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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 ln(x-2), which is 1/(x-2), as a tool to measure how the natural logarithm 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 ln(x-2) in detail.</p>
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<p>We use the derivative of ln(x-2), which is 1/(x-2), as a tool to measure how the natural logarithm 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 ln(x-2) in detail.</p>
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<h2>What is the Derivative of ln(x-2)?</h2>
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<h2>What is the Derivative of ln(x-2)?</h2>
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<p>We now understand the derivative<a>of</a>ln(x-2). It is commonly represented as d/dx [ln(x-2)] or [ln(x-2)]', and its value is 1/(x-2). The<a>function</a>ln(x-2) 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>ln(x-2). It is commonly represented as d/dx [ln(x-2)] or [ln(x-2)]', and its value is 1/(x-2). The<a>function</a>ln(x-2) 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>Natural Logarithm Function: ln(x-2).</p>
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<p>Natural Logarithm Function: ln(x-2).</p>
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<p>Chain Rule: Rule for differentiating ln(x-2).</p>
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<p>Chain Rule: Rule for differentiating ln(x-2).</p>
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<p>Reciprocal Rule: Derivative expressed as 1/(x-2).</p>
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<p>Reciprocal Rule: Derivative expressed as 1/(x-2).</p>
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<h2>Derivative of ln(x-2) Formula</h2>
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<h2>Derivative of ln(x-2) Formula</h2>
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<p>The derivative of ln(x-2) can be denoted as d/dx [ln(x-2)] or [ln(x-2)]'.</p>
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<p>The derivative of ln(x-2) can be denoted as d/dx [ln(x-2)] or [ln(x-2)]'.</p>
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<p>The<a>formula</a>we use to differentiate ln(x-2) is: d/dx [ln(x-2)] = 1/(x-2)</p>
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<p>The<a>formula</a>we use to differentiate ln(x-2) is: d/dx [ln(x-2)] = 1/(x-2)</p>
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<p>The formula applies to all x where x > 2.</p>
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<p>The formula applies to all x where x > 2.</p>
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<h2>Proofs of the Derivative of ln(x-2)</h2>
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<h2>Proofs of the Derivative of ln(x-2)</h2>
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<p>We can derive the derivative of ln(x-2) using proofs. To show this, we will use the rules of differentiation.</p>
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<p>We can derive the derivative of ln(x-2) 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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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Chain Rule</li>
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<li>Using Chain Rule</li>
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</ol><h3>By First Principle</h3>
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</ol><h3>By First Principle</h3>
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<p>The derivative of ln(x-2) 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 ln(x-2) 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 ln(x-2) using the first principle, we will consider f(x) = ln(x-2). Its derivative can be expressed as the following limit.</p>
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<p>To find the derivative of ln(x-2) using the first principle, we will consider f(x) = ln(x-2). Its derivative can be expressed as the following limit.</p>
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<p>f'(x) = limₕ→₀ [f(x + h) - f(x)] / h</p>
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<p>f'(x) = limₕ→₀ [f(x + h) - f(x)] / h</p>
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<p>Given that f(x) = ln(x-2), we write f(x + h) = ln((x + h) - 2).</p>
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<p>Given that f(x) = ln(x-2), we write f(x + h) = ln((x + h) - 2).</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [ln((x + h) - 2) - ln(x-2)] / h</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [ln((x + h) - 2) - ln(x-2)] / h</p>
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<p>Using properties of<a>logarithms</a>, f'(x) = limₕ→₀ ln[((x + h) - 2) / (x-2)] / h = limₕ→₀ 1/[(x + h) - 2] = 1/(x-2)</p>
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<p>Using properties of<a>logarithms</a>, f'(x) = limₕ→₀ ln[((x + h) - 2) / (x-2)] / h = limₕ→₀ 1/[(x + h) - 2] = 1/(x-2)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Chain Rule</h3>
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<h3>Using Chain Rule</h3>
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<p>To prove the differentiation of ln(x-2) using the chain rule, Consider f(x) = x-2 and g(x) = ln(x).</p>
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<p>To prove the differentiation of ln(x-2) using the chain rule, Consider f(x) = x-2 and g(x) = ln(x).</p>
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<p>So we get, ln(x-2) = g(f(x))</p>
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<p>So we get, ln(x-2) = g(f(x))</p>
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<p>By chain rule: d/dx [g(f(x))] = g'(f(x)) * f'(x)</p>
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<p>By chain rule: d/dx [g(f(x))] = g'(f(x)) * f'(x)</p>
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<p>Let’s substitute g(x) = ln(x) and f(x) = x-2, g'(x) = 1/x and f'(x) = 1 d/dx [ln(x-2)] = 1/(x-2) * 1 = 1/(x-2)</p>
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<p>Let’s substitute g(x) = ln(x) and f(x) = x-2, g'(x) = 1/x and f'(x) = 1 d/dx [ln(x-2)] = 1/(x-2) * 1 = 1/(x-2)</p>
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<h2>Higher-Order Derivatives of ln(x-2)</h2>
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<h2>Higher-Order Derivatives of ln(x-2)</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 ln(x-2).</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 ln(x-2).</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 ln(x-2), 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 ln(x-2), 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 = 2, the derivative is undefined because ln(x-2) has a vertical asymptote there. When x = 3, the derivative of ln(x-2) = 1/1 = 1.</p>
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<p>When x = 2, the derivative is undefined because ln(x-2) has a vertical asymptote there. When x = 3, the derivative of ln(x-2) = 1/1 = 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(x-2)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(x-2)</h2>
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<p>Students frequently make mistakes when differentiating ln(x-2). 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 ln(x-2). 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 ln(x-2)·(x-2).</p>
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<p>Calculate the derivative of ln(x-2)·(x-2).</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) = ln(x-2)·(x-2).</p>
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<p>Here, we have f(x) = ln(x-2)·(x-2).</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(x-2) and v = (x-2).</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(x-2) and v = (x-2).</p>
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<p>Let’s differentiate each term, u′ = d/dx [ln(x-2)] = 1/(x-2) v′ = d/dx (x-2) = 1</p>
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<p>Let’s differentiate each term, u′ = d/dx [ln(x-2)] = 1/(x-2) v′ = d/dx (x-2) = 1</p>
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<p>Substituting into the given equation, f'(x) = (1/(x-2))·(x-2) + ln(x-2)·1</p>
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<p>Substituting into the given equation, f'(x) = (1/(x-2))·(x-2) + ln(x-2)·1</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 1 + ln(x-2)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 1 + ln(x-2)</p>
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<p>Thus, the derivative of the specified function is 1 + ln(x-2).</p>
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<p>Thus, the derivative of the specified function is 1 + ln(x-2).</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 company monitors its revenue with the function R(x) = ln(x-2), where x represents the number of units sold. If the number of units sold is 10, find the rate of change of revenue.</p>
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<p>A company monitors its revenue with the function R(x) = ln(x-2), where x represents the number of units sold. If the number of units sold is 10, find the rate of change of revenue.</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 R(x) = ln(x-2) (revenue function)...(1)</p>
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<p>We have R(x) = ln(x-2) (revenue function)...(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 ln(x-2): dR/dx = 1/(x-2)</p>
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<p>Take the derivative ln(x-2): dR/dx = 1/(x-2)</p>
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<p>Given x = 10 (substitute this into the derivative)</p>
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<p>Given x = 10 (substitute this into the derivative)</p>
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<p>dR/dx = 1/(10-2) dR/dx = 1/8</p>
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<p>dR/dx = 1/(10-2) dR/dx = 1/8</p>
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<p>Hence, the rate of change of revenue when 10 units are sold is 1/8.</p>
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<p>Hence, the rate of change of revenue when 10 units are sold is 1/8.</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 revenue at x = 10 as 1/8, which means that for the 10th unit sold, the revenue is increasing at a rate of 1/8.</p>
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<p>We find the rate of change of revenue at x = 10 as 1/8, which means that for the 10th unit sold, the revenue is increasing at a rate of 1/8.</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 = ln(x-2).</p>
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<p>Derive the second derivative of the function y = ln(x-2).</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/(x-2)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 1/(x-2)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/(x-2)] = -1/(x-2)²</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/(x-2)] = -1/(x-2)²</p>
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<p>Therefore, the second derivative of the function y = ln(x-2) is -1/(x-2)².</p>
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<p>Therefore, the second derivative of the function y = ln(x-2) is -1/(x-2)².</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 reciprocal rule, we differentiate 1/(x-2). 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 reciprocal rule, we differentiate 1/(x-2). 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 [ln((x-2)²)] = 2/(x-2).</p>
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<p>Prove: d/dx [ln((x-2)²)] = 2/(x-2).</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 = ln((x-2)²) = 2 ln(x-2)</p>
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<p>Let’s start using the chain rule: Consider y = ln((x-2)²) = 2 ln(x-2)</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2 * d/dx [ln(x-2)]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2 * d/dx [ln(x-2)]</p>
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<p>Since the derivative of ln(x-2) is 1/(x-2), dy/dx = 2 * 1/(x-2) = 2/(x-2)</p>
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<p>Since the derivative of ln(x-2) is 1/(x-2), dy/dx = 2 * 1/(x-2) = 2/(x-2)</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 ln(x-2) with its derivative. As a final step, we substitute y = ln((x-2)²) 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 ln(x-2) with its derivative. As a final step, we substitute y = ln((x-2)²) 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 [ln(x-2)/x]</p>
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<p>Solve: d/dx [ln(x-2)/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 quotient rule: d/dx [ln(x-2)/x] = (d/dx [ln(x-2)]·x - ln(x-2)·d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx [ln(x-2)/x] = (d/dx [ln(x-2)]·x - ln(x-2)·d/dx(x))/x²</p>
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<p>We will substitute d/dx [ln(x-2)] = 1/(x-2) and d/dx (x) = 1 = (1/(x-2)·x - ln(x-2)·1)/x² = (x/(x-2) - ln(x-2))/x²</p>
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<p>We will substitute d/dx [ln(x-2)] = 1/(x-2) and d/dx (x) = 1 = (1/(x-2)·x - ln(x-2)·1)/x² = (x/(x-2) - ln(x-2))/x²</p>
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<p>Therefore, d/dx [ln(x-2)/x] = (x/(x-2) - ln(x-2))/x²</p>
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<p>Therefore, d/dx [ln(x-2)/x] = (x/(x-2) - ln(x-2))/x²</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we differentiate the given function using the product rule and 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 product rule and 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 ln(x-2)</h2>
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<h2>FAQs on the Derivative of ln(x-2)</h2>
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<h3>1.Find the derivative of ln(x-2).</h3>
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<h3>1.Find the derivative of ln(x-2).</h3>
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<p>Using the chain rule and reciprocal rule, d/dx [ln(x-2)] = 1/(x-2)</p>
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<p>Using the chain rule and reciprocal rule, d/dx [ln(x-2)] = 1/(x-2)</p>
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<h3>2.Can we use the derivative of ln(x-2) in real life?</h3>
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<h3>2.Can we use the derivative of ln(x-2) in real life?</h3>
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<p>Yes, we can use the derivative of ln(x-2) in real life in calculating the rate of change of any motion, especially in fields such as mathematics, physics, and economics.</p>
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<p>Yes, we can use the derivative of ln(x-2) in real life in calculating the rate of change of any motion, especially in fields such as mathematics, physics, and economics.</p>
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<h3>3.Is it possible to take the derivative of ln(x-2) at the point where x = 2?</h3>
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<h3>3.Is it possible to take the derivative of ln(x-2) at the point where x = 2?</h3>
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<p>No, x = 2 is a point where ln(x-2) 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 = 2 is a point where ln(x-2) 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 ln(x-2)/x?</h3>
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<h3>4.What rule is used to differentiate ln(x-2)/x?</h3>
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<p>We use the quotient rule to differentiate ln(x-2)/x, d/dx [ln(x-2)/x] = (x/(x-2) - ln(x-2))/x².</p>
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<p>We use the quotient rule to differentiate ln(x-2)/x, d/dx [ln(x-2)/x] = (x/(x-2) - ln(x-2))/x².</p>
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<h3>5.Can we find the derivative of the ln(x-2) formula?</h3>
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<h3>5.Can we find the derivative of the ln(x-2) formula?</h3>
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<p>To find, consider y = ln(x-2). We use the reciprocal rule: y' = 1/(x-2).</p>
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<p>To find, consider y = ln(x-2). We use the reciprocal rule: y' = 1/(x-2).</p>
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<h2>Important Glossaries for the Derivative of ln(x-2)</h2>
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<h2>Important Glossaries for the Derivative of ln(x-2)</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>Natural Logarithm:</strong>A logarithmic function denoted as ln(x), where the base is the mathematical constant e.</li>
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</ul><ul><li><strong>Natural Logarithm:</strong>A logarithmic function denoted as ln(x), where the base is the mathematical constant e.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus for differentiating the composition of two or more functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus for differentiating the composition of two or more functions.</li>
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</ul><ul><li><strong>Reciprocal Rule:</strong>A rule that involves taking the reciprocal of a function to find its derivative.</li>
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</ul><ul><li><strong>Reciprocal Rule:</strong>A rule that involves taking the reciprocal of a function to find its derivative.</li>
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</ul><ul><li><strong>Asymptote:</strong>The function approaches a line without intersecting or crossing it. This line is known as an asymptote.</li>
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</ul><ul><li><strong>Asymptote:</strong>The function approaches a line without intersecting or crossing it. This line is known as an asymptote.</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>