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2026-01-01
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>We use the derivative of 4/x, which is -4/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 discuss the derivative of 4/x in detail.</p>
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<p>We use the derivative of 4/x, which is -4/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 discuss the derivative of 4/x in detail.</p>
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<h2>What is the Derivative of 4/x?</h2>
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<h2>What is the Derivative of 4/x?</h2>
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<p>We now understand the derivative of 4/x.</p>
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<p>We now understand the derivative of 4/x.</p>
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<p>It is commonly represented as d/dx (4/x) or (4/x)', and its value is -4/x².</p>
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<p>It is commonly represented as d/dx (4/x) or (4/x)', and its value is -4/x².</p>
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<p>The<a>function</a>4/x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The<a>function</a>4/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: Reciprocal Function: (4/x is a<a>constant</a>divided by x).</p>
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<p>The key concepts are mentioned below: Reciprocal Function: (4/x is a<a>constant</a>divided by x).</p>
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<p>Power Rule: Rule for differentiating x raised to a<a>power</a>. Quotient Rule: Used for differentiating<a>expressions</a>like 4/x.</p>
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<p>Power Rule: Rule for differentiating x raised to a<a>power</a>. Quotient Rule: Used for differentiating<a>expressions</a>like 4/x.</p>
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<h2>Derivative of 4/x Formula</h2>
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<h2>Derivative of 4/x Formula</h2>
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<p>The derivative of 4/x can be denoted as d/dx (4/x) or (4/x)'.</p>
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<p>The derivative of 4/x can be denoted as d/dx (4/x) or (4/x)'.</p>
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<p>The<a>formula</a>we use to differentiate 4/x is: d/dx (4/x) = -4/x² The formula applies to all x ≠ 0.</p>
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<p>The<a>formula</a>we use to differentiate 4/x is: d/dx (4/x) = -4/x² The formula applies to all x ≠ 0.</p>
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<h2>Proofs of the Derivative of 4/x</h2>
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<h2>Proofs of the Derivative of 4/x</h2>
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<p>We can derive the derivative of 4/x using proofs. To show this, we will use basic differentiation rules.</p>
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<p>We can derive the derivative of 4/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: Using the Power Rule Using the Quotient Rule We will now demonstrate that the differentiation of 4/x results in -4/x² using the above-mentioned methods: Using the Power Rule The derivative of 4/x can be expressed by rewriting it as 4x⁻¹.</p>
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<p>There are several methods we use to prove this, such as: Using the Power Rule Using the Quotient Rule We will now demonstrate that the differentiation of 4/x results in -4/x² using the above-mentioned methods: Using the Power Rule The derivative of 4/x can be expressed by rewriting it as 4x⁻¹.</p>
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<p>To find the derivative, consider f(x) = 4x⁻¹. Using the power rule, which states that d/dx (xⁿ) = n·xⁿ⁻¹, f'(x) = 4(-1)x⁻² f'(x) = -4/x²</p>
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<p>To find the derivative, consider f(x) = 4x⁻¹. Using the power rule, which states that d/dx (xⁿ) = n·xⁿ⁻¹, f'(x) = 4(-1)x⁻² f'(x) = -4/x²</p>
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<p>Hence, proved. Using the Quotient Rule To prove the differentiation of 4/x using the<a>quotient</a>rule, We use the formula: d/dx (u/v) = (v·u' - u·v') / v² Let u = 4 and v = x, u' = d/dx (4) = 0 v' = d/dx (x) = 1</p>
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<p>Hence, proved. Using the Quotient Rule To prove the differentiation of 4/x using the<a>quotient</a>rule, We use the formula: d/dx (u/v) = (v·u' - u·v') / v² Let u = 4 and v = x, u' = d/dx (4) = 0 v' = d/dx (x) = 1</p>
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<p>Substituting these into the quotient rule, d/dx (4/x) = (x·0 - 4·1) / x² d/dx (4/x) = -4/x² Hence, proved.</p>
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<p>Substituting these into the quotient rule, d/dx (4/x) = (x·0 - 4·1) / x² d/dx (4/x) = -4/x² Hence, proved.</p>
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<h2>Higher-Order Derivatives of 4/x</h2>
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<h2>Higher-Order Derivatives of 4/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.</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.</p>
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<p>Higher-order derivatives make it easier to understand functions like 4/x. 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>Higher-order derivatives make it easier to understand functions like 4/x. 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 4/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 4/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 4/x is undefined at x = 0.</p>
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<p>When x is 0, the derivative is undefined because the function 4/x is undefined at x = 0.</p>
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<p>As x approaches infinity, the derivative of 4/x approaches 0, indicating the slope becomes flatter.</p>
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<p>As x approaches infinity, the derivative of 4/x approaches 0, indicating the slope becomes flatter.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 4/x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 4/x</h2>
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<p>Students frequently make mistakes when differentiating 4/x. These mistakes can be resolved by understanding the proper solutions.</p>
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<p>Students frequently make mistakes when differentiating 4/x. These mistakes can be resolved by understanding the proper solutions.</p>
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<p>Here are a few common mistakes and ways to solve them:</p>
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<p>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 (4/x)·x³.</p>
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<p>Calculate the derivative of (4/x)·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 f(x) = (4/x)·x³. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 4/x and v = x³. Let’s differentiate each term, u′ = d/dx (4/x) = -4/x² v′ = d/dx (x³) = 3x²</p>
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<p>Here, we have f(x) = (4/x)·x³. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 4/x and v = x³. Let’s differentiate each term, u′ = d/dx (4/x) = -4/x² v′ = d/dx (x³) = 3x²</p>
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<p>Substituting into the given equation, f'(x) = (-4/x²)·x³ + (4/x)·3x²</p>
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<p>Substituting into the given equation, f'(x) = (-4/x²)·x³ + (4/x)·3x²</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -4x + 12</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -4x + 12</p>
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<p>Thus, the derivative of the specified function is -4x + 12.</p>
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<p>Thus, the derivative of the specified function is -4x + 12.</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 emptied, and the rate at which the water level decreases is represented by y = 4/x, where y represents the rate of decrease in liters per minute, and x is time in minutes. If x = 2 minutes, find the rate of change of this rate.</p>
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<p>A water tank is being emptied, and the rate at which the water level decreases is represented by y = 4/x, where y represents the rate of decrease in liters per minute, and x is time in minutes. If x = 2 minutes, find the rate of change of this rate.</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 = 4/x (rate of water level decrease)...(1) Now, we will differentiate the equation (1)</p>
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<p>We have y = 4/x (rate of water level decrease)...(1) Now, we will differentiate the equation (1)</p>
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<p>Take the derivative of 4/x: dy/dx = -4/x² Given x = 2 (substitute this into the derivative) dy/dx = -4/(2)² = -1</p>
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<p>Take the derivative of 4/x: dy/dx = -4/x² Given x = 2 (substitute this into the derivative) dy/dx = -4/(2)² = -1</p>
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<p>Hence, the rate of change of the rate of decrease is -1 liter per minute per minute at x = 2 minutes.</p>
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<p>Hence, the rate of change of the rate of decrease is -1 liter per minute per minute at x = 2 minutes.</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 rate of decrease at x = 2 minutes as -1, which means that the decrease in water level slows down at that point.</p>
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<p>We find the rate of change of the rate of decrease at x = 2 minutes as -1, which means that the decrease in water level slows down at that point.</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 = 4/x.</p>
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<p>Derive the second derivative of the function y = 4/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 = -4/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-4/x²] Using the power rule, d²y/dx² = 8/x³ Therefore, the second derivative of the function y = 4/x is 8/x³.</p>
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<p>The first step is to find the first derivative, dy/dx = -4/x²...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-4/x²] Using the power rule, d²y/dx² = 8/x³ Therefore, the second derivative of the function y = 4/x is 8/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, starting with the first derivative. Using the power rule, we differentiate -4/x². We then simplify the terms to find the final answer.</p>
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<p>We use the step-by-step process, starting with the first derivative. Using the power rule, we differentiate -4/x². 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 (2(4/x)) = -8/x².</p>
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<p>Prove: d/dx (2(4/x)) = -8/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 constant multiple rule: Consider y = 2(4/x) = 8/x</p>
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<p>Let’s start using the constant multiple rule: Consider y = 2(4/x) = 8/x</p>
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<p>To differentiate, we use the power rule: dy/dx = -8/x² Hence, proved.</p>
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<p>To differentiate, we use the power rule: dy/dx = -8/x² 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 constant multiple rule to differentiate the equation.</p>
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<p>In this step-by-step process, we used the constant multiple rule to differentiate the equation.</p>
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<p>Then, we applied the power rule to find the derivative.</p>
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<p>Then, we applied the power rule to find the derivative.</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 (4/(x + 1))</p>
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<p>Solve: d/dx (4/(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>To differentiate the function, we use the quotient rule: d/dx (4/(x + 1)) = (d/dx (4)·(x + 1) - 4·d/dx (x + 1))/ (x + 1)²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (4/(x + 1)) = (d/dx (4)·(x + 1) - 4·d/dx (x + 1))/ (x + 1)²</p>
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<p>We will substitute d/dx (4) = 0 and d/dx (x + 1) = 1 = (0·(x + 1) - 4·1)/ (x + 1)² = -4/(x + 1)²</p>
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<p>We will substitute d/dx (4) = 0 and d/dx (x + 1) = 1 = (0·(x + 1) - 4·1)/ (x + 1)² = -4/(x + 1)²</p>
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<p>Therefore, d/dx (4/(x + 1)) = -4/(x + 1)².</p>
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<p>Therefore, d/dx (4/(x + 1)) = -4/(x + 1)².</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the quotient rule. 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 4/x</h2>
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<h2>FAQs on the Derivative of 4/x</h2>
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<h3>1.Find the derivative of 4/x.</h3>
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<h3>1.Find the derivative of 4/x.</h3>
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<p>Using the power rule for 4/x, written as 4x⁻¹, d/dx (4/x) = -4/x² (simplified).</p>
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<p>Using the power rule for 4/x, written as 4x⁻¹, d/dx (4/x) = -4/x² (simplified).</p>
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<h3>2.Can we use the derivative of 4/x in real life?</h3>
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<h3>2.Can we use the derivative of 4/x in real life?</h3>
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<p>Yes, we can use the derivative of 4/x in real life in calculating rates of change, especially in fields such as mathematics, physics, and engineering.</p>
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<p>Yes, we can use the derivative of 4/x in real life in calculating rates of change, especially in fields such as mathematics, physics, and engineering.</p>
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<h3>3.Is it possible to take the derivative of 4/x at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of 4/x at the point where x = 0?</h3>
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<p>No, x = 0 is a point where 4/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 4/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 4/(x + 1)?</h3>
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<h3>4.What rule is used to differentiate 4/(x + 1)?</h3>
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<p>We use the quotient rule to differentiate 4/(x + 1), d/dx (4/(x + 1)) = -4/(x + 1)².</p>
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<p>We use the quotient rule to differentiate 4/(x + 1), d/dx (4/(x + 1)) = -4/(x + 1)².</p>
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<h3>5.Are the derivatives of 4/x and 1/x the same?</h3>
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<h3>5.Are the derivatives of 4/x and 1/x the same?</h3>
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<p>No, they are different. The derivative of 4/x is -4/x², while the derivative of 1/x is -1/x².</p>
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<p>No, they are different. The derivative of 4/x is -4/x², while the derivative of 1/x is -1/x².</p>
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<h2>Important Glossaries for the Derivative of 4/x</h2>
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<h2>Important Glossaries for the Derivative of 4/x</h2>
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<ul><li>Derivative: 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>Derivative: 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>Reciprocal Function: A function of the form k/x, where k is a constant.</li>
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</ul><ul><li>Reciprocal Function: A function of the form k/x, where k is a constant.</li>
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</ul><ul><li>Power Rule: A basic differentiation rule used to find the derivative of x raised to a power.</li>
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</ul><ul><li>Power Rule: A basic differentiation rule used to find the derivative of x raised to a power.</li>
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</ul><ul><li>Quotient Rule: A rule used to differentiate functions expressed as the quotient of two other functions.</li>
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</ul><ul><li>Quotient Rule: A rule used to differentiate functions expressed as the quotient of two other functions.</li>
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</ul><ul><li>Undefined Points: Points at which a function does not exist or is not continuous.</li>
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</ul><ul><li>Undefined Points: Points at which a function does not exist or is not continuous.</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>