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2026-01-01
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>Last updated on<strong>September 15, 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 -lnx?</h2>
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<h2>What is the Derivative of -lnx?</h2>
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<p>We now understand the derivative<a>of</a>-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. The key concepts are mentioned below:</p>
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<p>We now understand the derivative<a>of</a>-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. The key concepts are mentioned below:</p>
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<p><strong>Natural Logarithm Function:</strong>ln(x) is the natural logarithm, which is the inverse of the exponential function.</p>
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<p><strong>Natural Logarithm Function:</strong>ln(x) is the natural logarithm, which is the inverse of the exponential function.</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</p>
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<p><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</p>
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<p><strong>Reciprocal Function:</strong>A function of the form 1/x.</p>
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<p><strong>Reciprocal Function:</strong>A function of the form 1/x.</p>
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<h2>Derivative of -lnx Formula</h2>
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<h2>Derivative of -lnx 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 -lnx</h2>
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<h2>Proofs of the Derivative of -lnx</h2>
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<p>We can derive the derivative of -ln(x) using proofs. To show this, we will use 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 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) - ln(x + h)] / h = limₕ→₀ ln[(x)/(x + h)] / 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) - ln(x + h)] / h = limₕ→₀ ln[(x)/(x + h)] / h</p>
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<p>We use the property ln(a/b) = ln a - ln b.</p>
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<p>We use the property ln(a/b) = ln a - ln b.</p>
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<p>Using the approximation ln(1 + u) ≈ u for small u, f'(x) = limₕ→₀ -h/x(x + h) / h = -1/x</p>
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<p>Using the approximation ln(1 + u) ≈ u for small u, f'(x) = limₕ→₀ -h/x(x + h) / h = -1/x</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) using the chain rule, Let y = -ln(x) = -1 * ln(x)</p>
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<p>To prove the differentiation of -ln(x) using the chain rule, Let y = -ln(x) = -1 * ln(x)</p>
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<p>Using the chain rule, we need to differentiate the outer and the inner functions.</p>
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<p>Using the chain rule, we need to differentiate the outer and the inner functions.</p>
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<p>The derivative of ln(x) is 1/x.</p>
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<p>The derivative of ln(x) is 1/x.</p>
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<p>Therefore, dy/dx = -1 * (1/x) = -1/x</p>
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<p>Therefore, dy/dx = -1 * (1/x) = -1/x</p>
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<h2>Higher-Order Derivatives of -lnx</h2>
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<h2>Higher-Order Derivatives of -lnx</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 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), 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 = 0, the derivative is undefined because ln(x) is not defined for non-positive values. When x = 1, the derivative of -ln(x) = -1/1 = -1.</p>
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<p>When x = 0, the derivative is undefined because ln(x) is not defined for non-positive values. When x = 1, the derivative of -ln(x) = -1/1 = -1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of -lnx</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of -lnx</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(x)·x²)</p>
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<p>Calculate the derivative of (-ln(x)·x²)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = -ln(x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = -ln(x) and v = x².</p>
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<p>Here, we have f(x) = -ln(x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = -ln(x) and v = x².</p>
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<p>Let’s differentiate each term, u′= d/dx (-ln(x)) = -1/x v′= d/dx (x²) = 2x</p>
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<p>Let’s differentiate each term, u′= d/dx (-ln(x)) = -1/x v′= d/dx (x²) = 2x</p>
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<p>Substituting into the given equation, f'(x) = (-1/x)·x² + (-ln(x))·2x</p>
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<p>Substituting into the given equation, f'(x) = (-1/x)·x² + (-ln(x))·2x</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -x + (-2x ln(x))</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -x + (-2x ln(x))</p>
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<p>Thus, the derivative of the specified function is -x - 2x ln(x).</p>
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<p>Thus, the derivative of the specified function is -x - 2x ln(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 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 is analyzing its profit margin, which is modeled as y = -ln(x) where y represents the profit decline as x increases. If x = 5, calculate the rate of profit decline.</p>
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<p>A company is analyzing its profit margin, which is modeled as y = -ln(x) where y represents the profit decline as x increases. If x = 5, calculate the rate of profit decline.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>We have y = -ln(x) (profit decline)...(1)</p>
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<p>We have y = -ln(x) (profit decline)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative -ln(x): dy/dx = -1/x</p>
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<p>Now, we will differentiate the equation (1) Take the derivative -ln(x): dy/dx = -1/x</p>
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<p>Given x = 5 (substitute this into the derivative) dy/dx = -1/5</p>
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<p>Given x = 5 (substitute this into the derivative) dy/dx = -1/5</p>
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<p>Hence, we get the rate of profit decline at x=5 as -1/5.</p>
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<p>Hence, we get the rate of profit decline at x=5 as -1/5.</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 profit decline at x=5 as -1/5, which means that at a given point, the profit declines by 1/5 as x increases.</p>
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<p>We find the rate of profit decline at x=5 as -1/5, which means that at a given point, the profit declines by 1/5 as x increases.</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]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/x]</p>
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<p>Here we use the power rule, d²y/dx² = 1/x²</p>
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<p>Here we use the power rule, d²y/dx² = 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. Using the power rule, we differentiate -1/x. We then substitute the identity and simplify the terms to find the final answer.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate -1/x. We then substitute the identity and 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 ln(x)/x.</p>
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<p>Prove: d/dx (-ln(x)²) = -2 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>Let’s start using the chain rule: Consider y = (-ln(x))²</p>
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<p>Let’s start using the chain rule: Consider y = (-ln(x))²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(-ln(x))·d/dx [-ln(x)]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(-ln(x))·d/dx [-ln(x)]</p>
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<p>Since the derivative of -ln(x) is -1/x, dy/dx = 2(-ln(x))·(-1/x) = -2 ln(x)/x Hence proved.</p>
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<p>Since the derivative of -ln(x) is -1/x, dy/dx = 2(-ln(x))·(-1/x) = -2 ln(x)/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 ln(x) with its derivative. As a final step, we substitute y = (-ln(x))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) with its derivative. As a final step, we substitute y = (-ln(x))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)/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))/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))/x² = (-1 + ln(x))/x²</p>
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<p>Therefore, d/dx (-ln(x)/x) = (ln(x) - 1)/x²</p>
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<p>Therefore, d/dx (-ln(x)/x) = (ln(x) - 1)/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 -lnx</h2>
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<h2>FAQs on the Derivative of -lnx</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, the derivative of -ln(x) is -1/x.</p>
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<p>Using the chain rule, the derivative of -ln(x) is -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, we can use the derivative of -ln(x) in real life in calculating the rate of change of any natural logarithm-related phenomena, 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 natural logarithm-related phenomena, 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 these points (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 these points (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) = (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) = (ln(x) - 1)/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 -lnx</h2>
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<h2>Important Glossaries for the Derivative of -lnx</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 function is the inverse of the exponential function, written as ln(x).</li>
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</ul><ul><li><strong>Natural Logarithm:</strong>The natural logarithm function is the inverse of the exponential function, written as ln(x).</li>
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</ul><ul><li><strong>Chain Rule:</strong>A differentiation rule for composite functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A differentiation rule for composite functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the division of two functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating the division of two 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><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>