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2026-01-01
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2026-02-28
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<p>178 Learners</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 logn, which is 1/n, as a measuring tool for how the logarithmic function changes in response to a slight change in n. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of logn in detail.</p>
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<h2>What is the Derivative of logn?</h2>
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<p>We now understand the derivative<a>of</a>logn. It is commonly represented as d/dn (logn) or (logn)', and its value is 1/n. The<a>function</a>logn 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>Logarithmic Function: (logn = ln(n)).</p>
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<p>Power Rule: Rule for differentiating logn (since it consists of ln(n)).</p>
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<p>Reciprocal Function: 1/n.</p>
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<h2>Derivative of logn Formula</h2>
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<p>The derivative of logn can be denoted as d/dn (logn) or (logn)'. The<a>formula</a>we use to differentiate logn is: d/dn (logn) = 1/n (or) (logn)' = 1/n The formula applies to all n where n > 0</p>
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<h2>Proofs of the Derivative of logn</h2>
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<p>We can derive the derivative of logn using proofs. To show this, we will use the logarithmic identities along with the rules of differentiation.</p>
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<p>We can derive the derivative of logn using proofs. To show this, we will use the logarithmic identities along with the rules of differentiation.</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>There are several methods we use to prove this, such as:</p>
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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Chain Rule</li>
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<li>Using Chain Rule</li>
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<li>Using Power Rule</li>
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<li>Using Power Rule</li>
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</ol><p>We will now demonstrate that the differentiation of logn results in 1/n using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of logn results in 1/n 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 logn 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 logn using the first principle, we will consider f(n) = logn.</p>
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<p>The derivative of logn 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 logn using the first principle, we will consider f(n) = logn.</p>
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<p>Its derivative can be expressed as the following limit. f'(n) = limₕ→₀ [f(n + h) - f(n)] / h … (1)</p>
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<p>Its derivative can be expressed as the following limit. f'(n) = limₕ→₀ [f(n + h) - f(n)] / h … (1)</p>
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<p>Given that f(n) = logn, we write f(n + h) =<a>log</a>(n + h).</p>
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<p>Given that f(n) = logn, we write f(n + h) =<a>log</a>(n + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(n) = limₕ→₀ [log(n + h) - logn] / h = limₕ→₀ [ln((n + h)/n)] / h = limₕ→₀ [ln(1 + h/n)] / h Using the formula ln(1 + u) ≈ u for small u, f'(n) = limₕ→₀ (h/n) / h = 1/n</p>
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<p>Substituting these into<a>equation</a>(1), f'(n) = limₕ→₀ [log(n + h) - logn] / h = limₕ→₀ [ln((n + h)/n)] / h = limₕ→₀ [ln(1 + h/n)] / h Using the formula ln(1 + u) ≈ u for small u, f'(n) = limₕ→₀ (h/n) / h = 1/n</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 logn using the chain rule, We use the formula: logn = ln(n) Consider g(n) = ln(n)</p>
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<p>To prove the differentiation of logn using the chain rule, We use the formula: logn = ln(n) Consider g(n) = ln(n)</p>
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<p>The derivative is 1/n Thus: d/dn (logn) = 1/n</p>
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<p>The derivative is 1/n Thus: d/dn (logn) = 1/n</p>
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<h3>Using Power Rule</h3>
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<h3>Using Power Rule</h3>
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<p>We will now prove the derivative of logn using the<a>power</a>rule. The step-by-step process is demonstrated below:</p>
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<p>We will now prove the derivative of logn using the<a>power</a>rule. The step-by-step process is demonstrated below:</p>
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<p>Here, we use the formula, logn = ln(n)</p>
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<p>Here, we use the formula, logn = ln(n)</p>
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<p>Given that, u = n</p>
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<p>Given that, u = n</p>
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<p>Using the power rule formula: d/dn (u^1) = 1 * u^(1-1) 1/n Thus: d/dn (logn) = 1/n</p>
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<p>Using the power rule formula: d/dn (u^1) = 1 * u^(1-1) 1/n Thus: d/dn (logn) = 1/n</p>
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<h2>Higher-Order Derivatives of logn</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>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 logn.</p>
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<p>For the first derivative of a function, we write f′(n), 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′′(n) Similarly, the third derivative, f′′′(n) is the result of the second derivative and this pattern continues.</p>
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<p>For the nth Derivative of logn, we generally use fⁿ(n) for the nth derivative of a function f(n) 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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<p>When n is 0, the derivative is undefined because logn is undefined at that point. When n is 1, the derivative of logn = 1/1, which is 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of logn</h2>
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<p>Students frequently make mistakes when differentiating logn. 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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<p>Calculate the derivative of (logn·n^2)</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(n) = logn·n². Using the product rule, f'(n) = u′v + uv′ In the given equation, u = logn and v = n². Let’s differentiate each term, u′= d/dn (logn) = 1/n v′= d/dn (n²) = 2n</p>
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<p>substituting into the given equation, f'(n) = (1/n) * n² + logn * 2n</p>
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<p>Let’s simplify terms to get the final answer, f'(n) = n + 2n logn</p>
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<p>Thus, the derivative of the specified function is n + 2n logn.</p>
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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>Well explained 👍</p>
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<h3>Problem 2</h3>
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<p>A company wants to optimize its production cost. The cost is represented by the function C = log(x) where C represents the cost at the production level x. If x = 100 units, measure the rate of change of cost.</p>
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<p>Okay, lets begin</p>
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<p>We have C = log(x) (rate of change of cost)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative log(x): dC/dx = 1/x</p>
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<p>Given x = 100 (substitute this into the derivative) dC/dx = 1/100</p>
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<p>Hence, we get the rate of change of cost at a production level x = 100 as 0.01.</p>
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<h3>Explanation</h3>
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<p>We find the rate of change of cost at x = 100 as 0.01, which means that at a given point, the cost would increase at a rate of 0.01 for each additional unit of production.</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = logn.</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/dn = 1/n...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dn² = d/dn [1/n]</p>
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<p>Here we use the power rule, d²y/dn² = -1/n²</p>
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<p>Therefore, the second derivative of the function y = logn is -1/n².</p>
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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 apply the power rule to differentiate 1/n. We 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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<h3>Problem 4</h3>
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<p>Prove: d/dn (logn^2) = 2 logn/n + 2/n.</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the chain rule: Consider y = logn^2 [logn^2 = ln(n^2)]</p>
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<p>To differentiate, we use the chain rule: dy/dn = d/dn [2 ln(n)]</p>
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<p>Since the derivative of ln(n) is 1/n, dy/dn = 2 * (1/n) + 2 logn * (1/n)</p>
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<p>Substituting y = logn^2, d/dn (logn^2) = 2 logn/n + 2/n</p>
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<p>Hence proved.</p>
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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>Then, we replace ln(n) with its derivative. As a final step, we substitute y = logn^2 to derive the equation.</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<p>Solve: d/dn (logn/n)</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/dn (logn/n) = (d/dn (logn) * n - logn * d/dn(n)) / n²</p>
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<p>We will substitute d/dn (logn) = 1/n and d/dn (n) = 1 = (1/n * n - logn * 1) / n² = (1 - logn) / n²</p>
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<p>Therefore, d/dn (logn/n) = (1 - logn) / n²</p>
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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>Well explained 👍</p>
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<h2>FAQs on the Derivative of logn</h2>
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<h3>1.Find the derivative of logn.</h3>
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<p>Using the power rule on logn gives 1/n, d/dn (logn) = 1/n (simplified)</p>
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<h3>2.Can we use the derivative of logn in real life?</h3>
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<p>Yes, we can use the derivative of logn in real life in calculating growth rates, especially in fields such as mathematics, biology, and economics.</p>
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<h3>3.Is it possible to take the derivative of logn at the point where n = 0?</h3>
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<p>No, n = 0 is a point where logn 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 logn/n?</h3>
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<p>We use the quotient rule to differentiate logn/n, d/dn (logn/n) = (1/n * n - logn * 1) / n².</p>
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<h3>5.Are the derivatives of logn and ln(n) the same?</h3>
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<p>Yes, they are the same. The derivative of logn is equal to 1/n, which is the same as the derivative of ln(n).</p>
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<h3>6.Can we find the derivative of the logn formula?</h3>
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<p>To find, consider y = logn. We use the power rule: y’ = d/dn (ln(n)) = 1/n.</p>
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<h2>Important Glossaries for the Derivative of logn</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 n.</li>
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</ul><ul><li><strong>Logarithmic Function:</strong>The logarithmic function is written as logn or ln(n).</li>
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</ul><ul><li><strong>Reciprocal Function:</strong>A function that is the reciprocal of n, represented as 1/n.</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 condition where a function does not exist at certain points, such as logn at n = 0.</li>
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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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<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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<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>