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<p>Last updated on<strong>September 26, 2025</strong></p>
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<p>Last updated on<strong>September 26, 2025</strong></p>
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<p>We use the derivative of ln(-x), which is -1/x, as a measuring tool for 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) in detail.</p>
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<p>We use the derivative of ln(-x), which is -1/x, as a measuring tool for 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) 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><strong>Natural Logarithm Function:</strong>ln(x) is the logarithm to the<a>base</a>e, where e is the Euler's<a>number</a>.</p>
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<p><strong>Natural Logarithm Function:</strong>ln(x) is the logarithm to the<a>base</a>e, where e is the Euler's<a>number</a>.</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating composite functions.</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating composite functions.</p>
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<p><strong>Negative Sign:</strong>The derivative introduces a negative sign due to the nature of the function.</p>
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<p><strong>Negative Sign:</strong>The derivative introduces a negative sign due to the nature of the function.</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))'.</p>
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<p>The derivative of ln(-x) can be denoted as d/dx (ln(-x)) or (ln(-x))'.</p>
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<p>The<a>formula</a>we use to differentiate ln(-x) is: d/dx (ln(-x)) = -1/x</p>
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<p>The<a>formula</a>we use to differentiate ln(-x) is: d/dx (ln(-x)) = -1/x</p>
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<p>The formula applies to all x where x < 0.</p>
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<p>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 chain rule 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 ln(-x) using proofs. To show this, we will use the chain rule along with the rules of differentiation. There are several methods we use to prove this, such as:</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>. To find the derivative of ln(-x)</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>. To find the derivative of ln(-x)</p>
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<p>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 Given that f(x) = ln(-x), we write f(x + h) = ln(-(x + h)).</p>
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<p>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 Given that f(x) = ln(-x), we write f(x + h) = ln(-(x + h)).</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [ln(-(x + h)) - ln(-x)] / h = limₕ→₀ [ln((x/x+h)/(x/x))] / h = limₕ→₀ [ln(1 + h/x)] / h</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [ln(-(x + h)) - ln(-x)] / h = limₕ→₀ [ln((x/x+h)/(x/x))] / h = limₕ→₀ [ln(1 + h/x)] / h</p>
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<p>Using the limit property ln(1 + y) ≈ y for small y, f'(x) = limₕ→₀ [(h/x) / h] = -1/x Hence, proved.</p>
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<p>Using the limit property ln(1 + y) ≈ y for small y, f'(x) = limₕ→₀ [(h/x) / h] = -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, We use the formula: ln(-x) = ln(u), where u = -x</p>
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<p>To prove the differentiation of ln(-x) using the chain rule, We use the formula: ln(-x) = ln(u), where u = -x</p>
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<p>By the chain rule: d/dx [ln(-x)] = (1/u) * (du/dx)</p>
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<p>By the chain rule: d/dx [ln(-x)] = (1/u) * (du/dx)</p>
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<p>Let’s substitute u = -x, d/dx (ln(-x)) = (1/(-x)) * (-1) = -1/x Hence, proved.</p>
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<p>Let’s substitute u = -x, d/dx (ln(-x)) = (1/(-x)) * (-1) = -1/x Hence, proved.</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 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).</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).</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 f n(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 ln(-x), we generally use f n(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 is 0, the derivative is undefined because ln(-x) is undefined there.</p>
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<p>When x is 0, the derivative is undefined because ln(-x) is undefined there.</p>
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<p>When x is -1, the derivative of ln(-x) = -1/(-1), which is 1.</p>
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<p>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(-3x).</p>
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<p>Calculate the derivative of ln(-3x).</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(-3x).</p>
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<p>Here, we have f(x) = ln(-3x).</p>
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<p>Using the chain rule, f'(x) = (1/(-3x)) * (-3) = -1/x</p>
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<p>Using the chain rule, f'(x) = (1/(-3x)) * (-3) = -1/x</p>
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<p>Thus, the derivative of ln(-3x) is -1/x.</p>
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<p>Thus, the derivative of ln(-3x) is -1/x.</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 applying the chain rule.</p>
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<p>We find the derivative of the given function by applying the chain rule.</p>
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<p>The first step is finding its derivative and then simplifying to get the final result.</p>
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<p>The first step is finding its derivative and then simplifying 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>The temperature T(x) of a chemical reaction is represented by T(x) = ln(-4x), where x is the time in minutes. Find the rate of change of temperature at x = -2 minutes.</p>
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<p>The temperature T(x) of a chemical reaction is represented by T(x) = ln(-4x), where x is the time in minutes. Find the rate of change of temperature at x = -2 minutes.</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 T(x) = ln(-4x) (temperature function)...(1)</p>
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<p>We have T(x) = ln(-4x) (temperature 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 of ln(-4x): dT/dx = -1/x</p>
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<p>Take the derivative of ln(-4x): dT/dx = -1/x</p>
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<p>Given x = -2 (substitute this into the derivative), dT/dx = -1/(-2) = 1/2</p>
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<p>Given x = -2 (substitute this into the derivative), dT/dx = -1/(-2) = 1/2</p>
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<p>Hence, the rate of change of temperature at x = -2 minutes is 1/2.</p>
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<p>Hence, the rate of change of temperature at x = -2 minutes is 1/2.</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 temperature at x = -2 as 1/2, which means that at that point, the temperature increases at a rate of 0.5 units per minute.</p>
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<p>We find the rate of change of temperature at x = -2 as 1/2, which means that at that point, the temperature increases at a rate of 0.5 units per minute.</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)(-1/x²) = 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)(-1/x²) = 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 get 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 get 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(-5x²)) = -2/x.</p>
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<p>Prove: d/dx (ln(-5x²)) = -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(-5x²) = ln(-(5x²))</p>
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<p>Let’s start using the chain rule: Consider y = ln(-5x²) = ln(-(5x²))</p>
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<p>To differentiate, we use the chain rule: dy/dx = (1/(-5x²)) * (-10x) = -2/x Hence proved.</p>
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<p>To differentiate, we use the chain rule: dy/dx = (1/(-5x²)) * (-10x) = -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.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation.</p>
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<p>Then, we replace the function with its derivative. As a final step, we simplify to derive the equation.</p>
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<p>Then, we replace the 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</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</p>
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<p>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</p>
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<p>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</p>
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<p>Therefore, d/dx (ln(-x)/x) = (-1 - ln(-x)) / x²</p>
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<p>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 chain rule for ln(-x), d/dx (ln(-x)) = -1/x (simplified).</p>
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<p>Using the chain rule for ln(-x), d/dx (ln(-x)) = -1/x (simplified).</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, we can use the derivative of ln(-x) in real life in calculating the rate of change of any logarithmic processes, especially in fields such as mathematics, physics, and economics.</p>
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<p>Yes, we can use the derivative of ln(-x) in real life in calculating the rate of change of any logarithmic processes, 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) 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) = (-1 - ln(-x)) / x².</p>
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<p>We use the quotient rule to differentiate ln(-x)/x, d/dx (ln(-x)/x) = (-1 - ln(-x)) / x².</p>
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<h3>5.Are the derivatives of ln(-x) and ln(x) the same?</h3>
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<h3>5.Are the derivatives of ln(-x) and ln(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(x) is 1/x.</p>
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<p>No, they are different. The derivative of ln(-x) is -1/x, while the derivative of ln(x) is 1/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 Function:</strong>The natural logarithm function is one of the primary logarithmic functions and is written as ln(x).</li>
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</ul><ul><li><strong>Natural Logarithm Function:</strong>The natural logarithm function is one of the primary logarithmic functions and is written as ln(x).</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus used to differentiate composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule in calculus used to differentiate composite functions.</li>
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</ul><ul><li><strong>First Derivative:</strong>It is the initial result of a function, which gives us the rate of change of a specific function.</li>
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</ul><ul><li><strong>First Derivative:</strong>It is the initial result of a function, which gives us the rate of change of a specific function.</li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a function does not have a value at a certain point.</li>
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</ul><ul><li><strong>Undefined:</strong>A term used when a function does not have a value at a certain point.</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>