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2026-01-01
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2026-02-28
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<p>305 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 ln(1-x), which is -1/(1-x), as a tool for understanding how the natural logarithm function changes in response to a slight change in x. Derivatives help us calculate rates of change in real-life situations. We will now discuss the derivative of ln(1-x) in detail.</p>
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<h2>What is the Derivative of ln(1-x)?</h2>
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<p>We now understand the derivative of ln(1-x). It is commonly represented as d/dx (ln(1-x)) or (ln(1-x))', and its value is -1/(1-x). The<a>function</a>ln(1-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 natural logarithm of x). Chain Rule: Rule for differentiating ln(1-x) (since it involves a composition<a>of functions</a>).</p>
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<p>Reciprocal Function: A function of the form 1/x.</p>
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<h2>Derivative of ln(1-x) Formula</h2>
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<p>The derivative of ln(1-x) can be denoted as d/dx (ln(1-x)) or (ln(1-x))'. The<a>formula</a>we use to differentiate ln(1-x) is: d/dx (ln(1-x)) = -1/(1-x) The formula applies to all x where 1-x > 0, i.e., x < 1.</p>
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<h2>Proofs of the Derivative of ln(1-x)</h2>
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<p>We can derive the derivative of ln(1-x) using proofs. To show this, we will use the basic rules of differentiation along with the chain rule</p>
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<p>We can derive the derivative of ln(1-x) using proofs. To show this, we will use the basic rules of differentiation along with the chain rule</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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<p>By First Principle Using Chain Rule We will now demonstrate that the differentiation of ln(1-x) results in -1/(1-x) using the above-mentioned methods:</p>
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<p>By First Principle Using Chain Rule We will now demonstrate that the differentiation of ln(1-x) results in -1/(1-x) using the above-mentioned methods:</p>
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<p>By First Principle The derivative of ln(1-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>By First Principle The derivative of ln(1-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(1-x) using the first principle, we will consider f(x) = ln(1-x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = ln(1-x), we write f(x + h) = ln(1-(x + h)).</p>
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<p>To find the derivative of ln(1-x) using the first principle, we will consider f(x) = ln(1-x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = ln(1-x), we write f(x + h) = ln(1-(x + h)).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(1-(x + h)) - ln(1-x)] / h = limₕ→₀ ln[(1-(x + h))/(1-x)] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(1-(x + h)) - ln(1-x)] / h = limₕ→₀ ln[(1-(x + h))/(1-x)] / h</p>
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<p>Using the property of<a>logarithms</a>ln(a/b) = ln(a) - ln(b), f'(x) = limₕ→₀ ln[(1-x-h)/(1-x)] / h = limₕ→₀ ln[1 - h/(1-x)] / h</p>
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<p>Using the property of<a>logarithms</a>ln(a/b) = ln(a) - ln(b), f'(x) = limₕ→₀ ln[(1-x-h)/(1-x)] / h = limₕ→₀ ln[1 - h/(1-x)] / h</p>
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<p>Using the approximation ln(1-u) ≈ -u for small u, f'(x) = limₕ→₀ [-h/(1-x)] / h = -1/(1-x)</p>
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<p>Using the approximation ln(1-u) ≈ -u for small u, f'(x) = limₕ→₀ [-h/(1-x)] / h = -1/(1-x)</p>
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<p>Hence, proved. Using Chain Rule To prove the differentiation of ln(1-x) using the chain rule, Consider f(x) = 1-x, then ln(1-x) = ln(f(x)).</p>
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<p>Hence, proved. Using Chain Rule To prove the differentiation of ln(1-x) using the chain rule, Consider f(x) = 1-x, then ln(1-x) = ln(f(x)).</p>
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<p>By the chain rule: d/dx [ln(f(x))] = (1/f(x)) * f'(x)</p>
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<p>By the chain rule: d/dx [ln(f(x))] = (1/f(x)) * f'(x)</p>
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<p>Let’s substitute f(x) = 1-x, then f'(x) = -1. d/dx (ln(1-x)) = (1/(1-x)) * (-1) = -1/(1-x)</p>
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<p>Let’s substitute f(x) = 1-x, then f'(x) = -1. d/dx (ln(1-x)) = (1/(1-x)) * (-1) = -1/(1-x)</p>
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<h2>Higher-Order Derivatives of ln(1-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>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(1-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 nth Derivative of ln(1-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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<p>When x approaches 1, the derivative is undefined because ln(1-x) is not defined for x ≥ 1. When x is 0, the derivative of ln(1-x) = -1/(1-0), which is -1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(1-x)</h2>
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<p>Students frequently make mistakes when differentiating ln(1-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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<p>Calculate the derivative of ln(1-x)·e^x</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = ln(1-x)·e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = ln(1-x) and v = e^x.</p>
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<p>Let’s differentiate each term, u′ = d/dx (ln(1-x)) = -1/(1-x) v′ = d/dx (e^x) = e^x</p>
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<p>Substituting into the given equation, f'(x) = (-1/(1-x))·e^x + ln(1-x)·e^x</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = e^x[-1/(1-x) + ln(1-x)]</p>
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<p>Thus, the derivative of the specified function is e^x[-1/(1-x) + ln(1-x)].</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>The height of a balloon is measured by the function y = ln(1-x), where y represents the height at a distance x from the starting point. If x = 1/2 meters, measure the rate of descent of the balloon.</p>
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<p>Okay, lets begin</p>
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<p>We have y = ln(1-x) (height of the balloon)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative ln(1-x): dy/dx = -1/(1-x)</p>
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<p>Given x = 1/2 (substitute this into the derivative) dy/dx = -1/(1-1/2) dy/dx = -1/(1/2) = -2</p>
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<p>Hence, we get the rate of descent of the balloon at a distance x = 1/2 as -2.</p>
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<h3>Explanation</h3>
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<p>We find the rate of descent of the balloon at x = 1/2 as -2, which means that at a given point, the height of the balloon decreases at a rate of 2 units per unit distance.</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 = ln(1-x).</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/(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/(1-x)]</p>
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<p>Here we use the chain rule, d²y/dx² = (-1) * d/dx [1/(1-x)] = (-1) * (1/(1-x)²) * (-1) = 1/(1-x)²</p>
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<p>Therefore, the second derivative of the function y = ln(1-x) is 1/(1-x)².</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. Using the chain rule, we differentiate -1/(1-x). We then substitute 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/dx (ln²(1-x)) = -2ln(1-x)/(1-x).</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the chain rule: Consider y = ln²(1-x) = [ln(1-x)]²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2ln(1-x) * d/dx [ln(1-x)]</p>
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<p>Since the derivative of ln(1-x) is -1/(1-x), dy/dx = 2ln(1-x) * (-1/(1-x))</p>
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<p>Substituting y = ln²(1-x), d/dx (ln²(1-x)) = -2ln(1-x)/(1-x)</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. Then, we replace ln(1-x) with its derivative. As a final step, we substitute y = ln²(1-x) 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/dx (ln(1-x)/x)</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(1-x)/x) = (d/dx (ln(1-x)) * x - ln(1-x) * d/dx(x))/x²</p>
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<p>We will substitute d/dx (ln(1-x)) = -1/(1-x) and d/dx (x) = 1 = (-1/(1-x) * x - ln(1-x) * 1) / x² = (-x/(1-x) - ln(1-x)) / x²</p>
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<p>Therefore, d/dx (ln(1-x)/x) = (-x/(1-x) - ln(1-x)) / x²</p>
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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>Well explained 👍</p>
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<h2>FAQs on the Derivative of ln(1-x)</h2>
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<h3>1.Find the derivative of ln(1-x).</h3>
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<p>Using the chain rule for ln(1-x), d/dx (ln(1-x)) = -1/(1-x)</p>
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<h3>2.Can we use the derivative of ln(1-x) in real life?</h3>
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<p>Yes, we can use the derivative of ln(1-x) in real life in calculating the rate of change of 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(1-x) at the point where x = 1?</h3>
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<p>No, ln(1-x) is not defined for x ≥ 1, 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(1-x)/x?</h3>
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<p>We use the quotient rule to differentiate ln(1-x)/x, d/dx (ln(1-x)/x) = (-x/(1-x) - ln(1-x)) / x².</p>
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<h3>5.Are the derivatives of ln(1-x) and ln(x) the same?</h3>
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<p>No, they are different. The derivative of ln(1-x) is -1/(1-x), while the derivative of ln(x) is 1/x.</p>
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<h2>Important Glossaries for the Derivative of ln(1-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><ul><li><strong>Natural Logarithm:</strong>A logarithm with base e, commonly written as ln(x).</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus for finding the derivative of a composite function.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A method used to differentiate functions that are the quotient of two other functions.</li>
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</ul><ul><li><strong>Domain:</strong>The set of all possible input values (x-values) for which the function is defined.</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>