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2026-01-01
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2026-02-28
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<p>150 Learners</p>
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<p>Last updated on<strong>September 9, 2025</strong></p>
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<p>Last updated on<strong>September 9, 2025</strong></p>
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<p>We use the derivative of -1/x, which is 1/x², as a tool for understanding how this 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 -1/x in detail.</p>
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<p>We use the derivative of -1/x, which is 1/x², as a tool for understanding how this 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 -1/x in detail.</p>
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<h2>What is the Derivative of -1/x?</h2>
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<h2>What is the Derivative of -1/x?</h2>
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<p>We now understand the derivative<a>of</a>-1/x.</p>
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<p>We now understand the derivative<a>of</a>-1/x.</p>
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<p>It is commonly represented as d/dx (-1/x) or (-1/x)', and its value is 1/x².</p>
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<p>It is commonly represented as d/dx (-1/x) or (-1/x)', and its value is 1/x².</p>
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<p>The<a>function</a>-1/x has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The<a>function</a>-1/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>Rational Function: (-1/x) is an example of a rational function.</p>
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<p>Rational Function: (-1/x) is an example of a rational function.</p>
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<p>Negative Exponent Rule: Used in simplifying derivatives of functions like -1/x.</p>
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<p>Negative Exponent Rule: Used in simplifying derivatives of functions like -1/x.</p>
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<p>Power Rule: A rule for differentiating<a>expressions</a>like x⁻¹.</p>
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<p>Power Rule: A rule for differentiating<a>expressions</a>like x⁻¹.</p>
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<h2>Derivative of -1/x Formula</h2>
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<h2>Derivative of -1/x Formula</h2>
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<p>The derivative of -1/x can be denoted as d/dx (-1/x) or (-1/x)'. The<a>formula</a>we use to differentiate -1/x is: d/dx (-1/x) = 1/x² The formula applies to all x where x ≠ 0.</p>
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<p>The derivative of -1/x can be denoted as d/dx (-1/x) or (-1/x)'. The<a>formula</a>we use to differentiate -1/x is: d/dx (-1/x) = 1/x² The formula applies to all x where x ≠ 0.</p>
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<h2>Proofs of the Derivative of -1/x</h2>
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<h2>Proofs of the Derivative of -1/x</h2>
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<p>We can derive the derivative of -1/x using proofs.</p>
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<p>We can derive the derivative of -1/x using proofs.</p>
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<p>To show this, we will use the differentiation rules.</p>
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<p>To show this, we will use the 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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<p>Using the Power Rule</p>
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<p>Using the Power Rule</p>
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<p>Using the Quotient Rule</p>
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<p>Using the Quotient Rule</p>
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<p>We will now demonstrate that the differentiation of -1/x results in 1/x² using the above-mentioned methods:</p>
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<p>We will now demonstrate that the differentiation of -1/x results in 1/x² using the above-mentioned methods:</p>
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<p>Using the Power Rule</p>
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<p>Using the Power Rule</p>
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<p>Consider the function f(x) = -1/x = -x⁻¹.</p>
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<p>Consider the function f(x) = -1/x = -x⁻¹.</p>
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<p>The<a>power</a>rule states that d/dx (xⁿ) = nxⁿ⁻¹.</p>
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<p>The<a>power</a>rule states that d/dx (xⁿ) = nxⁿ⁻¹.</p>
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<p>Applying the power rule, we have: f'(x) = d/dx (-x⁻¹) = -(-1)x⁻² = 1/x².</p>
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<p>Applying the power rule, we have: f'(x) = d/dx (-x⁻¹) = -(-1)x⁻² = 1/x².</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using the Quotient Rule</p>
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<p>Using the Quotient Rule</p>
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<p>To prove the differentiation of -1/x using the<a>quotient</a>rule, We use the formula: d/dx (u/v) = (v·u' - u·v')/v².</p>
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<p>To prove the differentiation of -1/x using the<a>quotient</a>rule, We use the formula: d/dx (u/v) = (v·u' - u·v')/v².</p>
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<p>Let u = -1 and v = x, then u' = 0 and v' = 1.</p>
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<p>Let u = -1 and v = x, then u' = 0 and v' = 1.</p>
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<p>Applying the quotient rule: d/dx (-1/x) = (x·0 - (-1)·1)/x² = 1/x².</p>
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<p>Applying the quotient rule: d/dx (-1/x) = (x·0 - (-1)·1)/x² = 1/x².</p>
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<p>Therefore, the derivative of -1/x is 1/x².</p>
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<p>Therefore, the derivative of -1/x is 1/x².</p>
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<h2>Higher-Order Derivatives of -1/x</h2>
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<h2>Higher-Order Derivatives of -1/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 -1/x.</p>
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<p>Higher-order derivatives make it easier to understand functions like -1/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).</p>
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<p>The second derivative is derived from the first derivative, which is denoted using f′′(x).</p>
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<p>Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>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 -1/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.</p>
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<p>For the nth Derivative of -1/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.</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 -1/x is undefined there. When x = 1, the derivative of -1/x = 1/(1²) = 1.</p>
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<p>When x = 0, the derivative is undefined because -1/x is undefined there. When x = 1, the derivative of -1/x = 1/(1²) = 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of -1/x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of -1/x</h2>
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<p>Students frequently make mistakes when differentiating -1/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 -1/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 (-1/x)·(x²).</p>
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<p>Calculate the derivative of (-1/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) = (-1/x)·x².</p>
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<p>Here, we have f(x) = (-1/x)·x².</p>
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<p>Using the product rule, f'(x) = u′v + uv′.</p>
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<p>Using the product rule, f'(x) = u′v + uv′.</p>
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<p>In the given equation, u = -1/x and v = x².</p>
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<p>In the given equation, u = -1/x and v = x².</p>
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<p>Let’s differentiate each term: u′ = d/dx (-1/x) = 1/x² v′ = d/dx (x²) = 2x</p>
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<p>Let’s differentiate each term: u′ = d/dx (-1/x) = 1/x² v′ = d/dx (x²) = 2x</p>
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<p>Substituting into the given equation, f'(x) = (1/x²)·(x²) + (-1/x)·(2x)</p>
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<p>Substituting into the given equation, f'(x) = (1/x²)·(x²) + (-1/x)·(2x)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 1 - 2 = -1.</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 1 - 2 = -1.</p>
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<p>Thus, the derivative of the specified function is -1.</p>
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<p>Thus, the derivative of the specified function is -1.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative of the given function by dividing the function into two parts. 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>We find the derivative of the given function by dividing the function into two parts. 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 draining at a rate represented by the function y = -1/x, where y represents the rate of water level change at time x. If x = 2 hours, measure the rate of change of the water level.</p>
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<p>A water tank is draining at a rate represented by the function y = -1/x, where y represents the rate of water level change at time x. If x = 2 hours, measure the rate of change of the water level.</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 = -1/x (rate of change of water level) …(1)</p>
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<p>We have y = -1/x (rate of change of water level) …(1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Take the derivative of -1/x: dy/dx = 1/x²</p>
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<p>Take the derivative of -1/x: dy/dx = 1/x²</p>
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<p>Given x = 2 (substitute this into the derivative) dy/dx = 1/2² = 1/4</p>
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<p>Given x = 2 (substitute this into the derivative) dy/dx = 1/2² = 1/4</p>
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<p>Hence, we get the rate of change of the water level at time x = 2 hours as 1/4.</p>
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<p>Hence, we get the rate of change of the water level at time x = 2 hours as 1/4.</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 water level at x = 2 hours as 1/4, which means that at a given point, the water level decreases at a rate of 1/4 the square of the time.</p>
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<p>We find the rate of change of the water level at x = 2 hours as 1/4, which means that at a given point, the water level decreases at a rate of 1/4 the square of the time.</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 = -1/x.</p>
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<p>Derive the second derivative of the function y = -1/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 = 1/x² …(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 1/x² …(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/x²]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/x²]</p>
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<p>Here we use the power rule, d²y/dx² = -2/x³</p>
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<p>Here we use the power rule, d²y/dx² = -2/x³</p>
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<p>Therefore, the second derivative of the function y = -1/x is -2/x³.</p>
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<p>Therefore, the second derivative of the function y = -1/x is -2/x³.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate 1/x². We then simplify the terms to find the final answer.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate 1/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 ((-1/x)²) = 2/x³.</p>
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<p>Prove: d/dx ((-1/x)²) = 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 chain rule: Consider y = (-1/x)² = (x⁻¹)²</p>
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<p>Let’s start using the chain rule: Consider y = (-1/x)² = (x⁻¹)²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(x⁻¹)·d/dx (x⁻¹)</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(x⁻¹)·d/dx (x⁻¹)</p>
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<p>Since the derivative of x⁻¹ is -1/x², dy/dx = 2(x⁻¹)(-1/x²) = 2/x³</p>
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<p>Since the derivative of x⁻¹ is -1/x², dy/dx = 2(x⁻¹)(-1/x²) = 2/x³</p>
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<p>Hence proved.</p>
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<p>Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace x⁻¹ with its derivative. As a final step, we simplify to derive the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace x⁻¹ with its derivative. 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 (-1/x²).</p>
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<p>Solve: d/dx (-1/x²).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we use the power rule:</p>
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<p>To differentiate the function, we use the power rule:</p>
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<p>Consider y = -1/x² = -x⁻²</p>
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<p>Consider y = -1/x² = -x⁻²</p>
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<p>Applying the power rule: dy/dx = 2x⁻³ = 2/x³</p>
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<p>Applying the power rule: dy/dx = 2x⁻³ = 2/x³</p>
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<p>Therefore, d/dx (-1/x²) = 2/x³.</p>
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<p>Therefore, d/dx (-1/x²) = 2/x³.</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 power 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 power 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 -1/x</h2>
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<h2>FAQs on the Derivative of -1/x</h2>
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<h3>1.Find the derivative of -1/x.</h3>
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<h3>1.Find the derivative of -1/x.</h3>
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<p>Using the power rule for -x⁻¹ gives, d/dx (-1/x) = 1/x².</p>
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<p>Using the power rule for -x⁻¹ gives, d/dx (-1/x) = 1/x².</p>
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<h3>2.Can we use the derivative of -1/x in real life?</h3>
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<h3>2.Can we use the derivative of -1/x in real life?</h3>
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<p>Yes, we can use the derivative of -1/x in real life in calculating rates of change in various contexts, such as physics and engineering.</p>
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<p>Yes, we can use the derivative of -1/x in real life in calculating rates of change in various contexts, such as physics and engineering.</p>
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<h3>3.Is it possible to take the derivative of -1/x at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of -1/x at the point where x = 0?</h3>
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<p>No, x = 0 is a point where -1/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 -1/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 -1/x²?</h3>
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<h3>4.What rule is used to differentiate -1/x²?</h3>
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<p>We use the power rule to differentiate -1/x², d/dx (-1/x²) = 2/x³.</p>
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<p>We use the power rule to differentiate -1/x², d/dx (-1/x²) = 2/x³.</p>
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<h3>5.Are the derivatives of -1/x and -x the same?</h3>
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<h3>5.Are the derivatives of -1/x and -x the same?</h3>
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<p>No, they are different. The derivative of -1/x is 1/x², while the derivative of -x is -1.</p>
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<p>No, they are different. The derivative of -1/x is 1/x², while the derivative of -x is -1.</p>
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<h3>6.Can we find the derivative of the -1/x formula?</h3>
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<h3>6.Can we find the derivative of the -1/x formula?</h3>
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<p>To find, consider y = -1/x. We use the power rule: y’ = d/dx (-x⁻¹) = 1/x².</p>
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<p>To find, consider y = -1/x. We use the power rule: y’ = d/dx (-x⁻¹) = 1/x².</p>
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<h2>Important Glossaries for the Derivative of -1/x</h2>
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<h2>Important Glossaries for the Derivative of -1/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 of calculus used to find the derivative of functions of the form xⁿ.</li>
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</ul><ul><li><strong>Power Rule:</strong>A fundamental rule of calculus used to find the derivative of functions of the form xⁿ.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used for differentiating functions that are divisions of two other functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used for differentiating functions that are divisions of two other functions.</li>
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</ul><ul><li><strong>Undefined Points:</strong>Points where a function does not exist or is not continuous.</li>
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</ul><ul><li><strong>Undefined Points:</strong>Points where a function does not exist or is not continuous.</li>
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</ul><ul><li><strong>Rational Function:</strong>A function that is the ratio of two polynomials, such as -1/x.</li>
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</ul><ul><li><strong>Rational Function:</strong>A function that is the ratio of two polynomials, such as -1/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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<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>