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2026-01-01
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<p>Last updated on<strong>October 8, 2025</strong></p>
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<p>Last updated on<strong>October 8, 2025</strong></p>
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<p>We use the derivative of a/x, which is -a/x², as a measuring tool for how the function a/x changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of a/x in detail.</p>
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<p>We use the derivative of a/x, which is -a/x², as a measuring tool for how the function a/x changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of a/x in detail.</p>
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<h2>What is the Derivative of a/x?</h2>
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<h2>What is the Derivative of a/x?</h2>
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<p>We now understand the derivative of a/x. It is commonly represented as d/dx (a/x) or (a/x)', and its value is -a/x². The<a>function</a>a/x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>We now understand the derivative of a/x. It is commonly represented as d/dx (a/x) or (a/x)', and its value is -a/x². The<a>function</a>a/x has a clearly defined derivative, indicating it 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><strong>Function:</strong>a/x is a simple rational function. </p>
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<p><strong>Function:</strong>a/x is a simple rational function. </p>
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<p><strong>Derivative Rule:</strong>Rule for differentiating a/x using the<a>power</a>rule. </p>
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<p><strong>Derivative Rule:</strong>Rule for differentiating a/x using the<a>power</a>rule. </p>
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<p><strong>Simplification:</strong>-a/x² is obtained by differentiating the function.</p>
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<p><strong>Simplification:</strong>-a/x² is obtained by differentiating the function.</p>
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<h2>Derivative of a/x Formula</h2>
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<h2>Derivative of a/x Formula</h2>
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<p>The derivative of a/x can be denoted as d/dx (a/x) or (a/x)'.</p>
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<p>The derivative of a/x can be denoted as d/dx (a/x) or (a/x)'.</p>
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<p>The<a>formula</a>we use to differentiate a/x is: d/dx (a/x) = -a/x²</p>
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<p>The<a>formula</a>we use to differentiate a/x is: d/dx (a/x) = -a/x²</p>
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<p>The formula applies to all x where x ≠ 0.</p>
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<p>The formula applies to all x where x ≠ 0.</p>
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<h2>Proofs of the Derivative of a/x</h2>
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<h2>Proofs of the Derivative of a/x</h2>
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<p>We can derive the derivative of a/x using proofs. To show this, we will use basic differentiation rules.</p>
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<p>We can derive the derivative of a/x using proofs. To show this, we will use basic differentiation rules.</p>
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<p>There are several methods we use to prove this, such as: </p>
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<p>There are several methods we use to prove this, such as: </p>
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<ul><li>By First Principle </li>
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<ul><li>By First Principle </li>
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<li>Using Power Rule</li>
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<li>Using Power Rule</li>
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</ul><p>We will now demonstrate that the differentiation of a/x results in -a/x² using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of a/x results in -a/x² using the above-mentioned methods:</p>
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<h2>By First Principle</h2>
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<h2>By First Principle</h2>
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<p>The derivative of a/x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of a/x using the first principle, we will consider f(x) = a/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>The derivative of a/x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of a/x using the first principle, we will consider f(x) = a/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = a/x, we write f(x + h) = a/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [a/(x + h) - a/x] / h = limₕ→₀ [a(x - (x + h))] / [h(x)(x + h)] = limₕ→₀ [-ah] / [h(x)(x + h)] = limₕ→₀ [-a] / [x(x + h)] = -a/x² as h approaches 0. Hence, proved.</p>
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<p>Given that f(x) = a/x, we write f(x + h) = a/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [a/(x + h) - a/x] / h = limₕ→₀ [a(x - (x + h))] / [h(x)(x + h)] = limₕ→₀ [-ah] / [h(x)(x + h)] = limₕ→₀ [-a] / [x(x + h)] = -a/x² as h approaches 0. Hence, proved.</p>
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<h2>Using Power Rule</h2>
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<h2>Using Power Rule</h2>
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<p>To prove the differentiation of a/x using the power rule, we rewrite the function: a/x = ax⁻¹ Differentiate using the power rule: d/dx(ax⁻¹) = a(-1)x⁻² = -a/x². Hence, the derivative is -a/x².</p>
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<p>To prove the differentiation of a/x using the power rule, we rewrite the function: a/x = ax⁻¹ Differentiate using the power rule: d/dx(ax⁻¹) = a(-1)x⁻² = -a/x². Hence, the derivative is -a/x².</p>
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<h2>Higher-Order Derivatives of a/x</h2>
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<h2>Higher-Order Derivatives of a/x</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. 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 a/x.</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. 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 a/x.</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 is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(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 f′(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 f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the nth Derivative of a/x, we generally use fⁿ(x) for the nth derivative of a function f(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 a/x, we generally use fⁿ(x) for the nth derivative of a function f(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 x = 0, the derivative is undefined because a/x has a vertical asymptote there. When a = 0, the derivative of a/x = -0/x², which is 0 for all x ≠ 0.</p>
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<p>When x = 0, the derivative is undefined because a/x has a vertical asymptote there. When a = 0, the derivative of a/x = -0/x², which is 0 for all x ≠ 0.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of a/x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of a/x</h2>
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<p>Students frequently make mistakes when differentiating a/x. 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 a/x. 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 (a/x)·(b/x).</p>
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<p>Calculate the derivative of (a/x)·(b/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>Here, we have f(x) = (a/x)·(b/x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = a/x and v = b/x. Let’s differentiate each term, u′ = d/dx (a/x) = -a/x² v′ = d/dx (b/x) = -b/x² Substituting into the given equation, f'(x) = (-a/x²)·(b/x) + (a/x)·(-b/x²) = -ab/x³ - ab/x³ = -2ab/x³. Thus, the derivative of the specified function is -2ab/x³.</p>
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<p>Here, we have f(x) = (a/x)·(b/x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = a/x and v = b/x. Let’s differentiate each term, u′ = d/dx (a/x) = -a/x² v′ = d/dx (b/x) = -b/x² Substituting into the given equation, f'(x) = (-a/x²)·(b/x) + (a/x)·(-b/x²) = -ab/x³ - ab/x³ = -2ab/x³. Thus, the derivative of the specified function is -2ab/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 of the given function by dividing the function into two parts.</p>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>The first step is finding its derivative and then combining them using the product 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 manufacturing company monitors its production efficiency, represented by the function y = a/x, where y represents the efficiency at a time x. If x = 5 hours, measure the change in efficiency.</p>
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<p>A manufacturing company monitors its production efficiency, represented by the function y = a/x, where y represents the efficiency at a time x. If x = 5 hours, measure the change in efficiency.</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 y = a/x (efficiency)...(1) Now, we will differentiate the equation (1) Take the derivative a/x: dy/dx = -a/x² Given x = 5 (substitute this into the derivative) dy/dx = -a/5² = -a/25 Hence, the change in efficiency at time x = 5 is -a/25.</p>
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<p>We have y = a/x (efficiency)...(1) Now, we will differentiate the equation (1) Take the derivative a/x: dy/dx = -a/x² Given x = 5 (substitute this into the derivative) dy/dx = -a/5² = -a/25 Hence, the change in efficiency at time x = 5 is -a/25.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the change in efficiency at x = 5 as -a/25, which means that at a given point, the efficiency decreases at this rate.</p>
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<p>We find the change in efficiency at x = 5 as -a/25, which means that at a given point, the efficiency decreases at this rate.</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 = a/x.</p>
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<p>Derive the second derivative of the function y = a/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>The first step is to find the first derivative, dy/dx = -a/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-a/x²] = 2a/x³. Therefore, the second derivative of the function y = a/x is 2a/x³.</p>
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<p>The first step is to find the first derivative, dy/dx = -a/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-a/x²] = 2a/x³. Therefore, the second derivative of the function y = a/x is 2a/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.</p>
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<p>We use the step-by-step process, where we start with the first derivative.</p>
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<p>Using basic differentiation rules, we differentiate -a/x².</p>
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<p>Using basic differentiation rules, we differentiate -a/x².</p>
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<p>We then simplify the terms to find the final answer.</p>
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<p>We then simplify the terms to find 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 ((a/x)²) = -2a²/x³.</p>
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<p>Prove: d/dx ((a/x)²) = -2a²/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>Let’s start using the chain rule: Consider y = (a/x)² To differentiate, we use the chain rule: dy/dx = 2(a/x)·d/dx [a/x] Since the derivative of a/x is -a/x², dy/dx = 2(a/x)·(-a/x²) = -2a²/x³. Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = (a/x)² To differentiate, we use the chain rule: dy/dx = 2(a/x)·d/dx [a/x] Since the derivative of a/x is -a/x², dy/dx = 2(a/x)·(-a/x²) = -2a²/x³. 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.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation.</p>
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<p>Then, we replace a/x with its derivative.</p>
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<p>Then, we replace a/x with its derivative.</p>
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<p>As a final step, we simplify to derive the equation.</p>
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<p>As a final step, we simplify 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 (a/x + x).</p>
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<p>Solve: d/dx (a/x + 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 sum rule: d/dx (a/x + x) = d/dx (a/x) + d/dx (x) We will substitute d/dx (a/x) = -a/x² and d/dx (x) = 1 = -a/x² + 1. Therefore, d/dx (a/x + x) = -a/x² + 1.</p>
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<p>To differentiate the function, we use the sum rule: d/dx (a/x + x) = d/dx (a/x) + d/dx (x) We will substitute d/dx (a/x) = -a/x² and d/dx (x) = 1 = -a/x² + 1. Therefore, d/dx (a/x + x) = -a/x² + 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 differentiate the given function using the sum rule.</p>
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<p>In this process, we differentiate the given function using the sum rule.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>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 a/x</h2>
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<h2>FAQs on the Derivative of a/x</h2>
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<h3>1.Find the derivative of a/x.</h3>
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<h3>1.Find the derivative of a/x.</h3>
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<p>Using the power rule for a/x gives ax⁻¹, d/dx (a/x) = -a/x² (simplified).</p>
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<p>Using the power rule for a/x gives ax⁻¹, d/dx (a/x) = -a/x² (simplified).</p>
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<h3>2.Can we use the derivative of a/x in real life?</h3>
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<h3>2.Can we use the derivative of a/x in real life?</h3>
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<p>Yes, we can use the derivative of a/x in real life to calculate rates of change in various fields such as physics, engineering, and economics.</p>
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<p>Yes, we can use the derivative of a/x in real life to calculate rates of change in various fields such as physics, engineering, and economics.</p>
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<h3>3.Is it possible to take the derivative of a/x at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of a/x at the point where x = 0?</h3>
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<p>No, x = 0 is a point where a/x is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<p>No, x = 0 is a point where a/x is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate a/x?</h3>
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<h3>4.What rule is used to differentiate a/x?</h3>
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<p>We use the power rule to differentiate a/x, rewritten as ax⁻¹.</p>
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<p>We use the power rule to differentiate a/x, rewritten as ax⁻¹.</p>
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<h3>5.Are the derivatives of a/x and x⁻¹ the same?</h3>
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<h3>5.Are the derivatives of a/x and x⁻¹ the same?</h3>
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<p>No, their derivatives are different. The derivative of a/x is -a/x², while the derivative of x⁻¹ is -1/x².</p>
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<p>No, their derivatives are different. The derivative of a/x is -a/x², while the derivative of x⁻¹ is -1/x².</p>
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<h2>Important Glossaries for the Derivative of a/x</h2>
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<h2>Important Glossaries for the Derivative of a/x</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>Power Rule</strong>: A fundamental rule in differentiation used to find derivatives of functions in the form of axⁿ.</li>
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</ul><ul><li><strong>Power Rule</strong>: A fundamental rule in differentiation used to find derivatives of functions in the form of axⁿ.</li>
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</ul><ul><li><strong>Rational Function:</strong>A function represented by a ratio of two polynomials, such as a/x.</li>
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</ul><ul><li><strong>Rational Function:</strong>A function represented by a ratio of two polynomials, such as a/x.</li>
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</ul><ul><li><strong>Undefined Points:</strong>Points in the domain where the function does not exist, such as x = 0 for a/x.</li>
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</ul><ul><li><strong>Undefined Points:</strong>Points in the domain where the function does not exist, such as x = 0 for a/x.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to compute the derivative of a composite function.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to compute the derivative of a composite function.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<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>