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2026-01-01
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<p>Last updated on<strong>September 26, 2025</strong></p>
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<p>Last updated on<strong>September 26, 2025</strong></p>
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<p>We use the derivative of 8/x, which is -8/x², as a measuring tool for how the function 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 8/x in detail.</p>
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<p>We use the derivative of 8/x, which is -8/x², as a measuring tool for how the function 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 8/x in detail.</p>
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<h2>What is the Derivative of 8/x?</h2>
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<h2>What is the Derivative of 8/x?</h2>
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<p>We now understand the derivative<a>of</a>8/x. It is commonly represented as d/dx (8/x) or (8/x)', and its value is -8/x². The<a>function</a>8/x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>We now understand the derivative<a>of</a>8/x. It is commonly represented as d/dx (8/x) or (8/x)', and its value is -8/x². The<a>function</a>8/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>Reciprocal Function: (8/x is a<a>constant</a>multiplied by 1/x).</p>
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<p>Reciprocal Function: (8/x is a<a>constant</a>multiplied by 1/x).</p>
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<p>Power Rule: Rule for differentiating 1/x (since it can be written as x⁻¹).</p>
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<p>Power Rule: Rule for differentiating 1/x (since it can be written as x⁻¹).</p>
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<p>Negative Exponent: x⁻¹ = 1/x.</p>
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<p>Negative Exponent: x⁻¹ = 1/x.</p>
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<h2>Derivative of 8/x Formula</h2>
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<h2>Derivative of 8/x Formula</h2>
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<p>The derivative of 8/x can be denoted as d/dx (8/x) or (8/x)'.</p>
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<p>The derivative of 8/x can be denoted as d/dx (8/x) or (8/x)'.</p>
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<p>The<a>formula</a>we use to differentiate 8/x is: d/dx (8/x) = -8/x² (or) (8/x)' = -8/x² The formula applies to all x where x ≠ 0.</p>
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<p>The<a>formula</a>we use to differentiate 8/x is: d/dx (8/x) = -8/x² (or) (8/x)' = -8/x² The formula applies to all x where x ≠ 0.</p>
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<h2>Proofs of the Derivative of 8/x</h2>
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<h2>Proofs of the Derivative of 8/x</h2>
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<p>We can derive the derivative of 8/x using proofs. To show this, we will use the rules of differentiation along with algebraic manipulation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of 8/x using proofs. To show this, we will use the rules of differentiation along with algebraic manipulation. 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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<li>Using Product Rule</li>
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<li>Using Product Rule</li>
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</ul><p>We will now demonstrate that the differentiation of 8/x results in -8/x² using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of 8/x results in -8/x² using the above-mentioned methods:</p>
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<h2><strong>By First Principle</strong></h2>
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<h2><strong>By First Principle</strong></h2>
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<p>The derivative of 8/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 8/x using the first principle, we will consider f(x) = 8/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 8/x, we write f(x + h) = 8/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [8/(x + h) - 8/x] / h = limₕ→₀ [8x - 8(x + h)] / [h(x + h)x] = limₕ→₀ [-8h] / [h(x + h)x] = limₕ→₀ [-8] / [(x + h)x] = -8/x² Hence, proved.</p>
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<p>The derivative of 8/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 8/x using the first principle, we will consider f(x) = 8/x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 8/x, we write f(x + h) = 8/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [8/(x + h) - 8/x] / h = limₕ→₀ [8x - 8(x + h)] / [h(x + h)x] = limₕ→₀ [-8h] / [h(x + h)x] = limₕ→₀ [-8] / [(x + h)x] = -8/x² Hence, proved.</p>
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<h2><strong>Using Power Rule</strong></h2>
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<h2><strong>Using Power Rule</strong></h2>
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<p>To prove the differentiation of 8/x using the<a>power</a>rule, We can write 8/x as 8·x⁻¹. The derivative of x⁻¹ is -x⁻². Therefore, d/dx (8·x⁻¹) = 8·(-x⁻²) = -8/x². Thus, the derivative of 8/x is -8/x².</p>
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<p>To prove the differentiation of 8/x using the<a>power</a>rule, We can write 8/x as 8·x⁻¹. The derivative of x⁻¹ is -x⁻². Therefore, d/dx (8·x⁻¹) = 8·(-x⁻²) = -8/x². Thus, the derivative of 8/x is -8/x².</p>
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<h2><strong>Using Product Rule</strong></h2>
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<h2><strong>Using Product Rule</strong></h2>
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<p>We will now prove the derivative of 8/x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, 8/x = 8·x⁻¹ Let u = 8 and v = x⁻¹. Using the product rule formula: d/dx [u·v] = u'·v + u·v' u' = 0 (since 8 is a constant) v' = d/dx (x⁻¹) = -x⁻² d/dx (8/x) = u'·v + u·v' = 0·x⁻¹ + 8·(-x⁻²) = -8/x² Thus, the derivative of 8/x is -8/x².</p>
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<p>We will now prove the derivative of 8/x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, 8/x = 8·x⁻¹ Let u = 8 and v = x⁻¹. Using the product rule formula: d/dx [u·v] = u'·v + u·v' u' = 0 (since 8 is a constant) v' = d/dx (x⁻¹) = -x⁻² d/dx (8/x) = u'·v + u·v' = 0·x⁻¹ + 8·(-x⁻²) = -8/x² Thus, the derivative of 8/x is -8/x².</p>
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<h2>Higher-Order Derivatives of 8/x</h2>
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<h2>Higher-Order Derivatives of 8/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.</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 8/x.</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 8/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.</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.</p>
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<p>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>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 8/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 8/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 is 0, the derivative is undefined because 8/x is undefined at x = 0. When x is 1, the derivative of 8/x = -8/1², which is -8.</p>
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<p>When x is 0, the derivative is undefined because 8/x is undefined at x = 0. When x is 1, the derivative of 8/x = -8/1², which is -8.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 8/x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 8/x</h2>
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<p>Students frequently make mistakes when differentiating 8/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 8/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 (8/x)·ln(x)</p>
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<p>Calculate the derivative of (8/x)·ln(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) = (8/x)·ln(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 8/x and v = ln(x). Let’s differentiate each term, u′ = d/dx (8/x) = -8/x² v′ = d/dx (ln(x)) = 1/x Substituting into the given equation, f'(x) = (-8/x²)·ln(x) + (8/x)·(1/x) Let’s simplify terms to get the final answer, f'(x) = -8ln(x)/x² + 8/x² Thus, the derivative of the specified function is -8ln(x)/x² + 8/x².</p>
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<p>Here, we have f(x) = (8/x)·ln(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 8/x and v = ln(x). Let’s differentiate each term, u′ = d/dx (8/x) = -8/x² v′ = d/dx (ln(x)) = 1/x Substituting into the given equation, f'(x) = (-8/x²)·ln(x) + (8/x)·(1/x) Let’s simplify terms to get the final answer, f'(x) = -8ln(x)/x² + 8/x² Thus, the derivative of the specified function is -8ln(x)/x² + 8/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.</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>Jupiter Company is tracking their inventory with a function y = 8/x, where y represents the inventory level at time x. If x = 2 hours, determine the rate of change of inventory.</p>
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<p>Jupiter Company is tracking their inventory with a function y = 8/x, where y represents the inventory level at time x. If x = 2 hours, determine the rate of change of inventory.</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 = 8/x (inventory level)...(1) Now, we will differentiate the equation (1) Take the derivative of 8/x: dy/dx = -8/x² Given x = 2 (substitute this into the derivative) dy/dx = -8/2² = -8/4 = -2 Hence, the rate of change of inventory at x = 2 hours is -2.</p>
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<p>We have y = 8/x (inventory level)...(1) Now, we will differentiate the equation (1) Take the derivative of 8/x: dy/dx = -8/x² Given x = 2 (substitute this into the derivative) dy/dx = -8/2² = -8/4 = -2 Hence, the rate of change of inventory at x = 2 hours is -2.</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 inventory at x = 2 hours as -2, which means that at that time, the inventory level is decreasing by 2 units per hour.</p>
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<p>We find the rate of change of inventory at x = 2 hours as -2, which means that at that time, the inventory level is decreasing by 2 units per hour.</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 = 8/x.</p>
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<p>Derive the second derivative of the function y = 8/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 = -8/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-8/x²] Using the power rule, d²y/dx² = 16/x³ Therefore, the second derivative of the function y = 8/x is 16/x³.</p>
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<p>The first step is to find the first derivative, dy/dx = -8/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-8/x²] Using the power rule, d²y/dx² = 16/x³ Therefore, the second derivative of the function y = 8/x is 16/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 the power rule, we differentiate -8/x².</p>
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<p>Using the power rule, we differentiate -8/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 (1/x²) = -2/x³.</p>
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<p>Prove: d/dx (1/x²) = -2/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 power rule: Consider y = 1/x² = x⁻² To differentiate, we use the power rule: dy/dx = -2x⁻³ Substituting y = 1/x², d/dx (1/x²) = -2/x³ Hence proved.</p>
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<p>Let’s start using the power rule: Consider y = 1/x² = x⁻² To differentiate, we use the power rule: dy/dx = -2x⁻³ Substituting y = 1/x², d/dx (1/x²) = -2/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 power rule to differentiate the equation.</p>
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<p>In this step-by-step process, we used the power rule to differentiate the equation.</p>
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<p>Then, we replace x⁻² with its derivative.</p>
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<p>Then, we replace x⁻² with its derivative.</p>
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<p>As a final step, we substitute y = 1/x² to derive the equation.</p>
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<p>As a final step, we substitute y = 1/x² 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 (8x/x²)</p>
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<p>Solve: d/dx (8x/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 the function, we simplify first: 8x/x² = 8/x Then differentiate: d/dx (8/x) = -8/x² Therefore, d/dx (8x/x²) = -8/x²</p>
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<p>To differentiate the function, we simplify first: 8x/x² = 8/x Then differentiate: d/dx (8/x) = -8/x² Therefore, d/dx (8x/x²) = -8/x²</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we simplify the given function before differentiating.</p>
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<p>In this process, we simplify the given function before differentiating.</p>
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<p>We then apply the power rule to obtain the final result.</p>
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<p>We then apply the power rule 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 8/x</h2>
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<h2>FAQs on the Derivative of 8/x</h2>
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<h3>1.Find the derivative of 8/x.</h3>
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<h3>1.Find the derivative of 8/x.</h3>
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<p>Using the power rule to differentiate 8/x gives 8·x⁻¹. d/dx (8/x) = -8/x² (simplified).</p>
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<p>Using the power rule to differentiate 8/x gives 8·x⁻¹. d/dx (8/x) = -8/x² (simplified).</p>
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<h3>2.Can we use the derivative of 8/x in real life?</h3>
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<h3>2.Can we use the derivative of 8/x in real life?</h3>
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<p>Yes, we can use the derivative of 8/x in real life to calculate the rate of change of any quantity that follows an inverse relationship, especially in fields such as mathematics, physics, and economics.</p>
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<p>Yes, we can use the derivative of 8/x in real life to calculate the rate of change of any quantity that follows an inverse relationship, especially in fields such as mathematics, physics, and economics.</p>
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<h3>3.Is it possible to take the derivative of 8/x at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of 8/x at the point where x = 0?</h3>
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<p>No, x = 0 is a point where 8/x 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, x = 0 is a point where 8/x 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 8/x²?</h3>
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<h3>4.What rule is used to differentiate 8/x²?</h3>
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<p>We use the power rule to differentiate 8/x², d/dx (8/x²) = -16/x³.</p>
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<p>We use the power rule to differentiate 8/x², d/dx (8/x²) = -16/x³.</p>
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<h3>5.Are the derivatives of 8/x and x⁻² the same?</h3>
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<h3>5.Are the derivatives of 8/x and x⁻² the same?</h3>
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<p>No, they are different. The derivative of 8/x is -8/x², while the derivative of x⁻² is -2x⁻³.</p>
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<p>No, they are different. The derivative of 8/x is -8/x², while the derivative of x⁻² is -2x⁻³.</p>
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<h3>6.Can we find the derivative of the 8/x formula?</h3>
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<h3>6.Can we find the derivative of the 8/x formula?</h3>
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<p>To find, consider y = 8/x. We use the power rule: y' = 8·d/dx(x⁻¹) = -8/x².</p>
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<p>To find, consider y = 8/x. We use the power rule: y' = 8·d/dx(x⁻¹) = -8/x².</p>
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<h2>Important Glossaries for the Derivative of 8/x</h2>
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<h2>Important Glossaries for the Derivative of 8/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>Reciprocal Function:</strong>A function of the form 1/x or any constant multiplied by 1/x.</li>
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</ul><ul><li><strong>Reciprocal Function:</strong>A function of the form 1/x or any constant multiplied by 1/x.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to find the derivative of a power function.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to find the derivative of a power function.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point where the function does not have a defined value.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point where the function does not have a defined value.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives that are obtained after the first derivative, indicating further rates of change.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives that are obtained after the first derivative, indicating further rates of change.</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>