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2026-01-01
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<p>Last updated on<strong>September 12, 2025</strong></p>
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<p>Last updated on<strong>September 12, 2025</strong></p>
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<p>We use the derivative of 1/r, which is -1/r², as a measuring tool to understand how the function changes in response to a slight change in r. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 1/r in detail.</p>
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<p>We use the derivative of 1/r, which is -1/r², as a measuring tool to understand how the function changes in response to a slight change in r. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 1/r in detail.</p>
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<h2>What is the Derivative of 1/r?</h2>
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<h2>What is the Derivative of 1/r?</h2>
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<p>We now understand the derivative<a>of</a>1/r. It is commonly represented as d/dr (1/r) or (1/r)', and its value is -1/r². The<a>function</a>1/r 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>1/r. It is commonly represented as d/dr (1/r) or (1/r)', and its value is -1/r². The<a>function</a>1/r 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: (1/r) is the reciprocal of r.</p>
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<p>Reciprocal Function: (1/r) is the reciprocal of r.</p>
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<p>Power Rule: Rule for differentiating r raised to a<a>power</a>.</p>
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<p>Power Rule: Rule for differentiating r raised to a<a>power</a>.</p>
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<p>Negative Exponent: 1/r can be expressed as r^(-1).</p>
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<p>Negative Exponent: 1/r can be expressed as r^(-1).</p>
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<h2>Derivative of 1/r Formula</h2>
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<h2>Derivative of 1/r Formula</h2>
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<p>The derivative of 1/r can be denoted as d/dr (1/r) or (1/r)'.</p>
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<p>The derivative of 1/r can be denoted as d/dr (1/r) or (1/r)'.</p>
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<p>The<a>formula</a>we use to differentiate 1/r is: d/dr (1/r) = -1/r² (or) (1/r)' = -1/r²</p>
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<p>The<a>formula</a>we use to differentiate 1/r is: d/dr (1/r) = -1/r² (or) (1/r)' = -1/r²</p>
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<p>The formula applies to all r where r ≠ 0.</p>
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<p>The formula applies to all r where r ≠ 0.</p>
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<h2>Proofs of the Derivative of 1/r</h2>
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<h2>Proofs of the Derivative of 1/r</h2>
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<p>We can derive the derivative of 1/r using proofs. To show this, we will use the algebraic manipulation along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of 1/r using proofs. To show this, we will use the algebraic manipulation along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<ol><li>By First Principle</li>
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<ol><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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</ol><p>We will now demonstrate that the differentiation of 1/r results in -1/r² using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of 1/r results in -1/r² using the above-mentioned methods:</p>
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<h3>By First Principle</h3>
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<h3>By First Principle</h3>
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<p>The derivative of 1/r can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>The derivative of 1/r can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>To find the derivative of 1/r using the first principle, we will consider f(r) = 1/r.</p>
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<p>To find the derivative of 1/r using the first principle, we will consider f(r) = 1/r.</p>
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<p>Its derivative can be expressed as the following limit. f'(r) = limₕ→₀ [f(r + h) - f(r)] / h … (1)</p>
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<p>Its derivative can be expressed as the following limit. f'(r) = limₕ→₀ [f(r + h) - f(r)] / h … (1)</p>
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<p>Given that f(r) = 1/r, we write f(r + h) = 1/(r + h).</p>
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<p>Given that f(r) = 1/r, we write f(r + h) = 1/(r + h).</p>
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<p>Substituting these into<a>equation</a>(1),</p>
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<p>Substituting these into<a>equation</a>(1),</p>
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<p>f'(r) = limₕ→₀ [1/(r + h) - 1/r] / h = limₕ→₀ [(r - (r + h)) / (r(r + h))] / h = limₕ→₀ [-h / (r² + rh)] / h = limₕ→₀ [-1 / (r² + rh)]</p>
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<p>f'(r) = limₕ→₀ [1/(r + h) - 1/r] / h = limₕ→₀ [(r - (r + h)) / (r(r + h))] / h = limₕ→₀ [-h / (r² + rh)] / h = limₕ→₀ [-1 / (r² + rh)]</p>
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<p>As h approaches 0, the<a>term</a>rh vanishes, resulting in: f'(r) = -1/r²</p>
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<p>As h approaches 0, the<a>term</a>rh vanishes, resulting in: f'(r) = -1/r²</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Power Rule</h3>
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<h3>Using Power Rule</h3>
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<p>To prove the differentiation of 1/r using the power rule, We express 1/r as r(-1). Consider f(r) = r(-1)</p>
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<p>To prove the differentiation of 1/r using the power rule, We express 1/r as r(-1). Consider f(r) = r(-1)</p>
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<p>Using the power rule for differentiation: f'(r) = -1 * r(-2) = -1/r²</p>
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<p>Using the power rule for differentiation: f'(r) = -1 * r(-2) = -1/r²</p>
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<p>Thus, by applying the power rule, we find that the derivative is -1/r².</p>
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<p>Thus, by applying the power rule, we find that the derivative is -1/r².</p>
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<h2>Higher-Order Derivatives of 1/r</h2>
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<h2>Higher-Order Derivatives of 1/r</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 1/r.</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 1/r.</p>
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<p>For the first derivative of a function, we write f′(r), 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′′(r). Similarly, the third derivative, f′′′(r), 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′(r), 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′′(r). Similarly, the third derivative, f′′′(r), is the result of the second derivative and this pattern continues.</p>
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<p>For the nth Derivative of 1/r, we generally use fⁿ(r) for the nth derivative of a function f(r), 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 1/r, we generally use fⁿ(r) for the nth derivative of a function f(r), 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 r is 0, the derivative is undefined because 1/r has a vertical asymptote there. When r is 1, the derivative of 1/r = -1/1², which is -1.</p>
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<p>When r is 0, the derivative is undefined because 1/r has a vertical asymptote there. When r is 1, the derivative of 1/r = -1/1², which is -1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 1/r</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 1/r</h2>
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<p>Students frequently make mistakes when differentiating 1/r. 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/r. 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/r · r²).</p>
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<p>Calculate the derivative of (1/r · r²).</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(r) = (1/r) · r². Using simplification, f(r) = r. The derivative of r is, f'(r) = 1. Thus, the derivative of the specified function is 1.</p>
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<p>Here, we have f(r) = (1/r) · r². Using simplification, f(r) = r. The derivative of r is, f'(r) = 1. 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 simplifying the function first and then differentiating the simplified function to get the final result.</p>
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<p>We find the derivative of the given function by simplifying the function first and then differentiating the simplified function 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 spherical balloon is being inflated, and its radius r increases as a function of time t. The volume V is inversely proportional to r, given by V = k/r. If the radius r is 5 meters, find the rate of change of the volume with respect to r.</p>
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<p>A spherical balloon is being inflated, and its radius r increases as a function of time t. The volume V is inversely proportional to r, given by V = k/r. If the radius r is 5 meters, find the rate of change of the volume with respect to r.</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 V = k/r (volume of the balloon)...(1)</p>
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<p>We have V = k/r (volume of the balloon)...(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 with respect to r: dV/dr = -k/r².</p>
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<p>Take the derivative with respect to r: dV/dr = -k/r².</p>
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<p>Given r = 5 meters (substitute this into the derivative),</p>
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<p>Given r = 5 meters (substitute this into the derivative),</p>
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<p>dV/dr = -k/5². dV/dr = -k/25.</p>
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<p>dV/dr = -k/5². dV/dr = -k/25.</p>
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<p>Hence, the rate of change of the volume with respect to r when r = 5 meters is -k/25.</p>
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<p>Hence, the rate of change of the volume with respect to r when r = 5 meters is -k/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 volume at r = 5 meters, which indicates that as the radius increases, the volume decreases at a rate proportional to 1/r².</p>
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<p>We find the rate of change of the volume at r = 5 meters, which indicates that as the radius increases, the volume decreases at a rate proportional to 1/r².</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 V = 1/r.</p>
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<p>Derive the second derivative of the function V = 1/r.</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, dV/dr = -1/r²...(1)</p>
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<p>The first step is to find the first derivative, dV/dr = -1/r²...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²V/dr² = d/dr [-1/r²]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²V/dr² = d/dr [-1/r²]</p>
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<p>Using the power rule, d²V/dr² = 2/r³.</p>
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<p>Using the power rule, d²V/dr² = 2/r³.</p>
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<p>Therefore, the second derivative of the function V = 1/r is 2/r³.</p>
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<p>Therefore, the second derivative of the function V = 1/r is 2/r³.</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/r² to obtain the second derivative.</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/r² to obtain the second derivative.</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/dr (1/r²) = -2/r³.</p>
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<p>Prove: d/dr (1/r²) = -2/r³.</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 V = 1/r² = r(-2)</p>
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<p>Let’s start using the power rule: Consider V = 1/r² = r(-2)</p>
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<p>To differentiate, we use the power rule: dV/dr = -2r(-3) = -2/r³</p>
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<p>To differentiate, we use the power rule: dV/dr = -2r(-3) = -2/r³</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 power rule to differentiate the equation, which involves multiplying by the exponent and reducing the power by 1.</p>
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<p>In this step-by-step process, we used the power rule to differentiate the equation, which involves multiplying by the exponent and reducing the power by 1.</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/dr (1/r + r).</p>
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<p>Solve: d/dr (1/r + r).</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 basic differentiation rules:</p>
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<p>To differentiate the function, we use basic differentiation rules:</p>
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<p>d/dr (1/r + r) = d/dr (1/r) + d/dr (r) = -1/r² + 1.</p>
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<p>d/dr (1/r + r) = d/dr (1/r) + d/dr (r) = -1/r² + 1.</p>
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<p>Therefore, d/dr (1/r + r) = -1/r² + 1.</p>
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<p>Therefore, d/dr (1/r + r) = -1/r² + 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 each term separately and then combine them to obtain the final result.</p>
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<p>In this process, we differentiate each term separately and then combine them 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/r</h2>
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<h2>FAQs on the Derivative of 1/r</h2>
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<h3>1.Find the derivative of 1/r.</h3>
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<h3>1.Find the derivative of 1/r.</h3>
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<p>Using the power rule for 1/r gives r^(-1), d/dr (1/r) = -1/r² (simplified).</p>
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<p>Using the power rule for 1/r gives r^(-1), d/dr (1/r) = -1/r² (simplified).</p>
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<h3>2.Can we use the derivative of 1/r in real life?</h3>
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<h3>2.Can we use the derivative of 1/r in real life?</h3>
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<p>Yes, the derivative of 1/r is used in real life to understand rates of change in processes such as physics and engineering where inverse relationships are present.</p>
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<p>Yes, the derivative of 1/r is used in real life to understand rates of change in processes such as physics and engineering where inverse relationships are present.</p>
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<h3>3.Is it possible to take the derivative of 1/r at the point where r = 0?</h3>
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<h3>3.Is it possible to take the derivative of 1/r at the point where r = 0?</h3>
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<p>No, r = 0 is a point where 1/r 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, r = 0 is a point where 1/r 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 1/r²?</h3>
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<h3>4.What rule is used to differentiate 1/r²?</h3>
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<p>We use the power rule to differentiate 1/r², d/dr (r^(-2)) = -2r^(-3) = -2/r³.</p>
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<p>We use the power rule to differentiate 1/r², d/dr (r^(-2)) = -2r^(-3) = -2/r³.</p>
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<h3>5.Are the derivatives of 1/r and r^(-1) the same?</h3>
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<h3>5.Are the derivatives of 1/r and r^(-1) the same?</h3>
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<p>Yes, they are the same because 1/r is equivalent to r^(-1), and their derivatives are both -1/r².</p>
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<p>Yes, they are the same because 1/r is equivalent to r^(-1), and their derivatives are both -1/r².</p>
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<h2>Important Glossaries for the Derivative of 1/r</h2>
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<h2>Important Glossaries for the Derivative of 1/r</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 r.</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 r.</li>
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</ul><ul><li><strong>Reciprocal Function:</strong>A function of the form 1/r, where r is a variable.</li>
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</ul><ul><li><strong>Reciprocal Function:</strong>A function of the form 1/r, where r is a variable.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to find the derivative of a function of the form rn.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to find the derivative of a function of the form rn.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point where a function does not have a value, such as r = 0 for 1/r.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point where a function does not have a value, such as r = 0 for 1/r.</li>
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</ul><ul><li><strong>Rate of Change:</strong>A measure of how a quantity changes with respect to another quantity, often time or space.</li>
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</ul><ul><li><strong>Rate of Change:</strong>A measure of how a quantity changes with respect to another quantity, often time or space.</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>