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2026-01-01
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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(f(x)), which is f'(x)/f(x), as a tool for understanding 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 discuss the derivative of ln(f(x)) in detail.</p>
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<p>We use the derivative of ln(f(x)), which is f'(x)/f(x), as a tool for understanding 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 discuss the derivative of ln(f(x)) in detail.</p>
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<h2>What is the Derivative of ln(f(x))?</h2>
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<h2>What is the Derivative of ln(f(x))?</h2>
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<p>We now understand the derivative of ln(f(x)). It is commonly represented as d/dx (ln(f(x))) or (ln(f(x)))', and its value is f'(x)/f(x). The<a>function</a>ln(f(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(f(x)). It is commonly represented as d/dx (ln(f(x))) or (ln(f(x)))', and its value is f'(x)/f(x). The<a>function</a>ln(f(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: Natural Logarithm Function: ln(x) is the logarithm to the<a>base</a>e. Chain Rule: A rule for differentiating composite functions like ln(f(x)), which requires the derivative of the inner function. Quotient Function: Involves the derivative being expressed as a<a>quotient</a><a>of functions</a>.</p>
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<p>The key concepts are mentioned below: Natural Logarithm Function: ln(x) is the logarithm to the<a>base</a>e. Chain Rule: A rule for differentiating composite functions like ln(f(x)), which requires the derivative of the inner function. Quotient Function: Involves the derivative being expressed as a<a>quotient</a><a>of functions</a>.</p>
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<h2>Derivative of ln(f(x)) Formula</h2>
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<h2>Derivative of ln(f(x)) Formula</h2>
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<p>The derivative of ln(f(x)) can be denoted as d/dx (ln(f(x))) or (ln(f(x)))'.</p>
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<p>The derivative of ln(f(x)) can be denoted as d/dx (ln(f(x))) or (ln(f(x)))'.</p>
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<p>The<a>formula</a>we use to differentiate ln(f(x)) is: d/dx (ln(f(x))) = f'(x)/f(x) The formula applies to all x for which f(x) > 0.</p>
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<p>The<a>formula</a>we use to differentiate ln(f(x)) is: d/dx (ln(f(x))) = f'(x)/f(x) The formula applies to all x for which f(x) > 0.</p>
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<h2>Proofs of the Derivative of ln(f(x))</h2>
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<h2>Proofs of the Derivative of ln(f(x))</h2>
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<p>We can derive the derivative of ln(f(x)) using proofs. To show this, we will use the properties of<a>logarithms</a>along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule We will now demonstrate that the differentiation of ln(f(x)) results in f'(x)/f(x) using the above-mentioned methods:</p>
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<p>We can derive the derivative of ln(f(x)) using proofs. To show this, we will use the properties of<a>logarithms</a>along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule We will now demonstrate that the differentiation of ln(f(x)) results in f'(x)/f(x) using the above-mentioned methods:</p>
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<p>By First Principle The derivative of ln(f(x)) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of ln(f(x)) using the first principle, we will consider g(x) = ln(f(x)). Its derivative can be expressed as the following limit. g'(x) = limₕ→₀ [g(x + h) - g(x)] / h … (1) Given that g(x) = ln(f(x)), we write g(x + h) = ln(f(x + h)). Substituting these into<a>equation</a>(1), g'(x) = limₕ→₀ [ln(f(x + h)) - ln(f(x))] / h = limₕ→₀ ln[(f(x + h)/f(x))] / h</p>
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<p>By First Principle The derivative of ln(f(x)) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of ln(f(x)) using the first principle, we will consider g(x) = ln(f(x)). Its derivative can be expressed as the following limit. g'(x) = limₕ→₀ [g(x + h) - g(x)] / h … (1) Given that g(x) = ln(f(x)), we write g(x + h) = ln(f(x + h)). Substituting these into<a>equation</a>(1), g'(x) = limₕ→₀ [ln(f(x + h)) - ln(f(x))] / h = limₕ→₀ ln[(f(x + h)/f(x))] / h</p>
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<p>Using the properties of logarithms, ln(a) - ln(b) = ln(a/b), g'(x) = limₕ→₀ ln[1 + (f(x + h) - f(x))/f(x)] / h Using the approximation ln(1 + u) ≈ u for small u, g'(x) = limₕ→₀ (f(x + h) - f(x))/(h f(x)) g'(x) = f'(x)/f(x) Hence, proved.</p>
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<p>Using the properties of logarithms, ln(a) - ln(b) = ln(a/b), g'(x) = limₕ→₀ ln[1 + (f(x + h) - f(x))/f(x)] / h Using the approximation ln(1 + u) ≈ u for small u, g'(x) = limₕ→₀ (f(x + h) - f(x))/(h f(x)) g'(x) = f'(x)/f(x) Hence, proved.</p>
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<p>Using Chain Rule To prove the differentiation of ln(f(x)) using the chain rule, We use the formula: ln(f(x)) = ln(u), where u = f(x) The derivative of ln(u) is 1/u * du/dx Let u = f(x) Then, d/du(ln(u)) = 1/u And, du/dx = f'(x) Therefore, d/dx (ln(f(x))) = 1/f(x) * f'(x) = f'(x)/f(x)</p>
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<p>Using Chain Rule To prove the differentiation of ln(f(x)) using the chain rule, We use the formula: ln(f(x)) = ln(u), where u = f(x) The derivative of ln(u) is 1/u * du/dx Let u = f(x) Then, d/du(ln(u)) = 1/u And, du/dx = f'(x) Therefore, d/dx (ln(f(x))) = 1/f(x) * f'(x) = f'(x)/f(x)</p>
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<h2>Higher-Order Derivatives of ln(f(x))</h2>
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<h2>Higher-Order Derivatives of ln(f(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. 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(f(x)).</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. 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(f(x)).</p>
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<p>For the first derivative of a function, we write g′(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 g′′(x). Similarly, the third derivative, g′′′(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 g′(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 g′′(x). Similarly, the third derivative, g′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>For the nth Derivative of ln(f(x)), we generally use gⁿ(x) for the nth derivative of a function g(x), which tells us the change in the rate of change.</p>
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<p>For the nth Derivative of ln(f(x)), we generally use gⁿ(x) for the nth derivative of a function g(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 f(x) is a<a>constant</a>, the derivative is zero because the natural logarithm function becomes a constant. If f(x) is e^x, then the derivative of ln(e^x) = x is 1.</p>
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<p>When f(x) is a<a>constant</a>, the derivative is zero because the natural logarithm function becomes a constant. If f(x) is e^x, then the derivative of ln(e^x) = x is 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(f(x))</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(f(x))</h2>
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<p>Students frequently make mistakes when differentiating ln(f(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(f(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) · e^x</p>
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<p>Calculate the derivative of ln(x² + 1) · e^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>Here, we have f(x) = ln(x² + 1) · ex. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(x² + 1) and v = ex. Let’s differentiate each term, u′ = d/dx (ln(x² + 1)) = (2x)/(x² + 1) v′ = d/dx (ex) = ex</p>
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<p>Here, we have f(x) = ln(x² + 1) · ex. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(x² + 1) and v = ex. Let’s differentiate each term, u′ = d/dx (ln(x² + 1)) = (2x)/(x² + 1) v′ = d/dx (ex) = ex</p>
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<p>Substituting into the given equation, f'(x) = [(2x)/(x² + 1)] · ex + ln(x² + 1) · ex Let’s simplify terms to get the final answer, f'(x) = (2x · ex)/(x² + 1) + ln(x² + 1) · ex</p>
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<p>Substituting into the given equation, f'(x) = [(2x)/(x² + 1)] · ex + ln(x² + 1) · ex Let’s simplify terms to get the final answer, f'(x) = (2x · ex)/(x² + 1) + ln(x² + 1) · ex</p>
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<p>Thus, the derivative of the specified function is (2x · ex)/(x² + 1) + ln(x² + 1) · ex.</p>
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<p>Thus, the derivative of the specified function is (2x · ex)/(x² + 1) + ln(x² + 1) · ex.</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 car's velocity is represented by the function v(x) = ln(x + 1) where x represents time in seconds. If x = 3 seconds, measure the rate of change of velocity.</p>
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<p>A car's velocity is represented by the function v(x) = ln(x + 1) where x represents time in seconds. If x = 3 seconds, measure the rate of change of velocity.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>We have v(x) = ln(x + 1) (velocity of the car)...(1)Now, we will differentiate the equation (1) Take the derivative ln(x + 1): dv/dx = 1/(x + 1)Given x = 3 (substitute this into the derivative) dv/dx = 1/(3 + 1) = 1/4Hence, we get the rate of change of velocity at x = 3 seconds as 1/4.</p>
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<p>We have v(x) = ln(x + 1) (velocity of the car)...(1)Now, we will differentiate the equation (1) Take the derivative ln(x + 1): dv/dx = 1/(x + 1)Given x = 3 (substitute this into the derivative) dv/dx = 1/(3 + 1) = 1/4Hence, we get the rate of change of velocity at x = 3 seconds as 1/4.</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 velocity at x = 3 seconds as 1/4, which means that at this point, the rate of change of velocity is 1/4 units per second.</p>
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<p>We find the rate of change of velocity at x = 3 seconds as 1/4, which means that at this point, the rate of change of velocity is 1/4 units per second.</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(x² + 1).</p>
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<p>Derive the second derivative of the function y = 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>The first step is to find the first derivative, dy/dx = (2x)/(x² + 1)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [(2x)/(x² + 1)] Here we use the quotient rule, d²y/dx² = [(x² + 1) · 2 - 2x · 2x]/(x² + 1)² = [2x² + 2 - 4x²]/(x² + 1)² = [-2x² + 2]/(x² + 1)² Therefore, the second derivative of the function y = ln(x² + 1) is [-2x² + 2]/(x² + 1)².</p>
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<p>The first step is to find the first derivative, dy/dx = (2x)/(x² + 1)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [(2x)/(x² + 1)] Here we use the quotient rule, d²y/dx² = [(x² + 1) · 2 - 2x · 2x]/(x² + 1)² = [2x² + 2 - 4x²]/(x² + 1)² = [-2x² + 2]/(x² + 1)² Therefore, the second derivative of the function y = ln(x² + 1) is [-2x² + 2]/(x² + 1)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.</p>
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<p>We use the step-by-step process, where we start with the first derivative.</p>
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<p>Using the quotient rule, we differentiate the first derivative.</p>
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<p>Using the quotient rule, we differentiate the first derivative.</p>
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<p>We then simplify the terms to find the final answer.</p>
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<p>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 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 = [ln(x)]² To differentiate, we use the chain rule: dy/dx = 2 ln(x) · d/dx [ln(x)] Since the derivative of ln(x) is 1/x, dy/dx = 2 ln(x) · (1/x) Substituting y = ln(x)², d/dx (ln(x)²) = 2 ln(x)/x Hence proved.</p>
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<p>Let’s start using the chain rule: Consider y = ln(x)² = [ln(x)]² To differentiate, we use the chain rule: dy/dx = 2 ln(x) · d/dx [ln(x)] Since the derivative of ln(x) is 1/x, dy/dx = 2 ln(x) · (1/x) Substituting y = ln(x)², d/dx (ln(x)²) = 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.</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 ln(x) with its derivative.</p>
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<p>Then, we replace ln(x) with its derivative.</p>
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<p>As a final step, we substitute y = ln(x)² to derive the equation.</p>
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<p>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 + 1)/x)</p>
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<p>Solve: d/dx (ln(x + 1)/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 + 1)/x) = (d/dx (ln(x + 1)) · x - ln(x + 1) · d/dx(x))/x² We will substitute d/dx (ln(x + 1)) = 1/(x + 1) and d/dx(x) = 1 = [(1/(x + 1)) · x - ln(x + 1) · 1]/x² = [x/(x + 1) - ln(x + 1)]/x² Therefore, d/dx (ln(x + 1)/x) = [x/(x + 1) - ln(x + 1)]/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (ln(x + 1)/x) = (d/dx (ln(x + 1)) · x - ln(x + 1) · d/dx(x))/x² We will substitute d/dx (ln(x + 1)) = 1/(x + 1) and d/dx(x) = 1 = [(1/(x + 1)) · x - ln(x + 1) · 1]/x² = [x/(x + 1) - ln(x + 1)]/x² Therefore, d/dx (ln(x + 1)/x) = [x/(x + 1) - 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.</p>
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<p>In this process, we differentiate the given function using the product rule and quotient rule.</p>
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<p>As a final step, we simplify the equation to obtain the final result.</p>
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<p>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(f(x))</h2>
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<h2>FAQs on the Derivative of ln(f(x))</h2>
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<h3>1.Find the derivative of ln(f(x)).</h3>
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<h3>1.Find the derivative of ln(f(x)).</h3>
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<p>Using the chain rule for ln(f(x)) gives f'(x)/f(x).</p>
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<p>Using the chain rule for ln(f(x)) gives f'(x)/f(x).</p>
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<h3>2.Can we use the derivative of ln(f(x)) in real life?</h3>
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<h3>2.Can we use the derivative of ln(f(x)) in real life?</h3>
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<p>Yes, we can use the derivative of ln(f(x)) in real life to calculate the rate of change of any process, especially in fields such as mathematics, physics, and economics.</p>
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<p>Yes, we can use the derivative of ln(f(x)) in real life to calculate the rate of change of any process, 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(f(x)) at the point where f(x) = 0?</h3>
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<h3>3.Is it possible to take the derivative of ln(f(x)) at the point where f(x) = 0?</h3>
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<p>No, ln(f(x)) is undefined for f(x) ≤ 0, 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, ln(f(x)) is undefined for f(x) ≤ 0, 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(f(x))/x?</h3>
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<h3>4.What rule is used to differentiate ln(f(x))/x?</h3>
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<p>We use the quotient rule to differentiate ln(f(x))/x, d/dx (ln(f(x))/x) = (x · f'(x)/f(x) - ln(f(x)))/x².</p>
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<p>We use the quotient rule to differentiate ln(f(x))/x, d/dx (ln(f(x))/x) = (x · f'(x)/f(x) - ln(f(x)))/x².</p>
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<h3>5.Are the derivatives of ln(f(x)) and ln⁻¹(x) the same?</h3>
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<h3>5.Are the derivatives of ln(f(x)) and ln⁻¹(x) the same?</h3>
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<p>No, they are different. The derivative of ln(f(x)) is f'(x)/f(x), while the derivative of ln⁻¹(x) is 1/x.</p>
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<p>No, they are different. The derivative of ln(f(x)) is f'(x)/f(x), while the derivative of ln⁻¹(x) is 1/x.</p>
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<h3>6.Can we find the derivative of the ln(f(x)) formula?</h3>
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<h3>6.Can we find the derivative of the ln(f(x)) formula?</h3>
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<p>To find, consider y = ln(f(x)). We use the chain rule: y' = f'(x)/f(x) (since ln(f(x)) is the natural log function).</p>
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<p>To find, consider y = ln(f(x)). We use the chain rule: y' = f'(x)/f(x) (since ln(f(x)) is the natural log function).</p>
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<h2>Important Glossaries for the Derivative of ln(f(x))</h2>
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<h2>Important Glossaries for the Derivative of ln(f(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>A logarithm to the base e, denoted as ln(x).</li>
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</ul><ul><li><strong>Natural Logarithm:</strong>A logarithm to the base e, denoted as ln(x).</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for finding the derivative of the composition of two or more functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for finding the derivative of the composition of two or more 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>Higher-Order Derivatives:</strong>Derivatives of a function taken multiple times, such as the second derivative or third derivative.</li>
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</ul><ul><li><strong>Higher-Order Derivatives:</strong>Derivatives of a function taken multiple times, such as the second derivative or third derivative.</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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<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>