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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>The derivative of y/x helps us understand how the ratio of y to x changes with a slight variation in x. This concept is crucial in applications like calculating rates, optimizing functions, and understanding real-world relationships between variables. We will now explore the derivative of y/x in detail.</p>
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<p>The derivative of y/x helps us understand how the ratio of y to x changes with a slight variation in x. This concept is crucial in applications like calculating rates, optimizing functions, and understanding real-world relationships between variables. We will now explore the derivative of y/x in detail.</p>
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<h2>What is the Derivative of y/x?</h2>
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<h2>What is the Derivative of y/x?</h2>
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<p>Understanding the derivative<a>of</a>y/x involves applying the<a>quotient</a>rule. It is commonly represented as d/dx (y/x) or (y/x)'. The derivative is calculated using the<a>formula</a>for the quotient of two<a>functions</a>, indicating that y/x is differentiable within its domain.</p>
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<p>Understanding the derivative<a>of</a>y/x involves applying the<a>quotient</a>rule. It is commonly represented as d/dx (y/x) or (y/x)'. The derivative is calculated using the<a>formula</a>for the quotient of two<a>functions</a>, indicating that y/x is differentiable 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>Quotient Rule: The primary rule for differentiating y/x, as it involves the<a>division</a>of two functions.</p>
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<p>Quotient Rule: The primary rule for differentiating y/x, as it involves the<a>division</a>of two functions.</p>
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<p>Derivatives: Measures of how a function changes as its input changes.</p>
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<p>Derivatives: Measures of how a function changes as its input changes.</p>
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<h2>Derivative of y/x Formula</h2>
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<h2>Derivative of y/x Formula</h2>
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<p>The derivative of y/x can be denoted as d/dx (y/x) or (y/x)'. The formula for differentiating y/x is: d/dx (y/x) = (x * d/dx(y) - y)/x² The formula applies to all x where x ≠ 0.</p>
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<p>The derivative of y/x can be denoted as d/dx (y/x) or (y/x)'. The formula for differentiating y/x is: d/dx (y/x) = (x * d/dx(y) - y)/x² The formula applies to all x where x ≠ 0.</p>
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<h2>Proofs of the Derivative of y/x</h2>
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<h2>Proofs of the Derivative of y/x</h2>
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<p>We can derive the derivative of y/x using proofs. To show this, we will use the quotient rule of differentiation.</p>
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<p>We can derive the derivative of y/x using proofs. To show this, we will use the quotient rule of differentiation.</p>
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<p>Several methods can be used to prove this, such as:</p>
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<p>Several methods can be used to prove this, such as:</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>Using the Product Rule</li>
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<li>Using the Product Rule</li>
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</ol><p>Let's demonstrate that the differentiation of y/x results in the formula (x * d/dx(y) - y)/x² using these methods:</p>
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</ol><p>Let's demonstrate that the differentiation of y/x results in the formula (x * d/dx(y) - y)/x² 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 differentiate y/x using the quotient rule, we consider y as the<a>numerator</a>and x as the<a>denominator</a>.</p>
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<p>To differentiate y/x using the quotient rule, we consider y as the<a>numerator</a>and x as the<a>denominator</a>.</p>
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<p>The quotient rule states: d/dx (u/v) = (v * u' - u * v')/v²</p>
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<p>The quotient rule states: d/dx (u/v) = (v * u' - u * v')/v²</p>
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<p>Applying this to y/x, where u = y and v = x, d/dx (y/x) = (x * d/dx(y) - y * d/dx(x))/x² Since d/dx(x) = 1,</p>
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<p>Applying this to y/x, where u = y and v = x, d/dx (y/x) = (x * d/dx(y) - y * d/dx(x))/x² Since d/dx(x) = 1,</p>
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<p>we have: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>we have: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<h3>Using the Product Rule</h3>
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<h3>Using the Product Rule</h3>
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<p>To prove the differentiation of y/x using the<a>product</a>rule, we can rewrite y/x as y * (1/x). Given that u = y and v = 1/x,</p>
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<p>To prove the differentiation of y/x using the<a>product</a>rule, we can rewrite y/x as y * (1/x). Given that u = y and v = 1/x,</p>
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<p>Using the product rule formula: d/dx [u * v] = u' * v + u * v' u' = d/dx(y) v = 1/x v' = d/dx(1/x) = -1/x²</p>
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<p>Using the product rule formula: d/dx [u * v] = u' * v + u * v' u' = d/dx(y) v = 1/x v' = d/dx(1/x) = -1/x²</p>
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<p>Using the product rule: d/dx (y/x) = d/dx(y) * (1/x) + y * (-1/x²)</p>
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<p>Using the product rule: d/dx (y/x) = d/dx(y) * (1/x) + y * (-1/x²)</p>
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<p>Simplifying, we get: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>Simplifying, we get: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>Thus, the derivative is proved.</p>
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<p>Thus, the derivative is proved.</p>
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<h2>Higher-Order Derivatives of y/x</h2>
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<h2>Higher-Order Derivatives of y/x</h2>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are known as higher-order derivatives. Higher-order derivatives can be complex. To understand them better, consider a car where the speed changes (first derivative) and the acceleration (second derivative) also changes. Higher-order derivatives help us understand functions like y/x more deeply.</p>
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<p>When a function is differentiated<a>multiple</a>times, the resulting derivatives are known as higher-order derivatives. Higher-order derivatives can be complex. To understand them better, consider a car where the speed changes (first derivative) and the acceleration (second derivative) also changes. Higher-order derivatives help us understand functions like y/x more deeply.</p>
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<p>For the first derivative of a function, we write f′(x), indicating how the function changes at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is derived from the second derivative, and this pattern continues.</p>
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<p>For the first derivative of a function, we write f′(x), indicating how the function changes at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is derived from the second derivative, and this pattern continues.</p>
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<p>For the nth derivative of y/x, we generally use fⁿ(x) to represent the nth derivative of a function f(x), which shows the change in the<a>rate</a>of change.</p>
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<p>For the nth derivative of y/x, we generally use fⁿ(x) to represent the nth derivative of a function f(x), which shows the change in the<a>rate</a>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 x = 0, the derivative is undefined because y/x has a<a>division by zero</a>. When y = 0, the derivative of y/x simplifies to 0, since the numerator is zero.</p>
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<p>When x = 0, the derivative is undefined because y/x has a<a>division by zero</a>. When y = 0, the derivative of y/x simplifies to 0, since the numerator is zero.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of y/x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of y/x</h2>
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<p>Students often make mistakes when differentiating y/x. These errors can be mitigated by following the correct procedures. Here are some common mistakes and how to solve them:</p>
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<p>Students often make mistakes when differentiating y/x. These errors can be mitigated by following the correct procedures. Here are some common mistakes and how 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 (y/x)²</p>
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<p>Calculate the derivative of (y/x)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, let f(x) = (y/x)².</p>
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<p>Here, let f(x) = (y/x)².</p>
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<p>Using the chain rule, f'(x) = 2 * (y/x) * d/dx(y/x)</p>
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<p>Using the chain rule, f'(x) = 2 * (y/x) * d/dx(y/x)</p>
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<p>We know d/dx(y/x) = (x * d/dx(y) - y)/x².</p>
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<p>We know d/dx(y/x) = (x * d/dx(y) - y)/x².</p>
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<p>Substitute this into the equation: f'(x) = 2 * (y/x) * ((x * d/dx(y) - y)/x²)</p>
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<p>Substitute this into the equation: f'(x) = 2 * (y/x) * ((x * d/dx(y) - y)/x²)</p>
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<p>Simplify to find the final answer: f'(x) = 2 * (y * (x * d/dx(y) - y))/x³.</p>
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<p>Simplify to find the final answer: f'(x) = 2 * (y * (x * d/dx(y) - y))/x³.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative by applying the chain rule to (y/x)². We then substitute the derivative of y/x and simplify to reach the final result.</p>
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<p>We find the derivative by applying the chain rule to (y/x)². We then substitute the derivative of y/x and simplify to reach 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's profit is given by the function y/x, where y is the revenue and x is the number of products sold. If revenue y increases at a rate of 5 units per product when 50 products are sold, calculate the rate of change of profit.</p>
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<p>A company's profit is given by the function y/x, where y is the revenue and x is the number of products sold. If revenue y increases at a rate of 5 units per product when 50 products are sold, calculate the rate of change of profit.</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 y/x = profit. Differentiate the function using the quotient rule: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>Given y/x = profit. Differentiate the function using the quotient rule: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>We know d/dx(y) = 5 (rate of increase of revenue).</p>
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<p>We know d/dx(y) = 5 (rate of increase of revenue).</p>
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<p>Substitute x = 50.</p>
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<p>Substitute x = 50.</p>
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<p>d/dx (y/x) = (50 * 5 - y)/50²</p>
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<p>d/dx (y/x) = (50 * 5 - y)/50²</p>
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<p>Assume y = 250 (for example), d/dx (y/x) = (250 - 250)/2500 = 0</p>
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<p>Assume y = 250 (for example), d/dx (y/x) = (250 - 250)/2500 = 0</p>
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<p>Therefore, the rate of change of profit is 0.</p>
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<p>Therefore, the rate of change of profit is 0.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We calculate the rate of change of profit by differentiating y/x and substituting the given rates and values to determine the result.</p>
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<p>We calculate the rate of change of profit by differentiating y/x and substituting the given rates and values to determine 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 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y/x.</p>
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<p>Derive the second derivative of the function 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>First, find the first derivative: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>First, find the first derivative: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>Now, differentiate this to find the second derivative: d²/dx² (y/x) = d/dx [(x * d/dx(y) - y)/x²]</p>
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<p>Now, differentiate this to find the second derivative: d²/dx² (y/x) = d/dx [(x * d/dx(y) - y)/x²]</p>
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<p>Use the quotient rule again: d²/dx² (y/x) = ((x² * d²/dx²(y) - 2x * d/dx(y) + 2y)/x⁴)</p>
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<p>Use the quotient rule again: d²/dx² (y/x) = ((x² * d²/dx²(y) - 2x * d/dx(y) + 2y)/x⁴)</p>
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<p>Therefore, the second derivative is ((x² * d²/dx²(y) - 2x * d/dx(y) + 2y)/x⁴).</p>
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<p>Therefore, the second derivative is ((x² * d²/dx²(y) - 2x * d/dx(y) + 2y)/x⁴).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use step-by-step differentiation, first finding the first derivative, then applying the quotient rule again to obtain the second derivative.</p>
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<p>We use step-by-step differentiation, first finding the first derivative, then applying the quotient rule again to obtain the second derivative.</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 (y²/x) = (2y * d/dx(y) * x - y²)/x².</p>
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<p>Prove: d/dx (y²/x) = (2y * d/dx(y) * x - y²)/x².</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Start using the quotient rule: Consider y²/x.</p>
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<p>Start using the quotient rule: Consider y²/x.</p>
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<p>Let u = y² and v = x. d/dx (u/v) = (v * d/dx(u) - u * d/dx(v))/v² d/dx (y²/x) = (x * d/dx(y²) - y² * 1)/x²</p>
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<p>Let u = y² and v = x. d/dx (u/v) = (v * d/dx(u) - u * d/dx(v))/v² d/dx (y²/x) = (x * d/dx(y²) - y² * 1)/x²</p>
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<p>Since d/dx(y²) = 2y * d/dx(y), d/dx (y²/x) = (2y * d/dx(y) * x - y²)/x²</p>
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<p>Since d/dx(y²) = 2y * d/dx(y), d/dx (y²/x) = (2y * d/dx(y) * x - y²)/x²</p>
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<p>Thus proved.</p>
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<p>Thus proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We apply the quotient rule to differentiate y²/x, substituting d/dx(y²) and simplifying to arrive at the proof.</p>
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<p>We apply the quotient rule to differentiate y²/x, substituting d/dx(y²) and simplifying to arrive at the proof.</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 (y/x²)</p>
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<p>Solve: d/dx (y/x²)</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 y/x², use the quotient rule: d/dx (y/x²) = (x² * d/dx(y) - y * 2x)/x⁴</p>
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<p>To differentiate y/x², use the quotient rule: d/dx (y/x²) = (x² * d/dx(y) - y * 2x)/x⁴</p>
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<p>Simplify: d/dx (y/x²) = (x² * d/dx(y) - 2xy)/x⁴</p>
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<p>Simplify: d/dx (y/x²) = (x² * d/dx(y) - 2xy)/x⁴</p>
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<p>Therefore, d/dx (y/x²) = (x * d/dx(y) - 2y)/x³.</p>
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<p>Therefore, d/dx (y/x²) = (x * d/dx(y) - 2y)/x³.</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, simplifying 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, simplifying 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 y/x</h2>
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<h2>FAQs on the Derivative of y/x</h2>
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<h3>1.Find the derivative of y/x.</h3>
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<h3>1.Find the derivative of y/x.</h3>
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<p>Using the quotient rule for y/x gives: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<p>Using the quotient rule for y/x gives: d/dx (y/x) = (x * d/dx(y) - y)/x²</p>
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<h3>2.Can the derivative of y/x be used in real life?</h3>
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<h3>2.Can the derivative of y/x be used in real life?</h3>
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<p>Yes, it can be used to calculate the rate of change of various relationships, such as efficiency, speed, or economic<a>ratios</a>.</p>
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<p>Yes, it can be used to calculate the rate of change of various relationships, such as efficiency, speed, or economic<a>ratios</a>.</p>
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<h3>3.Is it possible to take the derivative of y/x at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of y/x at the point where x = 0?</h3>
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<p>No, x = 0 is a point where y/x is undefined, so it is impossible to take the derivative at this point.</p>
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<p>No, x = 0 is a point where y/x is undefined, so it is impossible to take the derivative at this point.</p>
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<h3>4.What rule is used to differentiate y/x²?</h3>
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<h3>4.What rule is used to differentiate y/x²?</h3>
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<p>The quotient rule is used to differentiate y/x²: d/dx (y/x²) = (x² * d/dx(y) - 2xy)/x⁴.</p>
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<p>The quotient rule is used to differentiate y/x²: d/dx (y/x²) = (x² * d/dx(y) - 2xy)/x⁴.</p>
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<h3>5.Are the derivatives of y/x and x/y the same?</h3>
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<h3>5.Are the derivatives of y/x and x/y the same?</h3>
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<p>No, they are different. The derivative of y/x is (x * d/dx(y) - y)/x², while the derivative of x/y is (y - x * d/dx(y))/y².</p>
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<p>No, they are different. The derivative of y/x is (x * d/dx(y) - y)/x², while the derivative of x/y is (y - x * d/dx(y))/y².</p>
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<h3>6.Can we find the derivative of the y/x formula?</h3>
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<h3>6.Can we find the derivative of the y/x formula?</h3>
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<p>Yes, using the quotient rule: d/dx (y/x) = (x * d/dx(y) - y)/x².</p>
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<p>Yes, using the quotient rule: d/dx (y/x) = (x * d/dx(y) - y)/x².</p>
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<h2>Important Glossaries for the Derivative of y/x</h2>
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<h2>Important Glossaries for the Derivative of y/x</h2>
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<ul><li><strong>Derivative:</strong>A measure of how a function changes as its input changes.</li>
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<ul><li><strong>Derivative:</strong>A measure of how a function changes as its input changes.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A technique for differentiating functions that are divided by each other.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A technique for differentiating functions that are divided by each other.</li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a mathematical expression lacks meaning.</li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a mathematical expression lacks meaning.</li>
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</ul><ul><li><strong>Higher-order Derivative:</strong>A derivative obtained by differentiating a function multiple times.</li>
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</ul><ul><li><strong>Higher-order Derivative:</strong>A derivative obtained by differentiating a function multiple times.</li>
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</ul><ul><li><strong>Rate of Change:</strong>The speed at which one variable changes in relation to another.</li>
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</ul><ul><li><strong>Rate of Change:</strong>The speed at which one variable changes in relation to another.</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>