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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), which is 1/x, to understand how the natural logarithm function changes in response to a slight change in x. Derivatives are crucial in various applications, including calculating growth rates and decay in natural processes. We will now discuss the derivative of ln(x) in detail.</p>
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<p>We use the derivative of ln(x), which is 1/x, to understand how the natural logarithm function changes in response to a slight change in x. Derivatives are crucial in various applications, including calculating growth rates and decay in natural processes. We will now discuss the derivative of ln(x) in detail.</p>
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<h2>What is the Derivative of ln(x)?</h2>
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<h2>What is the Derivative of ln(x)?</h2>
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<p>We now understand the derivative of ln(x). It is commonly represented as d/dx (ln x) or (ln x)', and its value is 1/x. The<a>function</a>ln(x) has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>We now understand the derivative of ln(x). It is commonly represented as d/dx (ln x) or (ln x)', and its value is 1/x. The<a>function</a>ln(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>Logarithm Function: ln(x) is the natural logarithm of x.</p>
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<p>Logarithm Function: ln(x) is the natural logarithm of x.</p>
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<p>Chain Rule: A rule used to differentiate composite functions involving ln(x).</p>
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<p>Chain Rule: A rule used to differentiate composite functions involving ln(x).</p>
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<p>Quotient Rule: Used when ln(x) appears in a<a>fraction</a>.</p>
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<p>Quotient Rule: Used when ln(x) appears in a<a>fraction</a>.</p>
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<h2>Derivative of ln(x) Formula</h2>
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<h2>Derivative of ln(x) Formula</h2>
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<p>The derivative of ln(x) can be denoted as d/dx (ln x) or (ln x)'. The<a>formula</a>we use to differentiate ln(x) is: d/dx (ln x) = 1/x The formula applies to all x where x > 0.</p>
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<p>The derivative of ln(x) can be denoted as d/dx (ln x) or (ln x)'. The<a>formula</a>we use to differentiate ln(x) is: d/dx (ln x) = 1/x The formula applies to all x where x > 0.</p>
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<h2>Proofs of the Derivative of ln(x)</h2>
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<h2>Proofs of the Derivative of ln(x)</h2>
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<p>We can derive the derivative of ln(x) using proofs. To show this, we will use the definition of the natural logarithm and 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 ln(x) using proofs. To show this, we will use the definition of the natural logarithm and 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 Chain Rule</li>
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<li>Using Chain Rule</li>
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</ol><p>We will now demonstrate that the differentiation of ln(x) results in 1/x using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of ln(x) results in 1/x 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 ln(x) 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) 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) using the first principle, we will consider f(x) = ln(x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>To find the derivative of ln(x) using the first principle, we will consider f(x) = ln(x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = ln(x), we write f(x + h) = ln(x + h).</p>
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<p>Given that f(x) = ln(x), we write f(x + h) = ln(x + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(x + h) - ln(x)] / h = limₕ→₀ [ln((x + h)/x)] / h = limₕ→₀ [1/x] * [ln(1 + h/x) / (h/x)]</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(x + h) - ln(x)] / h = limₕ→₀ [ln((x + h)/x)] / h = limₕ→₀ [1/x] * [ln(1 + h/x) / (h/x)]</p>
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<p>As h approaches 0, ln(1 + h/x) / (h/x) approaches 1 (using the limit property of ln), f'(x) = 1/x Hence, proved.</p>
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<p>As h approaches 0, ln(1 + h/x) / (h/x) approaches 1 (using the limit property of ln), f'(x) = 1/x 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) using the chain rule, Consider that we have a composite function involving ln(x), such as ln(u(x)), where u(x) is a differentiable function. The chain rule states that: d/dx [ln(u(x))] = (1/u(x)) * u'(x) For the simplest case where u(x) = x, we have: d/dx [ln(x)] = (1/x) * 1 = 1/x</p>
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<p>To prove the differentiation of ln(x) using the chain rule, Consider that we have a composite function involving ln(x), such as ln(u(x)), where u(x) is a differentiable function. The chain rule states that: d/dx [ln(u(x))] = (1/u(x)) * u'(x) For the simplest case where u(x) = x, we have: d/dx [ln(x)] = (1/x) * 1 = 1/x</p>
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<h2>Higher-Order Derivatives of ln(x)</h2>
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<h2>Higher-Order Derivatives of ln(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 provide deeper insights into the behavior<a>of functions</a>. For example, the second derivative can tell us about the concavity of the function.</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 provide deeper insights into the behavior<a>of functions</a>. For example, the second derivative can tell us about the concavity of the function.</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), we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the<a>rate</a>of change.</p>
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<p>For the nth Derivative of ln(x), we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the<a>rate</a>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 is 0, the derivative is undefined because ln(x) is undefined at x = 0. When x is 1, the derivative of ln(x) = 1/1, which is 1.</p>
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<p>When x is 0, the derivative is undefined because ln(x) is undefined at x = 0. When x is 1, the derivative of ln(x) = 1/1, which is 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(x)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(x)</h2>
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<p>Students frequently make mistakes when differentiating ln(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 ln(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 ln(x² + 1).</p>
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<p>Calculate the derivative of ln(x² + 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>Here, we have f(x) = ln(x² + 1). Using the chain rule, f'(x) = 1/(x² + 1) * d/dx(x² + 1) = 1/(x² + 1) * 2x = 2x/(x² + 1) Thus, the derivative of the specified function is 2x/(x² + 1).</p>
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<p>Here, we have f(x) = ln(x² + 1). Using the chain rule, f'(x) = 1/(x² + 1) * d/dx(x² + 1) = 1/(x² + 1) * 2x = 2x/(x² + 1) Thus, the derivative of the specified function is 2x/(x² + 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 using the chain rule. The first step is finding the derivative of the inner function and then combining it using the chain rule to get the final result.</p>
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<p>We find the derivative of the given function using the chain rule. The first step is finding the derivative of the inner function and then combining it using the chain 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 population grows according to the function P(t) = ln(t + 1), where t is time in years. Calculate the rate of change of the population at t = 4 years.</p>
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<p>A population grows according to the function P(t) = ln(t + 1), where t is time in years. Calculate the rate of change of the population at t = 4 years.</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 P(t) = ln(t + 1) (population function)...(1)</p>
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<p>We have P(t) = ln(t + 1) (population function)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative: dP/dt = 1/(t + 1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative: dP/dt = 1/(t + 1)</p>
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<p>Substitute t = 4 into the derivative: dP/dt = 1/(4 + 1) = 1/5</p>
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<p>Substitute t = 4 into the derivative: dP/dt = 1/(4 + 1) = 1/5</p>
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<p>Hence, the rate of change of the population at t = 4 years is 1/5 or 0.2.</p>
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<p>Hence, the rate of change of the population at t = 4 years is 1/5 or 0.2.</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 population at t = 4 years as 1/5, which indicates the rate at which the population is increasing at that specific time.</p>
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<p>We find the rate of change of the population at t = 4 years as 1/5, which indicates the rate at which the population is increasing at that specific 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 = ln(x).</p>
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<p>Derive the second derivative of the function y = ln(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/x...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 1/x...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/x] = -1/x²</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/x] = -1/x²</p>
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<p>Therefore, the second derivative of the function y = ln(x) is -1/x².</p>
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<p>Therefore, the second derivative of the function y = ln(x) is -1/x².</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. We then differentiate 1/x to find the second derivative, simplifying 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. We then differentiate 1/x to find the second derivative, simplifying 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(x²)) = 2/x.</p>
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<p>Prove: d/dx (ln(x²)) = 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>Let’s start using the chain rule: Consider y = ln(x²) = ln(u), where u = x²</p>
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<p>Let’s start using the chain rule: Consider y = ln(x²) = ln(u), where u = x²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 1/u * du/dx = 1/(x²) * 2x = 2/x Hence proved.</p>
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<p>To differentiate, we use the chain rule: dy/dx = 1/u * du/dx = 1/(x²) * 2x = 2/x 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 the inner function 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 the inner function 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 (ln(x)/x)</p>
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<p>Solve: d/dx (ln(x)/x)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (ln(x)/x) = (d/dx (ln(x)) * x - ln(x) * d/dx(x)) / x² We will substitute d/dx (ln(x)) = 1/x and d/dx (x) = 1 = (1/x * x - ln(x) * 1) / x² = (1 - ln(x)) / x² Therefore, d/dx (ln(x)/x) = (1 - ln(x)) / x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (ln(x)/x) = (d/dx (ln(x)) * x - ln(x) * d/dx(x)) / x² We will substitute d/dx (ln(x)) = 1/x and d/dx (x) = 1 = (1/x * x - ln(x) * 1) / x² = (1 - ln(x)) / x² Therefore, d/dx (ln(x)/x) = (1 - ln(x)) / 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 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 ln(x)</h2>
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<h2>FAQs on the Derivative of ln(x)</h2>
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<h3>1.Find the derivative of ln(x).</h3>
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<h3>1.Find the derivative of ln(x).</h3>
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<p>Using the basic rule for ln(x), d/dx (ln x) = 1/x</p>
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<p>Using the basic rule for ln(x), d/dx (ln x) = 1/x</p>
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<h3>2.Can we use the derivative of ln(x) in real life?</h3>
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<h3>2.Can we use the derivative of ln(x) in real life?</h3>
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<p>Yes, the derivative of ln(x) is used in real life in fields such as biology, economics, and physics to model growth rates and decay processes.</p>
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<p>Yes, the derivative of ln(x) is used in real life in fields such as biology, economics, and physics to model growth rates and decay processes.</p>
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<h3>3.Is it possible to take the derivative of ln(x) at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of ln(x) at the point where x = 0?</h3>
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<p>No, x = 0 is a point where ln(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 ln(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 ln(x)/x?</h3>
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<h3>4.What rule is used to differentiate ln(x)/x?</h3>
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<p>We use the quotient rule to differentiate ln(x)/x, d/dx (ln(x)/x) = (x * 1/x - ln(x) * 1) / x².</p>
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<p>We use the quotient rule to differentiate ln(x)/x, d/dx (ln(x)/x) = (x * 1/x - ln(x) * 1) / x².</p>
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<h3>5.Are the derivatives of ln(x) and ln⁻¹(x) the same?</h3>
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<h3>5.Are the derivatives of ln(x) and ln⁻¹(x) the same?</h3>
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<p>No, they are different. The derivative of ln(x) is 1/x, while the derivative of ln⁻¹(x) is not standardly defined as ln⁻¹(x) typically refers to the<a>inverse function</a>, which is e^x.</p>
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<p>No, they are different. The derivative of ln(x) is 1/x, while the derivative of ln⁻¹(x) is not standardly defined as ln⁻¹(x) typically refers to the<a>inverse function</a>, which is e^x.</p>
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<h2>Important Glossaries for the Derivative of ln(x)</h2>
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<h2>Important Glossaries for the Derivative of ln(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>Natural Logarithm:</strong>The natural logarithm is a logarithm to the base e, where e is approximately 2.71828.</li>
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</ul><ul><li><strong>Natural Logarithm:</strong>The natural logarithm is a logarithm to the base e, where e is approximately 2.71828.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used for differentiating composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used for differentiating composite functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for finding the derivative of a quotient of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for finding the derivative of a quotient of two functions.</li>
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</ul><ul><li><strong>First Principle:</strong>The foundational method for deriving the derivative of a function using limits.</li>
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</ul><ul><li><strong>First Principle:</strong>The foundational method for deriving the derivative of a function using limits.</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>