy/dx) + 2x(d²y/dx²)</p>
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<p>The second derivative involves differentiating the first derivative: d²/dx² (2xy) = 2(dy/dx) + 2x(d²y/dx²)</p>
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<p>Higher-order derivatives help in understanding the<a>rate</a>of change of the function concerning x and y.</p>
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<p>Higher-order derivatives help in understanding the<a>rate</a>of change of the function concerning x and y.</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 reduces to 2y, representing the slope of a line parallel to the x-axis.</p>
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<p>When y is constant, the derivative reduces to 2y, representing the slope of a line parallel to the x-axis.</p>
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<p>When x is constant, the derivative reduces to 2x(dy/dx), indicating the rate of change concerning y.</p>
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<p>When x is constant, the derivative reduces to 2x(dy/dx), indicating the rate of change concerning y.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 2xy</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 2xy</h2>
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<p>Students frequently make mistakes when differentiating 2xy.</p>
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<p>Students frequently make mistakes when differentiating 2xy.</p>
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<p>These mistakes can be resolved by understanding the proper solutions.</p>
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<p>These mistakes can be resolved by understanding the proper solutions.</p>
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<p>Here are a few common mistakes and ways to solve them:</p>
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<p>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 (2xy²).</p>
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<p>Calculate the derivative of (2xy²).</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) = 2xy².</p>
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<p>Here, we have f(x, y) = 2xy².</p>
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<p>Using the product rule and treating y² as a function of y, f'(x) = 2y² + 2xy(2y)(dy/dx)</p>
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<p>Using the product rule and treating y² as a function of y, f'(x) = 2y² + 2xy(2y)(dy/dx)</p>
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<p>Simplifying, we get: f'(x) = 2y² + 4xy(dy/dx) Thus, the derivative of the specified function is 2y² + 4xy(dy/dx).</p>
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<p>Simplifying, we get: f'(x) = 2y² + 4xy(dy/dx) Thus, the derivative of the specified function is 2y² + 4xy(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 and considering y² as a function of y. This involves differentiating with respect to both x and y.</p>
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<p>We find the derivative of the given function by applying the product rule and considering y² as a function of y. This involves differentiating with respect to both x and y.</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 rectangular field has an area represented by A = 2xy, where x is the length, and y is the width in meters. If x = 5 meters and y = 3 meters, find the rate of change of area with respect to x.</p>
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<p>A rectangular field has an area represented by A = 2xy, where x is the length, and y is the width in meters. If x = 5 meters and y = 3 meters, find the rate of change of area 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>Given A = 2xy, To find the rate of change concerning x, differentiate A with respect to x: dA/dx = 2y + 2x(dy/dx)</p>
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<p>Given A = 2xy, To find the rate of change concerning x, differentiate A with respect to x: dA/dx = 2y + 2x(dy/dx)</p>
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<p>Substitute x = 5, y = 3, and assume dy/dx = 0 (y is constant): dA/dx = 2(3) + 2(5)(0) dA/dx = 6</p>
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<p>Substitute x = 5, y = 3, and assume dy/dx = 0 (y is constant): dA/dx = 2(3) + 2(5)(0) dA/dx = 6</p>
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<p>Hence, the rate of change of the area with respect to x is 6 square meters per meter.</p>
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<p>Hence, the rate of change of the area with respect to x is 6 square meters per meter.</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 the area by differentiating the area function with respect to x, treating y as a constant. Substituting the given values provides the final result.</p>
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<p>We find the rate of change of the area by differentiating the area function with respect to x, treating y as a constant. Substituting the given values provides the final result.</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 f(x, y) = 2xy.</p>
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<p>Derive the second derivative of the function f(x, y) = 2xy.</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, d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>The first step is to find the first derivative, d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>Now, differentiate the first derivative to get the second derivative: d²/dx² (2xy) = 2(dy/dx) + 2x(d²y/dx²)</p>
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<p>Now, differentiate the first derivative to get the second derivative: d²/dx² (2xy) = 2(dy/dx) + 2x(d²y/dx²)</p>
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<p>Therefore, the second derivative of the function f(x, y) = 2xy is 2(dy/dx) + 2x(d²y/dx²).</p>
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<p>Therefore, the second derivative of the function f(x, y) = 2xy is 2(dy/dx) + 2x(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 to find the first derivative and then differentiate it again to obtain the second derivative concerning x.</p>
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<p>We use the step-by-step process to find the first derivative and then differentiate it again to obtain the second derivative concerning x.</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 (2x²y) = 4xy + 2x²(dy/dx).</p>
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<p>Prove: d/dx (2x²y) = 4xy + 2x²(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 f(x, y) = 2x²y To differentiate, we use the product rule: df/dx = d/dx (2x²)y + 2x²(dy/dx) = 4xy + 2x²(dy/dx) Hence, proved.</p>
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<p>Let's start using the product rule: Consider f(x, y) = 2x²y To differentiate, we use the product rule: df/dx = d/dx (2x²)y + 2x²(dy/dx) = 4xy + 2x²(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 used the product rule to differentiate 2x²y. We broke down the function into parts, differentiated, and then combined the results.</p>
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<p>In this step-by-step process, we used the product rule to differentiate 2x²y. We broke down the function into parts, differentiated, and then combined the results.</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 (2x/y).</p>
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<p>Solve: d/dx (2x/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, we use the quotient rule: d/dx (2x/y) = [d/dx (2x) * y - 2x * d/dx (y)] / y² = [2y - 2x(dy/dx)] / y² Therefore, d/dx (2x/y) = (2y - 2x(dy/dx)) / y²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (2x/y) = [d/dx (2x) * y - 2x * d/dx (y)] / y² = [2y - 2x(dy/dx)] / y² Therefore, d/dx (2x/y) = (2y - 2x(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. We simplify the expression to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the quotient rule. We simplify the expression 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 2xy</h2>
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<h2>FAQs on the Derivative of 2xy</h2>
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<h3>1.Find the derivative of 2xy.</h3>
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<h3>1.Find the derivative of 2xy.</h3>
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<p>Using the product rule to differentiate 2xy, we get: d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<p>Using the product rule to differentiate 2xy, we get: d/dx (2xy) = 2y + 2x(dy/dx)</p>
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<h3>2.Can the derivative of 2xy be used in real-life applications?</h3>
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<h3>2.Can the derivative of 2xy be used in real-life applications?</h3>
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<p>Yes, the derivative of 2xy can be used to determine rates of change in engineering, physics, and economics, especially when dealing with variables that depend on each other.</p>
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<p>Yes, the derivative of 2xy can be used to determine rates of change in engineering, physics, and economics, especially when dealing with variables that depend on each other.</p>
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<h3>3.Is it possible to take the derivative of 2xy when y is a function of x?</h3>
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<h3>3.Is it possible to take the derivative of 2xy when y is a function of x?</h3>
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<p>Yes, when y is a function of x, the derivative involves both dy/dx and x, allowing us to capture the rate of change concerning x.</p>
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<p>Yes, when y is a function of x, the derivative involves both dy/dx and x, allowing us to capture the rate of change concerning x.</p>
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<h3>4.What rule is used to differentiate 2x/y?</h3>
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<h3>4.What rule is used to differentiate 2x/y?</h3>
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<p>We use the<a>quotient</a>rule to differentiate 2x/y: d/dx (2x/y) = (2y - 2x(dy/dx)) / y²</p>
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<p>We use the<a>quotient</a>rule to differentiate 2x/y: d/dx (2x/y) = (2y - 2x(dy/dx)) / y²</p>
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<h3>5.Are the derivatives of 2xy and 2yx the same?</h3>
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<h3>5.Are the derivatives of 2xy and 2yx the same?</h3>
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<p>Yes, since<a>multiplication</a>is commutative, the derivatives of 2xy and 2yx are the same.</p>
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<p>Yes, since<a>multiplication</a>is commutative, the derivatives of 2xy and 2yx are the same.</p>
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<h2>Important Glossaries for the Derivative of 2xy</h2>
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<h2>Important Glossaries for the Derivative of 2xy</h2>
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<ul><li>Derivative: The derivative of a function indicates how the function changes concerning a slight change in variables.</li>
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<ul><li>Derivative: The derivative of a function indicates how the function changes concerning a slight change in variables.</li>
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</ul><ul><li>Product Rule: A rule used to differentiate the product of two functions.</li>
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</ul><ul><li>Product Rule: A rule used to differentiate the product of two functions.</li>
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</ul><ul><li>Partial Derivative: The derivative of a function of multiple variables with respect to one variable, keeping others constant.</li>
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</ul><ul><li>Partial Derivative: The derivative of a function of multiple variables with respect to one variable, keeping others constant.</li>
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</ul><ul><li>Quotient Rule: A rule used to differentiate the ratio of two functions.</li>
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</ul><ul><li>Quotient Rule: A rule used to differentiate the ratio of two functions.</li>
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</ul><ul><li>Second Derivative: The derivative of the first derivative, indicating the change in the rate of change of a function.</li>
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</ul><ul><li>Second Derivative: The derivative of the first derivative, indicating the change in the rate of change of a function.</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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<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>