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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 x+y, which is 1 for both x and y, as a measuring tool for how the function changes in response to a slight change in either x or y. Derivatives help us calculate changes in various scenarios in real-life situations. We will now talk about the derivative of x+y in detail.</p>
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<p>We use the derivative of x+y, which is 1 for both x and y, as a measuring tool for how the function changes in response to a slight change in either x or y. Derivatives help us calculate changes in various scenarios in real-life situations. We will now talk about the derivative of x+y in detail.</p>
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<h2>What is the Derivative of x+y?</h2>
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<h2>What is the Derivative of x+y?</h2>
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<p>We now understand the derivative of x+y. It is commonly represented as d/dx (x+y) or (x+y)'. The derivative of x is 1, and the derivative of y is 1, assuming both x and y are differentiable with respect to their respective<a>variables</a>.</p>
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<p>We now understand the derivative of x+y. It is commonly represented as d/dx (x+y) or (x+y)'. The derivative of x is 1, and the derivative of y is 1, assuming both x and y are differentiable with respect to their respective<a>variables</a>.</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>Simple Addition: The<a>function</a>x+y is the<a>sum</a>of two variables.</p>
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<p>Simple Addition: The<a>function</a>x+y is the<a>sum</a>of two variables.</p>
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<p>Derivative of a Constant: The derivative of a<a>constant</a>is 0.</p>
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<p>Derivative of a Constant: The derivative of a<a>constant</a>is 0.</p>
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<p>Linearity of Differentiation: Differentiation distributes over<a>addition</a>.</p>
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<p>Linearity of Differentiation: Differentiation distributes over<a>addition</a>.</p>
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<h2>Derivative of x+y Formula</h2>
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<h2>Derivative of x+y Formula</h2>
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<p>The derivative of x+y can be denoted as d/dx (x+y) or (x+y)'.</p>
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<p>The derivative of x+y can be denoted as d/dx (x+y) or (x+y)'.</p>
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<p>The<a>formula</a>we use to differentiate x+y is: d/dx (x+y) = 1 + 0 = 1 (or) (x+y)' = 1</p>
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<p>The<a>formula</a>we use to differentiate x+y is: d/dx (x+y) = 1 + 0 = 1 (or) (x+y)' = 1</p>
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<p>Similarly, d/dy (x+y) = 0 + 1 = 1</p>
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<p>Similarly, d/dy (x+y) = 0 + 1 = 1</p>
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<p>This formula holds for all x and y where they are differentiable.</p>
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<p>This formula holds for all x and y where they are differentiable.</p>
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<h2>Proofs of the Derivative of x+y</h2>
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<h2>Proofs of the Derivative of x+y</h2>
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<p>We can derive the derivative of x+y using basic differentiation rules. To show this, we will use the rules of differentiation, including:</p>
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<p>We can derive the derivative of x+y using basic differentiation rules. To show this, we will use the rules of differentiation, including:</p>
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<ol><li>By Linearity of Differentiation</li>
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<ol><li>By Linearity of Differentiation</li>
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<li>Using Basic Derivative Rules</li>
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<li>Using Basic Derivative Rules</li>
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</ol><h3>By Linearity of Differentiation</h3>
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</ol><h3>By Linearity of Differentiation</h3>
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<p>The derivative of x+y can be proved using the linearity of differentiation, which states that the derivative of a sum is the sum of the derivatives.</p>
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<p>The derivative of x+y can be proved using the linearity of differentiation, which states that the derivative of a sum is the sum of the derivatives.</p>
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<p>To find the derivative of x+y, we consider f(x, y) = x+y. Its derivative with respect to x can be expressed as: f'(x) = d/dx (x+y) = d/dx (x) + d/dx (y) = 1 + 0 = 1</p>
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<p>To find the derivative of x+y, we consider f(x, y) = x+y. Its derivative with respect to x can be expressed as: f'(x) = d/dx (x+y) = d/dx (x) + d/dx (y) = 1 + 0 = 1</p>
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<p>Similarly, the derivative with respect to y is: f'(y) = d/dy (x+y) = d/dy (x) + d/dy (y) = 0 + 1 = 1</p>
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<p>Similarly, the derivative with respect to y is: f'(y) = d/dy (x+y) = d/dy (x) + d/dy (y) = 0 + 1 = 1</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Basic Derivative Rules</h3>
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<h3>Using Basic Derivative Rules</h3>
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<p>To prove the differentiation of x+y using basic rules, Consider the functions f(x) = x and g(y) = y.</p>
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<p>To prove the differentiation of x+y using basic rules, Consider the functions f(x) = x and g(y) = y.</p>
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<p>The derivatives are straightforward: d/dx (x) = 1 and d/dy (y) = 1 Thus, d/dx (x+y) = 1 and d/dy (x+y) = 1</p>
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<p>The derivatives are straightforward: d/dx (x) = 1 and d/dy (y) = 1 Thus, d/dx (x+y) = 1 and d/dy (x+y) = 1</p>
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<p>Therefore, the derivative of x+y is 1 with respect to both x and y.</p>
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<p>Therefore, the derivative of x+y is 1 with respect to both x and y.</p>
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<h2>Higher-Order Derivatives of x+y</h2>
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<h2>Higher-Order Derivatives of x+y</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be less complex for linear functions like x+y.</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be less complex for linear functions like x+y.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative, f′′(x), would be 0 for linear functions like x+y, as they don't have curvature. Similarly, the third derivative, f′′′(x), is also 0, and this pattern continues.</p>
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<p>For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative, f′′(x), would be 0 for linear functions like x+y, as they don't have curvature. Similarly, the third derivative, f′′′(x), is also 0, and this pattern continues.</p>
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<p>For the nth Derivative of x+y, we generally use fⁿ(x), where n ≥ 2, resulting in 0, indicating no change in curvature.</p>
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<p>For the nth Derivative of x+y, we generally use fⁿ(x), where n ≥ 2, resulting in 0, indicating no change in curvature.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When x or y is a constant, the derivative of that constant is 0. The derivative of x or y with respect to itself is always 1.</p>
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<p>When x or y is a constant, the derivative of that constant is 0. The derivative of x or y with respect to itself is always 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x+y</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x+y</h2>
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<p>Students frequently make mistakes when differentiating x+y. 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 x+y. 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 (x+y)²</p>
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<p>Calculate the derivative of (x+y)²</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) = (x+y)². Using the chain rule, f'(x, y) = 2(x+y) * (d/dx (x) + d/dx (y)) In the given equation, differentiate with respect to x: = 2(x+y) * (1 + 0) = 2(x+y)</p>
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<p>Here, we have f(x, y) = (x+y)². Using the chain rule, f'(x, y) = 2(x+y) * (d/dx (x) + d/dx (y)) In the given equation, differentiate with respect to x: = 2(x+y) * (1 + 0) = 2(x+y)</p>
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<p>Similarly, with respect to y: = 2(x+y) * (0 + 1) = 2(x+y)</p>
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<p>Similarly, with respect to y: = 2(x+y) * (0 + 1) = 2(x+y)</p>
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<p>Thus, the derivative of the specified function is 2(x+y) for both x and y.</p>
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<p>Thus, the derivative of the specified function is 2(x+y) for both x and y.</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 chain rule. The first step is to differentiate with respect to each variable and then multiply by the derivative of the inner function to get the final result.</p>
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<p>We find the derivative of the given function by applying the chain rule. The first step is to differentiate with respect to each variable and then multiply by the derivative of the inner function to get 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 2</h3>
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<h3>Problem 2</h3>
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<p>A company manufactures a product where the cost is represented by the function C = x + y, where x is the material cost and y is the labor cost. If x = 100 units and y = 50 units, measure the rate of change of total cost with respect to material cost.</p>
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<p>A company manufactures a product where the cost is represented by the function C = x + y, where x is the material cost and y is the labor cost. If x = 100 units and y = 50 units, measure the rate of change of total cost with respect to material cost.</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 C = x + y (cost function)...(1)</p>
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<p>We have C = x + y (cost function)...(1)</p>
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<p>Now, we will differentiate the equation (1) with respect to x: dC/dx = 1</p>
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<p>Now, we will differentiate the equation (1) with respect to x: dC/dx = 1</p>
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<p>Given x = 100 units and y = 50 units,</p>
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<p>Given x = 100 units and y = 50 units,</p>
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<p>the rate of change of total cost with respect to material cost is 1, meaning the cost increases by 1 unit for every unit increase in material cost.</p>
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<p>the rate of change of total cost with respect to material cost is 1, meaning the cost increases by 1 unit for every unit increase in material cost.</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 total cost with respect to material cost as 1, which means that for every additional unit of material cost, the total cost increases by 1 unit.</p>
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<p>We find the rate of change of total cost with respect to material cost as 1, which means that for every additional unit of material cost, the total cost increases by 1 unit.</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 C = x + y.</p>
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<p>Derive the second derivative of the function C = 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>The first step is to find the first derivative, dC/dx = 1 and dC/dy = 1</p>
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<p>The first step is to find the first derivative, dC/dx = 1 and dC/dy = 1</p>
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<p>Now we will differentiate again to get the second derivative: d²C/dx² = 0 and d²C/dy² = 0</p>
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<p>Now we will differentiate again to get the second derivative: d²C/dx² = 0 and d²C/dy² = 0</p>
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<p>Therefore, the second derivative of the function C = x + y is 0 with respect to both x and y.</p>
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<p>Therefore, the second derivative of the function C = x + y is 0 with respect to both x and y.</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, where we start with the first derivative. Since x+y is a linear function, the second derivative is 0, indicating no curvature.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Since x+y is a linear function, the second derivative is 0, indicating no curvature.</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 ((x+y)²) = 2(x+y).</p>
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<p>Prove: d/dx ((x+y)²) = 2(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>Let’s start using the chain rule: Consider f(x, y) = (x+y)²</p>
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<p>Let’s start using the chain rule: Consider f(x, y) = (x+y)²</p>
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<p>To differentiate, we use the chain rule: df/dx = 2(x+y) * d/dx (x+y)</p>
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<p>To differentiate, we use the chain rule: df/dx = 2(x+y) * d/dx (x+y)</p>
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<p>Since the derivative of x+y with respect to x is 1, df/dx = 2(x+y) * 1</p>
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<p>Since the derivative of x+y with respect to x is 1, df/dx = 2(x+y) * 1</p>
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<p>Substituting f(x, y) = (x+y)², d/dx ((x+y)²) = 2(x+y)</p>
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<p>Substituting f(x, y) = (x+y)², d/dx ((x+y)²) = 2(x+y)</p>
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<p>Hence proved.</p>
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<p>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 chain rule to differentiate the equation. Then, we replaced x+y with its derivative. As a final step, we substituted f(x, y) = (x+y)² to derive the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replaced x+y with its derivative. As a final step, we substituted f(x, y) = (x+y)² 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)/x)</p>
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<p>Solve: d/dx ((x+y)/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>To differentiate the function, we use the quotient rule: d/dx ((x+y)/x) = (d/dx (x+y) * x - (x+y) * d/dx (x)) / x² We substitute d/dx (x+y) = 1 and d/dx (x) = 1: = (1 * x - (x+y) * 1) / x² = (x - x - y) / x² = -y / x² Therefore, d/dx ((x+y)/x) = -y / x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx ((x+y)/x) = (d/dx (x+y) * x - (x+y) * d/dx (x)) / x² We substitute d/dx (x+y) = 1 and d/dx (x) = 1: = (1 * x - (x+y) * 1) / x² = (x - x - y) / x² = -y / x² Therefore, d/dx ((x+y)/x) = -y / x²</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. As a final step, we 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. As a final step, we 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 x+y</h2>
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<h2>FAQs on the Derivative of x+y</h2>
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<h3>1.Find the derivative of x+y.</h3>
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<h3>1.Find the derivative of x+y.</h3>
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<p>The derivative of x+y with respect to x is 1, and the derivative with respect to y is also 1.</p>
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<p>The derivative of x+y with respect to x is 1, and the derivative with respect to y is also 1.</p>
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<h3>2.Can we use the derivative of x+y in real life?</h3>
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<h3>2.Can we use the derivative of x+y in real life?</h3>
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<p>Yes, the derivative of x+y can be used in real life to understand how changes in variables affect the overall outcome, such as in cost analysis or optimization problems.</p>
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<p>Yes, the derivative of x+y can be used in real life to understand how changes in variables affect the overall outcome, such as in cost analysis or optimization problems.</p>
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<h3>3.Is it possible to take the derivative of x+y if x or y is a constant?</h3>
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<h3>3.Is it possible to take the derivative of x+y if x or y is a constant?</h3>
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<p>Yes, if either x or y is a constant, its derivative will be 0, and the derivative of the other variable will be 1.</p>
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<p>Yes, if either x or y is a constant, its derivative will be 0, and the derivative of the other variable will be 1.</p>
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<h3>4.What rule is used to differentiate (x+y)/x?</h3>
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<h3>4.What rule is used to differentiate (x+y)/x?</h3>
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<p>We use the<a>quotient</a>rule to differentiate (x+y)/x: d/dx ((x+y)/x) = (x - x - y) / x² = -y / x².</p>
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<p>We use the<a>quotient</a>rule to differentiate (x+y)/x: d/dx ((x+y)/x) = (x - x - y) / x² = -y / x².</p>
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<h3>5.Are the derivatives of x+y and x-y the same?</h3>
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<h3>5.Are the derivatives of x+y and x-y the same?</h3>
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<p>Not exactly, they are similar but differ in sign. The derivative of x+y is 1 for each variable, while the derivative of x-y is 1 for x and -1 for y.</p>
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<p>Not exactly, they are similar but differ in sign. The derivative of x+y is 1 for each variable, while the derivative of x-y is 1 for x and -1 for y.</p>
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<h2>Important Glossaries for the Derivative of x+y</h2>
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<h2>Important Glossaries for the Derivative of x+y</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in its variables.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in its variables.</li>
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</ul><ul><li><strong>Linearity:</strong>The principle that the derivative of a sum is the sum of the derivatives.</li>
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</ul><ul><li><strong>Linearity:</strong>The principle that the derivative of a sum is the sum of the derivatives.</li>
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</ul><ul><li><strong>Constant:</strong>A value that does not change and whose derivative is 0.</li>
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</ul><ul><li><strong>Constant:</strong>A value that does not change and whose derivative is 0.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate expressions that are the quotient of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate expressions that are the quotient of two functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate compositions of 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>