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2026-01-01
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<p>Last updated on<strong>December 15, 2025</strong></p>
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<p>Last updated on<strong>December 15, 2025</strong></p>
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<p>We use the derivative of 1/(1-x)² to understand how this function changes in response to small changes in x. Derivatives are crucial in many fields, such as physics and economics, for determining rates of change. We will now discuss the derivative of 1/(1-x)² in detail.</p>
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<p>We use the derivative of 1/(1-x)² to understand how this function changes in response to small changes in x. Derivatives are crucial in many fields, such as physics and economics, for determining rates of change. We will now discuss the derivative of 1/(1-x)² in detail.</p>
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<h2>What is the Derivative of 1/(1-x)²?</h2>
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<h2>What is the Derivative of 1/(1-x)²?</h2>
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<p>The derivative<a>of</a>the<a>function</a>1/(1-x)² is found using basic differentiation techniques.</p>
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<p>The derivative<a>of</a>the<a>function</a>1/(1-x)² is found using basic differentiation techniques.</p>
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<p>It is represented as d/dx (1/(1-x)²) or (1/(1-x)²)'.</p>
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<p>It is represented as d/dx (1/(1-x)²) or (1/(1-x)²)'.</p>
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<p>The function is differentiable within its domain (x ≠ 1).</p>
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<p>The function is differentiable within its domain (x ≠ 1).</p>
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<p>The key concepts include:</p>
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<p>The key concepts include:</p>
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<p><strong>Power Rule</strong>: A basic rule for differentiating functions of the form x^n.</p>
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<p><strong>Power Rule</strong>: A basic rule for differentiating functions of the form x^n.</p>
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<p><strong>Chain Rule</strong>: A rule used when differentiating a composition of functions.</p>
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<p><strong>Chain Rule</strong>: A rule used when differentiating a composition of functions.</p>
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<p><strong>Negative Exponent</strong>: Expressing the function as (1-x)^-2 for differentiation.</p>
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<p><strong>Negative Exponent</strong>: Expressing the function as (1-x)^-2 for differentiation.</p>
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<h2>Derivative of 1/(1-x)² Formula</h2>
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<h2>Derivative of 1/(1-x)² Formula</h2>
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<p>The derivative of 1/(1-x)², denoted as d/dx (1/(1-x)²) or (1/(1-x)²)', can be determined using the chain rule and<a>power</a>rule: d/dx (1/(1-x)²) = 2/(1-x)³.</p>
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<p>The derivative of 1/(1-x)², denoted as d/dx (1/(1-x)²) or (1/(1-x)²)', can be determined using the chain rule and<a>power</a>rule: d/dx (1/(1-x)²) = 2/(1-x)³.</p>
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<p>This<a>formula</a>is valid for all x, where x ≠ 1.</p>
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<p>This<a>formula</a>is valid for all x, where x ≠ 1.</p>
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<h2>Proofs of the Derivative of 1/(1-x)²</h2>
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<h2>Proofs of the Derivative of 1/(1-x)²</h2>
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<p>We can derive the derivative of 1/(1-x)² using proofs.</p>
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<p>We can derive the derivative of 1/(1-x)² using proofs.</p>
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<p>We'll use the chain rule, considering the function as a composite function.</p>
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<p>We'll use the chain rule, considering the function as a composite function.</p>
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<p>Here are the methods: Using the Chain Rule.</p>
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<p>Here are the methods: Using the Chain Rule.</p>
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<p>To prove the differentiation of 1/(1-x)² using the chain rule: Express the function as (1-x)^-2.</p>
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<p>To prove the differentiation of 1/(1-x)² using the chain rule: Express the function as (1-x)^-2.</p>
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<p>Consider u(x) = (1-x), then f(u) = u^-2.</p>
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<p>Consider u(x) = (1-x), then f(u) = u^-2.</p>
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<p>By the chain rule: d/dx [f(u(x))] = f'(u) * u'(x).</p>
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<p>By the chain rule: d/dx [f(u(x))] = f'(u) * u'(x).</p>
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<p>Differentiate f(u) = u^-2: f'(u) = -2u^-3.</p>
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<p>Differentiate f(u) = u^-2: f'(u) = -2u^-3.</p>
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<p>Differentiate u(x) = (1-x): u'(x) = -1.</p>
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<p>Differentiate u(x) = (1-x): u'(x) = -1.</p>
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<p>Substitute these into the chain rule: d/dx [(1-x)^-2] = -2(1-x)^-3 * (-1) = 2/(1-x)³.</p>
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<p>Substitute these into the chain rule: d/dx [(1-x)^-2] = -2(1-x)^-3 * (-1) = 2/(1-x)³.</p>
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<p>Hence, the derivative is 2/(1-x)³.</p>
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<p>Hence, the derivative is 2/(1-x)³.</p>
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<h2>Higher-Order Derivatives of 1/(1-x)²</h2>
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<h2>Higher-Order Derivatives of 1/(1-x)²</h2>
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<p>When a function is differentiated several times, the resulting derivatives are called higher-order derivatives.</p>
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<p>When a function is differentiated several times, the resulting derivatives are called higher-order derivatives.</p>
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<p>Higher-order derivatives can be more complex to handle.</p>
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<p>Higher-order derivatives can be more complex to handle.</p>
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<p>For example, the second derivative of 1/(1-x)² is concerned with the<a>rate</a>of change of the rate of change.</p>
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<p>For example, the second derivative of 1/(1-x)² is concerned with the<a>rate</a>of change of the rate of change.</p>
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<p>For the first derivative, we have f′(x), which shows the slope or rate of change of the function.</p>
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<p>For the first derivative, we have f′(x), which shows the slope or rate of change of the function.</p>
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<p>The second derivative is derived from the first derivative, indicated by f′′(x).</p>
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<p>The second derivative is derived from the first derivative, indicated by f′′(x).</p>
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<p>In general, for the nth derivative of a function f(x), we use f n(x) to denote the nth derivative, which helps in analyzing the behavior of the function further.</p>
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<p>In general, for the nth derivative of a function f(x), we use f n(x) to denote the nth derivative, which helps in analyzing the behavior of the function further.</p>
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<h2>Special Cases</h2>
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<h2>Special Cases</h2>
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<p>When x approaches 1, the derivative becomes undefined because 1/(1-x)² has a vertical asymptote there.</p>
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<p>When x approaches 1, the derivative becomes undefined because 1/(1-x)² has a vertical asymptote there.</p>
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<p>When x is 0, the derivative of 1/(1-x)² is 2, since 1/(1-0)² = 1.</p>
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<p>When x is 0, the derivative of 1/(1-x)² is 2, since 1/(1-0)² = 1.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 1/(1-x)²</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 1/(1-x)²</h2>
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<p>Students often make mistakes when differentiating 1/(1-x)².</p>
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<p>Students often make mistakes when differentiating 1/(1-x)².</p>
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<p>Understanding the correct methods can help avoid these errors.</p>
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<p>Understanding the correct methods can help avoid these errors.</p>
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<p>Here are a few common mistakes and solutions:</p>
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<p>Here are a few common mistakes and solutions:</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 (1/(1-x)²)·(1-x).</p>
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<p>Calculate the derivative of (1/(1-x)²)·(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>Here, we have f(x) = (1/(1-x)²)·(1-x).</p>
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<p>Here, we have f(x) = (1/(1-x)²)·(1-x).</p>
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<p>Using the product rule, f'(x) = u′v + uv′.</p>
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<p>Using the product rule, f'(x) = u′v + uv′.</p>
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<p>In the given equation, u = 1/(1-x)² and v = (1-x).</p>
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<p>In the given equation, u = 1/(1-x)² and v = (1-x).</p>
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<p>Differentiate each term: u′ = d/dx (1/(1-x)²) = 2/(1-x)³.</p>
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<p>Differentiate each term: u′ = d/dx (1/(1-x)²) = 2/(1-x)³.</p>
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<p>v′ = d/dx (1-x) = -1.</p>
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<p>v′ = d/dx (1-x) = -1.</p>
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<p>Substituting into the product rule: f'(x) = (2/(1-x)³)·(1-x) + (1/(1-x)²)·(-1).</p>
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<p>Substituting into the product rule: f'(x) = (2/(1-x)³)·(1-x) + (1/(1-x)²)·(-1).</p>
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<p>Simplify to get: f'(x) = 2/(1-x)² - 1/(1-x)² = 1/(1-x)².</p>
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<p>Simplify to get: f'(x) = 2/(1-x)² - 1/(1-x)² = 1/(1-x)².</p>
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<p>Thus, the derivative of the specified function is 1/(1-x)².</p>
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<p>Thus, the derivative of the specified function is 1/(1-x)².</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 expressing it in two parts.</p>
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<p>We find the derivative of the given function by expressing it in two parts.</p>
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<p>We differentiate each part and then use the product rule to combine them for the final result.</p>
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<p>We differentiate each part and then use the product rule to combine them for 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 modeling a process where the rate of change is given by y = 1/(1-x)². If x = 0.5, find the rate of change.</p>
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<p>A company is modeling a process where the rate of change is given by y = 1/(1-x)². If x = 0.5, find the rate of change.</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 = 1/(1-x)² (rate of change)...(1) Differentiate equation (1) dy/dx = 2/(1-x)³.</p>
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<p>We have y = 1/(1-x)² (rate of change)...(1) Differentiate equation (1) dy/dx = 2/(1-x)³.</p>
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<p>Substitute x = 0.5 into the derivative: dy/dx = 2/(1-0.5)³ = 2/0.5³ = 2/(0.125) = 16.</p>
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<p>Substitute x = 0.5 into the derivative: dy/dx = 2/(1-0.5)³ = 2/0.5³ = 2/(0.125) = 16.</p>
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<p>Hence, the rate of change at x = 0.5 is 16.</p>
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<p>Hence, the rate of change at x = 0.5 is 16.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>By substituting x = 0.5 into the derivative, we determine the rate of change of the process at that specific point, showing how sensitive the process is around that value.</p>
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<p>By substituting x = 0.5 into the derivative, we determine the rate of change of the process at that specific point, showing how sensitive the process is around that value.</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 = 1/(1-x)².</p>
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<p>Derive the second derivative of the function y = 1/(1-x)².</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>First, find the first derivative: dy/dx = 2/(1-x)³...(1) Differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2/(1-x)³] Apply the chain rule: d²y/dx² = -6/(1-x)⁴ * (-1) = 6/(1-x)⁴.</p>
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<p>First, find the first derivative: dy/dx = 2/(1-x)³...(1) Differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2/(1-x)³] Apply the chain rule: d²y/dx² = -6/(1-x)⁴ * (-1) = 6/(1-x)⁴.</p>
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<p>Therefore, the second derivative of the function y = 1/(1-x)² is 6/(1-x)⁴.</p>
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<p>Therefore, the second derivative of the function y = 1/(1-x)² is 6/(1-x)⁴.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We start by finding the first derivative and apply the chain rule again to find the second derivative, showing the acceleration or curvature of the function.</p>
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<p>We start by finding the first derivative and apply the chain rule again to find the second derivative, showing the acceleration or curvature of the function.</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 ((1-x)⁻³) = 3(1-x)⁻⁴.</p>
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<p>Prove: d/dx ((1-x)⁻³) = 3(1-x)⁻⁴.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Start using the chain rule: Consider y = (1-x)⁻³.</p>
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<p>Start using the chain rule: Consider y = (1-x)⁻³.</p>
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<p>Differentiate using the chain rule: dy/dx = -3(1-x)⁻⁴ * (-1).</p>
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<p>Differentiate using the chain rule: dy/dx = -3(1-x)⁻⁴ * (-1).</p>
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<p>Simplify: dy/dx = 3(1-x)⁻⁴.</p>
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<p>Simplify: dy/dx = 3(1-x)⁻⁴.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We used the chain rule to differentiate the function, simplifying by multiplying by the derivative of the inner function and verifying the result.</p>
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<p>We used the chain rule to differentiate the function, simplifying by multiplying by the derivative of the inner function and verifying the result.</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 (1/(1-x)² + x).</p>
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<p>Solve: d/dx (1/(1-x)² + x).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, differentiate each term separately: d/dx (1/(1-x)²) = 2/(1-x)³. d/dx (x) = 1.</p>
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<p>To differentiate the function, differentiate each term separately: d/dx (1/(1-x)²) = 2/(1-x)³. d/dx (x) = 1.</p>
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<p>Combine the derivatives: d/dx (1/(1-x)² + x) = 2/(1-x)³ + 1.</p>
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<p>Combine the derivatives: d/dx (1/(1-x)² + x) = 2/(1-x)³ + 1.</p>
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<p>Therefore, the derivative is 2/(1-x)³ + 1.</p>
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<p>Therefore, the derivative is 2/(1-x)³ + 1.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We differentiate each term of the function individually and combine the results for the final derivative.</p>
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<p>We differentiate each term of the function individually and combine the results for the final derivative.</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 1/(1-x)²</h2>
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<h2>FAQs on the Derivative of 1/(1-x)²</h2>
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<h3>1.Find the derivative of 1/(1-x)².</h3>
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<h3>1.Find the derivative of 1/(1-x)².</h3>
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<p>Using the chain rule on 1/(1-x)², expressed as (1-x)⁻², gives: d/dx (1/(1-x)²) = 2/(1-x)³.</p>
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<p>Using the chain rule on 1/(1-x)², expressed as (1-x)⁻², gives: d/dx (1/(1-x)²) = 2/(1-x)³.</p>
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<h3>2.Can the derivative of 1/(1-x)² be used in real life?</h3>
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<h3>2.Can the derivative of 1/(1-x)² be used in real life?</h3>
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<p>Yes, it can be applied in fields like physics and engineering to model processes with rates of change related to geometric or physical constraints.</p>
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<p>Yes, it can be applied in fields like physics and engineering to model processes with rates of change related to geometric or physical constraints.</p>
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<h3>3.Is it possible to take the derivative of 1/(1-x)² at x = 1?</h3>
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<h3>3.Is it possible to take the derivative of 1/(1-x)² at x = 1?</h3>
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<p>No, at x = 1, the function 1/(1-x)² is undefined due to<a>division by zero</a>, making the derivative impossible to compute at this point.</p>
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<p>No, at x = 1, the function 1/(1-x)² is undefined due to<a>division by zero</a>, making the derivative impossible to compute at this point.</p>
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<h3>4.What rule is used to differentiate (1-x)^-2?</h3>
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<h3>4.What rule is used to differentiate (1-x)^-2?</h3>
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<p>The chain rule is used to differentiate (1-x)^-2, considering it as a composition of functions.</p>
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<p>The chain rule is used to differentiate (1-x)^-2, considering it as a composition of functions.</p>
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<h3>5.Are the derivatives of 1/(1-x)² and (1-x)⁻² the same?</h3>
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<h3>5.Are the derivatives of 1/(1-x)² and (1-x)⁻² the same?</h3>
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<p>Yes, they are the same. The function 1/(1-x)² is equivalent to (1-x)⁻², and their derivatives are both 2/(1-x)³.</p>
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<p>Yes, they are the same. The function 1/(1-x)² is equivalent to (1-x)⁻², and their derivatives are both 2/(1-x)³.</p>
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<h2>Important Glossaries for the Derivative of 1/(1-x)²</h2>
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<h2>Important Glossaries for the Derivative of 1/(1-x)²</h2>
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<ul><li><strong>Derivative</strong>: The derivative of a function measures how the function changes as its input changes.</li>
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<ul><li><strong>Derivative</strong>: The derivative of a function measures how the function changes as its input changes.</li>
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</ul><ul><li><strong>Chain Rule</strong>: A rule for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule</strong>: A rule for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Power Rule</strong>: A basic rule used to find the derivative of x raised to a power.</li>
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</ul><ul><li><strong>Power Rule</strong>: A basic rule used to find the derivative of x raised to a power.</li>
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</ul><ul><li><strong>Negative Exponent</strong>: Expressing a function with a negative exponent for simplification in differentiation.</li>
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</ul><ul><li><strong>Negative Exponent</strong>: Expressing a function with a negative exponent for simplification in differentiation.</li>
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</ul><ul><li><strong>Asymptote</strong>: A line that a graph approaches but never touches.</li>
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</ul><ul><li><strong>Asymptote</strong>: A line that a graph approaches but never touches.</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>