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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 f/g, which involves applying the quotient rule, as a tool to understand how the ratio of two functions changes in response to a slight change in x. Derivatives help us calculate rates of change in various real-life situations. We will now talk about the derivative of f/g in detail.</p>
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<p>We use the derivative of f/g, which involves applying the quotient rule, as a tool to understand how the ratio of two functions changes in response to a slight change in x. Derivatives help us calculate rates of change in various real-life situations. We will now talk about the derivative of f/g in detail.</p>
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<h2>What is the Derivative of f/g?</h2>
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<h2>What is the Derivative of f/g?</h2>
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<p>We now understand the derivative<a>of</a>f/g. It is commonly represented as d/dx (f/g) or (f/g)', and its value is given by the<a>quotient</a>rule: (f'g - fg')/g². The<a>function</a>f/g has a clearly defined derivative, indicating it is differentiable where g(x) ≠ 0. The key concepts are mentioned below:</p>
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<p>We now understand the derivative<a>of</a>f/g. It is commonly represented as d/dx (f/g) or (f/g)', and its value is given by the<a>quotient</a>rule: (f'g - fg')/g². The<a>function</a>f/g has a clearly defined derivative, indicating it is differentiable where g(x) ≠ 0. The key concepts are mentioned below:</p>
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<p><strong>Quotient Rule:</strong>A rule for differentiating a<a>ratio</a>of two functions (f/g).</p>
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<p><strong>Quotient Rule:</strong>A rule for differentiating a<a>ratio</a>of two functions (f/g).</p>
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<p><strong>Differentiation:</strong>The process of finding the derivative.</p>
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<p><strong>Differentiation:</strong>The process of finding the derivative.</p>
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<p><strong>Function:</strong>A relation that uniquely associates members of one<a>set</a>with members of another set.</p>
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<p><strong>Function:</strong>A relation that uniquely associates members of one<a>set</a>with members of another set.</p>
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<h2>Derivative of f/g Formula</h2>
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<h2>Derivative of f/g Formula</h2>
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<p>The derivative of f/g can be denoted as d/dx (f/g) or (f/g)'. The<a>formula</a>we use to differentiate f/g is: d/dx (f/g) = (f'g - fg')/g² The formula applies to all x where g(x) ≠ 0.</p>
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<p>The derivative of f/g can be denoted as d/dx (f/g) or (f/g)'. The<a>formula</a>we use to differentiate f/g is: d/dx (f/g) = (f'g - fg')/g² The formula applies to all x where g(x) ≠ 0.</p>
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<h2>Proofs of the Derivative of f/g</h2>
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<h2>Proofs of the Derivative of f/g</h2>
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<p>We can derive the derivative of f/g using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of f/g using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:</p>
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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Quotient Rule</li>
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<li>Using Quotient Rule</li>
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</ol><p>We will now demonstrate that the differentiation of f/g results in (f'g - fg')/g² using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of f/g results in (f'g - fg')/g² using the above-mentioned methods:</p>
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<h3>By First Principle</h3>
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<h3>By First Principle</h3>
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<p>The derivative of f/g can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.</p>
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<p>The derivative of f/g can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.</p>
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<p>To find the derivative of f/g using the first principle, we will consider h(x) = f(x)/g(x). Its derivative can be expressed as the following limit. h'(x) = limₕ→₀ [h(x + h) - h(x)] / h … (1) Given that h(x) = f(x)/g(x), we write h(x + h) = f(x + h)/g(x + h).</p>
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<p>To find the derivative of f/g using the first principle, we will consider h(x) = f(x)/g(x). Its derivative can be expressed as the following limit. h'(x) = limₕ→₀ [h(x + h) - h(x)] / h … (1) Given that h(x) = f(x)/g(x), we write h(x + h) = f(x + h)/g(x + h).</p>
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<p>Substituting these into<a>equation</a>(1), h'(x) = limₕ→₀ [f(x + h)/g(x + h) - f(x)/g(x)] / h = limₕ→₀ [f(x + h)g(x) - f(x)g(x + h)] / [h g(x)g(x + h)]</p>
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<p>Substituting these into<a>equation</a>(1), h'(x) = limₕ→₀ [f(x + h)/g(x + h) - f(x)/g(x)] / h = limₕ→₀ [f(x + h)g(x) - f(x)g(x + h)] / [h g(x)g(x + h)]</p>
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<p>We now use the formula f(x + h)g(x) - f(x)g(x + h) = f'(x)g(x) - f(x)g'(x), h'(x) = limₕ→₀ [f'(x)g(x) - f(x)g'(x)] / [g²(x)]</p>
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<p>We now use the formula f(x + h)g(x) - f(x)g(x + h) = f'(x)g(x) - f(x)g'(x), h'(x) = limₕ→₀ [f'(x)g(x) - f(x)g'(x)] / [g²(x)]</p>
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<p>As h approaches 0, we have, h'(x) = (f'g - fg')/g²</p>
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<p>As h approaches 0, we have, h'(x) = (f'g - fg')/g²</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Quotient Rule</h3>
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<h3>Using Quotient Rule</h3>
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<p>To prove the differentiation of f/g using the quotient rule, We use the formula: h(x) = f(x)/g(x)</p>
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<p>To prove the differentiation of f/g using the quotient rule, We use the formula: h(x) = f(x)/g(x)</p>
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<p>By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]²… (1)</p>
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<p>By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]²… (1)</p>
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<p>Let’s substitute f(x) and g(x) in equation (1), d/dx (f/g) = [(f'g - fg')]/g²</p>
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<p>Let’s substitute f(x) and g(x) in equation (1), d/dx (f/g) = [(f'g - fg')]/g²</p>
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<p>Thus, the derivative of f/g is (f'g - fg')/g².</p>
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<p>Thus, the derivative of f/g is (f'g - fg')/g².</p>
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<h2>Higher-Order Derivatives of f/g</h2>
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<h2>Higher-Order Derivatives of f/g</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 a little tricky.</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 a little tricky.</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like f/g.</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like f/g.</p>
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<p>For the first derivative of a function, we write h′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using h′′(x) Similarly, the third derivative, h′′′(x) is the result of the second derivative and this pattern continues.</p>
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<p>For the first derivative of a function, we write h′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using h′′(x) Similarly, the third derivative, h′′′(x) is the result of the second derivative and this pattern continues.</p>
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<p>For the nth Derivative of f/g, we generally use hⁿ(x) for the nth derivative of a function h(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).</p>
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<p>For the nth Derivative of f/g, we generally use hⁿ(x) for the nth derivative of a function h(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When g(x) = 0, the derivative is undefined because h(x) becomes undefined. When the x is such that g(x) ≠ 0, the derivative of f/g is (f'g - fg')/g², ensuring differentiability.</p>
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<p>When g(x) = 0, the derivative is undefined because h(x) becomes undefined. When the x is such that g(x) ≠ 0, the derivative of f/g is (f'g - fg')/g², ensuring differentiability.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of f/g</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of f/g</h2>
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<p>Students frequently make mistakes when differentiating f/g. 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 f/g. 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 (x²/x³).</p>
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<p>Calculate the derivative of (x²/x³).</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 h(x) = x²/x³.</p>
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<p>Here, we have h(x) = x²/x³.</p>
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<p>Using the quotient rule, h'(x) = (f'g - fg')/g² In the given equation, f(x) = x² and g(x) = x³.</p>
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<p>Using the quotient rule, h'(x) = (f'g - fg')/g² In the given equation, f(x) = x² and g(x) = x³.</p>
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<p>Let’s differentiate each term, f′= d/dx (x²) = 2x g′= d/dx (x³) = 3x²</p>
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<p>Let’s differentiate each term, f′= d/dx (x²) = 2x g′= d/dx (x³) = 3x²</p>
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<p>substituting into the given equation, h'(x) = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²</p>
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<p>substituting into the given equation, h'(x) = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²</p>
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<p>Thus, the derivative of the specified function is -1/x².</p>
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<p>Thus, the derivative of the specified function is -1/x².</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 dividing the function into two parts. The first step is finding its derivative using the quotient rule to get the final result.</p>
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<p>We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative using the quotient rule 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 is analyzing the cost efficiency of two products, where the costs are represented by the functions C1(x) = x² and C2(x) = x³. Calculate the rate of change of the cost efficiency ratio C1(x)/C2(x) at x = 1.</p>
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<p>A company is analyzing the cost efficiency of two products, where the costs are represented by the functions C1(x) = x² and C2(x) = x³. Calculate the rate of change of the cost efficiency ratio C1(x)/C2(x) at x = 1.</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) = C1(x)/C2(x) = x²/x³...(1)</p>
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<p>We have C(x) = C1(x)/C2(x) = x²/x³...(1)</p>
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<p>Now, we will differentiate the equation (1) Using the quotient rule: dC/dx = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²</p>
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<p>Now, we will differentiate the equation (1) Using the quotient rule: dC/dx = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²</p>
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<p>Given x = 1, substitute this into the derivative: dC/dx = -1/1² = -1</p>
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<p>Given x = 1, substitute this into the derivative: dC/dx = -1/1² = -1</p>
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<p>Hence, the rate of change of the cost efficiency ratio at x = 1 is -1.</p>
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<p>Hence, the rate of change of the cost efficiency ratio at x = 1 is -1.</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 cost efficiency ratio at x = 1 as -1, which indicates a decrease in efficiency as x increases.</p>
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<p>We find the rate of change of the cost efficiency ratio at x = 1 as -1, which indicates a decrease in efficiency as x increases.</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 = x²/x³.</p>
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<p>Derive the second derivative of the function y = x²/x³.</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, dy/dx = -1/x²...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = -1/x²...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/x²] = 2/x³</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/x²] = 2/x³</p>
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<p>Therefore, the second derivative of the function y = x²/x³ is 2/x³.</p>
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<p>Therefore, the second derivative of the function y = x²/x³ is 2/x³.</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. Then, we differentiate again to find the second derivative and simplify to get the final answer.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Then, we differentiate again to find the second derivative and simplify to get the final answer.</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/x³) = -1/x².</p>
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<p>Prove: d/dx (x²/x³) = -1/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>Let’s start using the quotient rule: Consider h(x) = x²/x³</p>
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<p>Let’s start using the quotient rule: Consider h(x) = x²/x³</p>
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<p>To differentiate, we use the quotient rule: h'(x) = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²</p>
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<p>To differentiate, we use the quotient rule: h'(x) = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²</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 quotient rule to differentiate the equation. We then simplified the resulting expression to derive the equation.</p>
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<p>In this step-by-step process, we used the quotient rule to differentiate the equation. We then simplified the resulting expression 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 (x²/x).</p>
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<p>Solve: d/dx (x²/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 simplify first: x²/x = x</p>
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<p>To differentiate the function, we simplify first: x²/x = x</p>
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<p>Now, differentiate x: d/dx(x) = 1 Therefore, d/dx (x²/x) = 1.</p>
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<p>Now, differentiate x: d/dx(x) = 1 Therefore, d/dx (x²/x) = 1.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we simplify the given function and then differentiate using basic differentiation rules to obtain the final result.</p>
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<p>In this process, we simplify the given function and then differentiate using basic differentiation rules 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 f/g</h2>
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<h2>FAQs on the Derivative of f/g</h2>
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<h3>1.Find the derivative of f/g.</h3>
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<h3>1.Find the derivative of f/g.</h3>
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<p>Using the quotient rule for f/g gives: d/dx (f/g) = (f'g - fg')/g².</p>
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<p>Using the quotient rule for f/g gives: d/dx (f/g) = (f'g - fg')/g².</p>
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<h3>2.Can we use the derivative of f/g in real life?</h3>
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<h3>2.Can we use the derivative of f/g in real life?</h3>
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<p>Yes, we can use the derivative of f/g in real life in calculating rates of change, especially in fields such as science, engineering, and economics.</p>
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<p>Yes, we can use the derivative of f/g in real life in calculating rates of change, especially in fields such as science, engineering, and economics.</p>
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<h3>3.Is it possible to take the derivative of f/g at the point where g(x) = 0?</h3>
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<h3>3.Is it possible to take the derivative of f/g at the point where g(x) = 0?</h3>
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<p>No, g(x) = 0 is a point where f/g is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
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<p>No, g(x) = 0 is a point where f/g is undefined, so it is impossible to take the derivative at these points (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate f/g?</h3>
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<h3>4.What rule is used to differentiate f/g?</h3>
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<p>We use the quotient rule to differentiate f/g, d/dx (f/g) = (f'g - fg')/g².</p>
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<p>We use the quotient rule to differentiate f/g, d/dx (f/g) = (f'g - fg')/g².</p>
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<h3>5.Are the derivatives of f/g and g/f the same?</h3>
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<h3>5.Are the derivatives of f/g and g/f the same?</h3>
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<p>No, they are not the same. The derivative of f/g is (f'g - fg')/g², while the derivative of g/f is (g'f - gf')/f².</p>
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<p>No, they are not the same. The derivative of f/g is (f'g - fg')/g², while the derivative of g/f is (g'f - gf')/f².</p>
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<h2>Important Glossaries for the Derivative of f/g</h2>
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<h2>Important Glossaries for the Derivative of f/g</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 x.</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 x.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate a ratio of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate a ratio of two functions.</li>
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</ul><ul><li><strong>Differentiation:</strong>The process of finding the derivative of a function.</li>
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</ul><ul><li><strong>Differentiation:</strong>The process of finding the derivative of a function.</li>
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</ul><ul><li><strong>Function:</strong>A relation that uniquely associates members of one set with members of another set.</li>
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</ul><ul><li><strong>Function:</strong>A relation that uniquely associates members of one set with members of another set.</li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a function does not have a meaningful value at a certain point. </li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a function does not have a meaningful value at a certain point. </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>