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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 7/x, which is -7/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 7/x in detail.</p>
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<p>We use the derivative of 7/x, which is -7/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 7/x in detail.</p>
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<h2>What is the Derivative of 7/x?</h2>
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<h2>What is the Derivative of 7/x?</h2>
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<p>We now understand the derivative<a>of</a>7/x. It is commonly represented as d/dx (7/x) or (7/x)', and its value is -7/x². The<a>function</a>7/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>7/x. It is commonly represented as d/dx (7/x) or (7/x)', and its value is -7/x². The<a>function</a>7/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>Fractional Function: (7/x is a<a>fraction</a>with a<a>constant</a><a>numerator</a>).</p>
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<p>Fractional Function: (7/x is a<a>fraction</a>with a<a>constant</a><a>numerator</a>).</p>
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<p>Quotient Rule: Rule for differentiating fractions (since it consists of a constant over x).</p>
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<p>Quotient Rule: Rule for differentiating fractions (since it consists of a constant over x).</p>
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<p>Power Rule: Used as an alternative method by expressing 7/x as 7x^-1.</p>
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<p>Power Rule: Used as an alternative method by expressing 7/x as 7x^-1.</p>
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<h2>Derivative of 7/x Formula</h2>
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<h2>Derivative of 7/x Formula</h2>
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<p>The derivative of 7/x can be denoted as d/dx (7/x) or (7/x)'.</p>
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<p>The derivative of 7/x can be denoted as d/dx (7/x) or (7/x)'.</p>
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<p>The<a>formula</a>we use to differentiate 7/x is: d/dx (7/x) = -7/x² (or) (7/x)' = -7/x²</p>
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<p>The<a>formula</a>we use to differentiate 7/x is: d/dx (7/x) = -7/x² (or) (7/x)' = -7/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 7/x</h2>
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<h2>Proofs of the Derivative of 7/x</h2>
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<p>We can derive the derivative of 7/x using proofs. To show this, we will use the rules of differentiation.</p>
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<p>We can derive the derivative of 7/x using proofs. To show this, we will use the rules of differentiation.</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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<li>Using Quotient Rule</li>
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<li>Using Quotient Rule</li>
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</ul><p>We will now demonstrate that the differentiation of 7/x results in -7/x² using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of 7/x results in -7/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 7/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 7/x using the first principle, we will consider f(x) = 7/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) = 7/x, we write f(x + h) = 7/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [7/(x + h) - 7/x] / h = limₕ→₀ [7x - 7(x + h)] / [h(x + h)x] = limₕ→₀ [-7h] / [h(x + h)x] = limₕ→₀ [-7] / [(x + h)x] = -7/x² (as h approaches 0) Hence, proved.</p>
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<p>The derivative of 7/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 7/x using the first principle, we will consider f(x) = 7/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) = 7/x, we write f(x + h) = 7/(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [7/(x + h) - 7/x] / h = limₕ→₀ [7x - 7(x + h)] / [h(x + h)x] = limₕ→₀ [-7h] / [h(x + h)x] = limₕ→₀ [-7] / [(x + h)x] = -7/x² (as h approaches 0) 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 7/x using the<a>power</a>rule, We express it as 7x^-1. The derivative of x^n is n*x^(n-1). So, the derivative of 7x^-1 is: d/dx (7x^-1) = 7 * (-1)x^(-1-1) = -7x^-2 = -7/x² Hence, this proves the derivative using the power rule.</p>
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<p>To prove the differentiation of 7/x using the<a>power</a>rule, We express it as 7x^-1. The derivative of x^n is n*x^(n-1). So, the derivative of 7x^-1 is: d/dx (7x^-1) = 7 * (-1)x^(-1-1) = -7x^-2 = -7/x² Hence, this proves the derivative using the power rule.</p>
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<h2><strong>Using Quotient Rule</strong></h2>
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<h2><strong>Using Quotient Rule</strong></h2>
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<p>We will now prove the derivative of 7/x using the quotient rule. The step-by-step process is demonstrated below: Consider f(x) = 7 and g(x) = x So we get, 7/x = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]² … (1) Let’s substitute f(x) = 7 and g(x) = x in equation (1), d/dx (7/x) = [0 * x - 7 * 1] / x² = -7/x² Thus, by applying the quotient rule, we get the derivative as -7/x².</p>
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<p>We will now prove the derivative of 7/x using the quotient rule. The step-by-step process is demonstrated below: Consider f(x) = 7 and g(x) = x So we get, 7/x = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]² … (1) Let’s substitute f(x) = 7 and g(x) = x in equation (1), d/dx (7/x) = [0 * x - 7 * 1] / x² = -7/x² Thus, by applying the quotient rule, we get the derivative as -7/x².</p>
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<h2>Higher-Order Derivatives of 7/x</h2>
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<h2>Higher-Order Derivatives of 7/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 7/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 7/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 7/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 7/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 the function 7/x has a vertical asymptote there. When x is 1, the derivative of 7/x = -7/1², which is -7.</p>
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<p>When x is 0, the derivative is undefined because the function 7/x has a vertical asymptote there. When x is 1, the derivative of 7/x = -7/1², which is -7.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 7/x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 7/x</h2>
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<p>Students frequently make mistakes when differentiating 7/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 7/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 (7/x · x²)</p>
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<p>Calculate the derivative of (7/x · 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>Here, we have f(x) = (7/x) · x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 7/x and v = x². Let’s differentiate each term, u′ = d/dx (7/x) = -7/x² v′ = d/dx (x²) = 2x substituting into the given equation, f'(x) = (-7/x²) · (x²) + (7/x) · (2x) Let’s simplify terms to get the final answer, f'(x) = -7 + 14 Thus, the derivative of the specified function is 7.</p>
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<p>Here, we have f(x) = (7/x) · x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 7/x and v = x². Let’s differentiate each term, u′ = d/dx (7/x) = -7/x² v′ = d/dx (x²) = 2x substituting into the given equation, f'(x) = (-7/x²) · (x²) + (7/x) · (2x) Let’s simplify terms to get the final answer, f'(x) = -7 + 14 Thus, the derivative of the specified function is 7.</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 water tank is being filled at a rate represented by the function y = 7/x, where y represents the rate of flow at a time x. If x = 5 minutes, measure the rate of change of the flow.</p>
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<p>A water tank is being filled at a rate represented by the function y = 7/x, where y represents the rate of flow at a time x. If x = 5 minutes, measure the rate of change of the flow.</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 = 7/x (rate of flow)...(1) Now, we will differentiate the equation (1) Take the derivative 7/x: dy/dx = -7/x² Given x = 5, substitute this into the derivative dy/dx = -7/(5)² dy/dx = -7/25 Hence, we get the rate of change of the flow at time x = 5 minutes as -7/25.</p>
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<p>We have y = 7/x (rate of flow)...(1) Now, we will differentiate the equation (1) Take the derivative 7/x: dy/dx = -7/x² Given x = 5, substitute this into the derivative dy/dx = -7/(5)² dy/dx = -7/25 Hence, we get the rate of change of the flow at time x = 5 minutes as -7/25.</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 flow at x = 5 minutes as -7/25, which means that at this point, the rate of flow is decreasing at a rate of 7/25 per minute.</p>
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<p>We find the rate of change of the flow at x = 5 minutes as -7/25, which means that at this point, the rate of flow is decreasing at a rate of 7/25 per minute.</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 = 7/x.</p>
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<p>Derive the second derivative of the function y = 7/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 = -7/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-7/x²] Here we use the power rule, d²y/dx² = 14/x³ Therefore, the second derivative of the function y = 7/x is 14/x³.</p>
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<p>The first step is to find the first derivative, dy/dx = -7/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-7/x²] Here we use the power rule, d²y/dx² = 14/x³ Therefore, the second derivative of the function y = 7/x is 14/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 -7/x².</p>
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<p>Using the power rule, we differentiate -7/x².</p>
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<p>We then substitute and simplify the terms to find the final answer.</p>
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<p>We then substitute and 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 ((7/x)²) = -14/x³.</p>
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<p>Prove: d/dx ((7/x)²) = -14/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 = (7/x)² = [7(x^-1)]² To differentiate, we use the chain rule: dy/dx = 2[7(x^-1)] · d/dx [7(x^-1)] Since the derivative of 7(x^-1) is -7x^-2, dy/dx = 2[7(x^-1)] · [-7x^-2] = -14/x³ Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = (7/x)² = [7(x^-1)]² To differentiate, we use the chain rule: dy/dx = 2[7(x^-1)] · d/dx [7(x^-1)] Since the derivative of 7(x^-1) is -7x^-2, dy/dx = 2[7(x^-1)] · [-7x^-2] = -14/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 7(x^-1) with its derivative.</p>
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<p>Then, we replace 7(x^-1) with its derivative.</p>
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<p>As a final step, we simplify the terms to derive the equation.</p>
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<p>As a final step, we simplify the terms 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 (7/x + x)</p>
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<p>Solve: d/dx (7/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 (7/x + x) = d/dx (7/x) + d/dx (x) We will substitute d/dx (7/x) = -7/x² and d/dx (x) = 1 = -7/x² + 1 Therefore, d/dx (7/x + x) = -7/x² + 1</p>
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<p>To differentiate the function, we use the sum rule: d/dx (7/x + x) = d/dx (7/x) + d/dx (x) We will substitute d/dx (7/x) = -7/x² and d/dx (x) = 1 = -7/x² + 1 Therefore, d/dx (7/x + x) = -7/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 7/x</h2>
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<h2>FAQs on the Derivative of 7/x</h2>
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<h3>1.Find the derivative of 7/x.</h3>
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<h3>1.Find the derivative of 7/x.</h3>
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<p>Using the power rule applied to 7x^-1, d/dx (7/x) = -7/x² (simplified)</p>
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<p>Using the power rule applied to 7x^-1, d/dx (7/x) = -7/x² (simplified)</p>
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<h3>2.Can we use the derivative of 7/x in real life?</h3>
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<h3>2.Can we use the derivative of 7/x in real life?</h3>
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<p>Yes, we can use the derivative of 7/x in real life for calculating the rate of change of any process, especially in fields such as mathematics, physics, and economics.</p>
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<p>Yes, we can use the derivative of 7/x in real life for calculating the rate of change of any process, especially in fields such as mathematics, physics, and economics.</p>
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<h3>3.Is it possible to take the derivative of 7/x at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of 7/x at the point where x = 0?</h3>
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<p>No, x = 0 is a point where 7/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 7/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 7/x?</h3>
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<h3>4.What rule is used to differentiate 7/x?</h3>
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<p>We can use the power rule or quotient rule to differentiate 7/x. Using the power rule: d/dx (7x^-1) = -7/x².</p>
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<p>We can use the power rule or quotient rule to differentiate 7/x. Using the power rule: d/dx (7x^-1) = -7/x².</p>
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<h3>5.Are the derivatives of 7/x and 1/x the same?</h3>
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<h3>5.Are the derivatives of 7/x and 1/x the same?</h3>
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<p>No, they are different. The derivative of 7/x is -7/x², while the derivative of 1/x is -1/x².</p>
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<p>No, they are different. The derivative of 7/x is -7/x², while the derivative of 1/x is -1/x².</p>
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<h2>Important Glossaries for the Derivative of 7/x</h2>
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<h2>Important Glossaries for the Derivative of 7/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>Fractional Function:</strong>A function expressed as a fraction, such as 7/x, with a constant numerator.</li>
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</ul><ul><li><strong>Fractional Function:</strong>A function expressed as a fraction, such as 7/x, with a constant numerator.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate functions that are expressed as a quotient of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate functions that are expressed as a quotient of two functions.</li>
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</ul><ul><li><strong>Power Rule:</strong>A rule used to differentiate functions in the form of x^n.</li>
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</ul><ul><li><strong>Power Rule:</strong>A rule used to differentiate functions in the form of x^n.</li>
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</ul><ul><li><strong>Undefined:</strong>Points where the function does not exist, such as x = 0 for 7/x.</li>
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</ul><ul><li><strong>Undefined:</strong>Points where the function does not exist, such as x = 0 for 7/x.</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>