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2026-01-01
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>We use the derivative of -x/y to understand how this function changes with respect to a slight change in x or y. Derivatives are essential tools in calculus that help us solve problems in various fields, including physics and economics. We will now discuss the derivative of -x/y in detail.</p>
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<p>We use the derivative of -x/y to understand how this function changes with respect to a slight change in x or y. Derivatives are essential tools in calculus that help us solve problems in various fields, including physics and economics. We will now discuss 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 explore the derivative<a>of</a>-x/y. It is represented as d/dx (-x/y) or (-x/y)'. The derivative of -x/y is determined using the<a>quotient</a>rule, as it is a<a>ratio</a>of<a>functions</a>. This derivative indicates how the function changes within its domain. The key concepts are as follows:</p>
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<p>We now explore the derivative<a>of</a>-x/y. It is represented as d/dx (-x/y) or (-x/y)'. The derivative of -x/y is determined using the<a>quotient</a>rule, as it is a<a>ratio</a>of<a>functions</a>. This derivative indicates how the function changes within its domain. The key concepts are as follows:</p>
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<p><strong>Function Representation:</strong>(-x/y) = -1 * (x/y).</p>
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<p><strong>Function Representation:</strong>(-x/y) = -1 * (x/y).</p>
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<p><strong>Quotient Rule:</strong>Rule for differentiating -x/y.</p>
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<p><strong>Quotient Rule:</strong>Rule for differentiating -x/y.</p>
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<p><strong>Negative Constant:</strong>Consideration of the negative<a>constant</a><a>factor</a>in differentiation.</p>
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<p><strong>Negative Constant:</strong>Consideration of the negative<a>constant</a><a>factor</a>in differentiation.</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 with respect to x can be denoted as d/dx (-x/y) or (-x/y)'.</p>
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<p>The derivative of -x/y with respect to x 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/y) * (y - x(dy/dx))/y²</p>
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<p>The<a>formula</a>we use to differentiate -x/y is: d/dx (-x/y) = (-1/y) * (y - x(dy/dx))/y²</p>
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<p>This applies to all values where y ≠ 0.</p>
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<p>This applies to all values where y ≠ 0.</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 derive the derivative of -x/y using proofs. This involves applying the quotient rule and understanding the role of constants. Here are the methods used to prove this:</p>
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<p>We derive the derivative of -x/y using proofs. This involves applying the quotient rule and understanding the role of constants. Here are the methods used to prove this:</p>
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<ol><li>Using the Quotient Rule</li>
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<ol><li>Using the Quotient Rule</li>
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<li>Considering Constant Multiplication</li>
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<li>Considering Constant Multiplication</li>
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</ol><p>We will now demonstrate that the differentiation of -x/y results in the formula mentioned above using these methods:</p>
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</ol><p>We will now demonstrate that the differentiation of -x/y results in the formula mentioned above using these methods:</p>
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<h3>Using the Quotient Rule</h3>
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<h3>Using the Quotient Rule</h3>
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<p>To find the derivative of -x/y using the quotient rule, consider the function f(x) = -x/y.</p>
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<p>To find the derivative of -x/y using the quotient rule, consider the function f(x) = -x/y.</p>
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<p>The derivative can be expressed as:</p>
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<p>The derivative can be expressed as:</p>
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<p>f'(x) = d/dx (-x/y) = d/dx (-1 * (x/y)) = -1 * [ (y * d/dx(x) - x * d/dx(y)) / y² ] = -1 * [ (y - x(dy/dx)) / y² ] = (-1/y) * (y - x(dy/dx))/y²</p>
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<p>f'(x) = d/dx (-x/y) = d/dx (-1 * (x/y)) = -1 * [ (y * d/dx(x) - x * d/dx(y)) / y² ] = -1 * [ (y - x(dy/dx)) / y² ] = (-1/y) * (y - x(dy/dx))/y²</p>
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<p>This result demonstrates the application of the quotient rule and the effect of the negative constant.</p>
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<p>This result demonstrates the application of the quotient rule and the effect of the negative constant.</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<a>multiple</a>times, the results are known as higher-order derivatives. These can become more complex as the order increases. Consider the analogy of a car where speed (first derivative) changes, and the<a>rate</a>of this change (second derivative) also varies. Higher-order derivatives aid in understanding functions like -x/y.</p>
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<p>When a function is differentiated<a>multiple</a>times, the results are known as higher-order derivatives. These can become more complex as the order increases. Consider the analogy of a car where speed (first derivative) changes, and the<a>rate</a>of this change (second derivative) also varies. Higher-order derivatives aid in understanding functions like -x/y.</p>
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<p>For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a point. The second derivative, f′′(x), is derived from the first derivative, and this pattern continues for higher orders.</p>
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<p>For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a point. The second derivative, f′′(x), is derived from the first derivative, and this pattern continues for higher orders.</p>
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<p>For the nth derivative of -x/y, we use f⁽ⁿ⁾(x) to represent the change in the rate of change.</p>
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<p>For the nth derivative of -x/y, we use f⁽ⁿ⁾(x) to represent 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>At points where y = 0, the derivative is undefined because -x/y is not defined there. When x = 0, the derivative of -x/y depends on the value of y, and it simplifies to 0 when y is constant.</p>
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<p>At points where y = 0, the derivative is undefined because -x/y is not defined there. When x = 0, the derivative of -x/y depends on the value of y, and it simplifies to 0 when y is constant.</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 often make mistakes when differentiating -x/y. These errors can be resolved by understanding the correct methods. Here are some common mistakes and ways to address them:</p>
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<p>Students often make mistakes when differentiating -x/y. These errors can be resolved by understanding the correct methods. Here are some common mistakes and ways to address 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 (-2x/y).</p>
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<p>Calculate the derivative of (-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>Here, we have f(x) = -2x/y.</p>
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<p>Here, we have f(x) = -2x/y.</p>
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<p>Using the quotient rule, f'(x) = d/dx (-2x/y) = -2 * [ (y - x(dy/dx)) / y² ] = -2(y - x(dy/dx))/y²</p>
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<p>Using the quotient rule, f'(x) = d/dx (-2x/y) = -2 * [ (y - x(dy/dx)) / y² ] = -2(y - x(dy/dx))/y²</p>
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<p>Thus, the derivative of the specified function is -2(y - x(dy/dx))/y².</p>
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<p>Thus, the derivative of the specified function is -2(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>We find the derivative of the given function by applying the quotient rule, considering the negative constant factor, and simplifying the expression to obtain the result.</p>
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<p>We find the derivative of the given function by applying the quotient rule, considering the negative constant factor, and simplifying the expression to obtain the 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 water tank is being drained at a rate represented by the function h = -x/y, where h is the height of the water and x is the time. If y = 5 seconds, find the rate of change of the water height when x = 10 seconds.</p>
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<p>A water tank is being drained at a rate represented by the function h = -x/y, where h is the height of the water and x is the time. If y = 5 seconds, find the rate of change of the water height when x = 10 seconds.</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 = -x/y (rate of change of water height)...(1)</p>
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<p>We have h = -x/y (rate of change of water height)...(1)</p>
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<p>Now, we will differentiate the equation (1) with respect to x. dh/dx = d/dx (-x/5) = -1/5 * (5 - x(dy/dx))/5² = -1/5</p>
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<p>Now, we will differentiate the equation (1) with respect to x. dh/dx = d/dx (-x/5) = -1/5 * (5 - x(dy/dx))/5² = -1/5</p>
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<p>Hence, the rate of change of the water height at x = 10 seconds is -1/5.</p>
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<p>Hence, the rate of change of the water height at x = 10 seconds is -1/5.</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 water height by differentiating the function with respect to time and substituting the given values into the derivative.</p>
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<p>We find the rate of change of the water height by differentiating the function with respect to time and substituting the given values into the derivative.</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 = -x/y.</p>
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<p>Derive the second derivative of the function h = -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, dh/dx = (-y - x(dy/dx))/y²...(1)</p>
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<p>The first step is to find the first derivative, dh/dx = (-y - x(dy/dx))/y²...(1)</p>
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<p>Now, we will differentiate equation (1) to get the second derivative: d²h/dx² = d/dx [(-y - x(dy/dx))/y²] = d/dx (-1/y² * (y - x(dy/dx))) = (2/y³) * (y - x(dy/dx))</p>
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<p>Now, we will differentiate equation (1) to get the second derivative: d²h/dx² = d/dx [(-y - x(dy/dx))/y²] = d/dx (-1/y² * (y - x(dy/dx))) = (2/y³) * (y - x(dy/dx))</p>
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<p>Therefore, the second derivative of the function h = -x/y is (2/y³) * (y - x(dy/dx)).</p>
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<p>Therefore, the second derivative of the function h = -x/y is (2/y³) * (y - x(dy/dx)).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use a step-by-step process, starting with the first derivative.</p>
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<p>We use a step-by-step process, starting with the first derivative.</p>
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<p>Using the quotient rule, we differentiate further to find the second derivative, considering the constant factors.</p>
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<p>Using the quotient rule, we differentiate further to find the second derivative, considering the constant factors.</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 (-x²/y) = (-2x/y) - (x²/y²)(dy/dx).</p>
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<p>Prove: d/dx (-x²/y) = (-2x/y) - (x²/y²)(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 by using the quotient rule: Consider h = -x²/y</p>
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<p>Let's start by using the quotient rule: Consider h = -x²/y</p>
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<p>To differentiate, we apply the quotient rule: dh/dx = d/dx (-x²/y) = (-y * 2x - x² * dy/dx)/y² = (-2xy - x²(dy/dx))/y²</p>
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<p>To differentiate, we apply the quotient rule: dh/dx = d/dx (-x²/y) = (-y * 2x - x² * dy/dx)/y² = (-2xy - x²(dy/dx))/y²</p>
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<p>Therefore, d/dx (-x²/y) = (-2x/y) - (x²/y²)(dy/dx), hence proved.</p>
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<p>Therefore, d/dx (-x²/y) = (-2x/y) - (x²/y²)(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 apply the quotient rule to differentiate the equation, substitute values, and simplify to derive the equation as required.</p>
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<p>In this step-by-step process, we apply the quotient rule to differentiate the equation, substitute values, and simplify to derive the equation as required.</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 (-3x/y).</p>
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<p>Solve: d/dx (-3x/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 (-3x/y) = -3 * [(y - x(dy/dx))/y²] = -3(y - x(dy/dx))/y² Therefore, d/dx (-3x/y) = -3(y - x(dy/dx))/y².</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (-3x/y) = -3 * [(y - x(dy/dx))/y²] = -3(y - x(dy/dx))/y² Therefore, d/dx (-3x/y) = -3(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, accounting for the negative constant, and 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, accounting for the negative constant, and 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 -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>Using the quotient rule for -x/y gives: d/dx (-x/y) = (-y - x(dy/dx))/y².</p>
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<p>Using the quotient rule for -x/y gives: d/dx (-x/y) = (-y - x(dy/dx))/y².</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 applications such as determining rates of change in physics and engineering.</p>
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<p>Yes, the derivative of -x/y can be used in real-life applications such as determining rates of change in physics and engineering.</p>
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<h3>3.Is it possible to take the derivative of -x/y when y = 0?</h3>
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<h3>3.Is it possible to take the derivative of -x/y when y = 0?</h3>
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<p>No, the derivative is undefined when y = 0 because -x/y is not defined at this point.</p>
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<p>No, the derivative is undefined when y = 0 because -x/y is not defined at this point.</p>
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<h3>4.What rule is used to differentiate -x²/y?</h3>
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<h3>4.What rule is used to differentiate -x²/y?</h3>
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<p>We use the quotient rule to differentiate -x²/y, resulting in: d/dx (-x²/y) = (-2x/y) - (x²/y²)(dy/dx).</p>
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<p>We use the quotient rule to differentiate -x²/y, resulting in: d/dx (-x²/y) = (-2x/y) - (x²/y²)(dy/dx).</p>
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<h3>5.Does the derivative of -x/y differ from that of a positive x/y?</h3>
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<h3>5.Does the derivative of -x/y differ from that of a positive x/y?</h3>
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<p>Yes, the derivative of -x/y includes a negative factor, affecting the overall result compared to the derivative of x/y.</p>
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<p>Yes, the derivative of -x/y includes a negative factor, affecting the overall result compared to the derivative of x/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 measures how the function changes concerning a change in its variables.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function measures how the function changes concerning a change in its variables.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used for differentiating a ratio of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used for differentiating a ratio of two functions.</li>
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</ul><ul><li><strong>Constant Factor:</strong>A fixed value that affects the differentiation of a function when multiplied by it.</li>
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</ul><ul><li><strong>Constant Factor:</strong>A fixed value that affects the differentiation of a function when multiplied by it.</li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a function or derivative does not exist at a certain point or value.</li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a function or derivative does not exist at a certain point or value.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>The result of differentiating a function multiple times, providing insights into its behavior and changes.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>The result of differentiating a function multiple times, providing insights into its behavior and changes.</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>