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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 1/2x², which is x, as a measuring tool for how this quadratic function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 1/2x² in detail.</p>
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<p>We use the derivative of 1/2x², which is x, as a measuring tool for how this quadratic function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 1/2x² in detail.</p>
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<h2>What is the Derivative of 1/2x²?</h2>
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<h2>What is the Derivative of 1/2x²?</h2>
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<p>We now understand the derivative<a>of</a>1/2x².</p>
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<p>We now understand the derivative<a>of</a>1/2x².</p>
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<p>It is commonly represented as d/dx (1/2x²) or (1/2x²)', and its value is x.</p>
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<p>It is commonly represented as d/dx (1/2x²) or (1/2x²)', and its value is x.</p>
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<p>The<a>function</a>1/2x² has a clearly defined derivative, indicating it is differentiable within its domain.</p>
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<p>The<a>function</a>1/2x² 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>The key concepts are mentioned below:</p>
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<p><strong>Quadratic Function</strong>: (1/2x² is a simple quadratic function).</p>
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<p><strong>Quadratic Function</strong>: (1/2x² is a simple quadratic function).</p>
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<p><strong>Power Rule</strong>: Rule for differentiating 1/2x².</p>
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<p><strong>Power Rule</strong>: Rule for differentiating 1/2x².</p>
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<h2>Derivative of 1/2x² Formula</h2>
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<h2>Derivative of 1/2x² Formula</h2>
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<p>The derivative of 1/2x² can be denoted as d/dx (1/2x²) or (1/2x²)'.</p>
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<p>The derivative of 1/2x² can be denoted as d/dx (1/2x²) or (1/2x²)'.</p>
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<p>The<a>formula</a>we use to differentiate 1/2x² is: d/dx (1/2x²) = x The formula applies to all x.</p>
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<p>The<a>formula</a>we use to differentiate 1/2x² is: d/dx (1/2x²) = x The formula applies to all x.</p>
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<h2>Proofs of the Derivative of 1/2x²</h2>
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<h2>Proofs of the Derivative of 1/2x²</h2>
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<p>We can derive the derivative of 1/2x² using proofs. To show this, we will use the rules of differentiation.</p>
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<p>We can derive the derivative of 1/2x² using proofs. To show this, we will use the rules of differentiation.</p>
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<p>There are several methods we use to prove this, such as: By First Principle Using Power Rule We will now demonstrate that the differentiation of 1/2x² results in x using the above-mentioned methods: By First Principle The derivative of 1/2x² 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>There are several methods we use to prove this, such as: By First Principle Using Power Rule We will now demonstrate that the differentiation of 1/2x² results in x using the above-mentioned methods: By First Principle The derivative of 1/2x² 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 1/2x² using the first principle, we will consider f(x) = 1/2x².</p>
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<p>To find the derivative of 1/2x² using the first principle, we will consider f(x) = 1/2x².</p>
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<p>Its derivative can be expressed as the following limit.</p>
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<p>Its derivative can be expressed as the following limit.</p>
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<p>f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = 1/2x², we write f(x + h) = 1/2(x + h)².</p>
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<p>Given that f(x) = 1/2x², we write f(x + h) = 1/2(x + h)².</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/2(x + h)² - 1/2x²] / h = limₕ→₀ [1/2(x² + 2xh + h²) - 1/2x²] / h = limₕ→₀ [1/2(2xh + h²)] / h = limₕ→₀ [xh + 1/2h²] / h = limₕ→₀ [x + 1/2h] As h approaches 0, the second<a>term</a>vanishes, so we have, f'(x) = x. Hence, proved.</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/2(x + h)² - 1/2x²] / h = limₕ→₀ [1/2(x² + 2xh + h²) - 1/2x²] / h = limₕ→₀ [1/2(2xh + h²)] / h = limₕ→₀ [xh + 1/2h²] / h = limₕ→₀ [x + 1/2h] As h approaches 0, the second<a>term</a>vanishes, so we have, f'(x) = x. Hence, proved.</p>
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<p>Using Power Rule To prove the differentiation of 1/2x² using the<a>power</a>rule, We use the formula: d/dx [xⁿ] = n*xⁿ⁻¹ Let n = 2 and a<a>constant</a><a>multiplier</a>of 1/2, d/dx (1/2x²) = 1/2 * d/dx (x²) = 1/2 * 2x = x Thus, d/dx (1/2x²) = x.</p>
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<p>Using Power Rule To prove the differentiation of 1/2x² using the<a>power</a>rule, We use the formula: d/dx [xⁿ] = n*xⁿ⁻¹ Let n = 2 and a<a>constant</a><a>multiplier</a>of 1/2, d/dx (1/2x²) = 1/2 * d/dx (x²) = 1/2 * 2x = x Thus, d/dx (1/2x²) = x.</p>
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<h2>Higher-Order Derivatives of 1/2x²</h2>
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<h2>Higher-Order Derivatives of 1/2x²</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives.</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives.</p>
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<p>Higher-order derivatives can be a little tricky.</p>
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<p>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.</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.</p>
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<p>Higher-order derivatives make it easier to understand functions like 1/2x².</p>
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<p>Higher-order derivatives make it easier to understand functions like 1/2x².</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.</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.</p>
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<p>The second derivative is derived from the first derivative, which is denoted using f′′(x).</p>
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<p>The second derivative is derived from the first derivative, which is denoted using f′′(x).</p>
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<p>Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.</p>
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<p>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 1/2x², we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change. (continuing for higher-order derivatives).</p>
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<p>For the nth derivative of 1/2x², we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change. (continuing for higher-order derivatives).</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>In quadratic functions like 1/2x², the higher-order derivatives eventually become zero.</p>
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<p>In quadratic functions like 1/2x², the higher-order derivatives eventually become zero.</p>
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<p>The second derivative of 1/2x² is 1, which represents a constant rate of change in the slope.</p>
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<p>The second derivative of 1/2x² is 1, which represents a constant rate of change in the slope.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 1/2x²</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 1/2x²</h2>
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<p>Students frequently make mistakes when differentiating 1/2x².</p>
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<p>Students frequently make mistakes when differentiating 1/2x².</p>
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<p>These mistakes can be resolved by understanding the proper solutions.</p>
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<p>These mistakes can be resolved by understanding the proper solutions.</p>
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<p>Here are a few common mistakes and ways to solve them:</p>
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<p>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 (1/2x²·x).</p>
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<p>Calculate the derivative of (1/2x²·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) = 1/2x²·x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 1/2x² and v = x.</p>
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<p>Here, we have f(x) = 1/2x²·x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 1/2x² and v = x.</p>
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<p>Let’s differentiate each term, u′ = d/dx (1/2x²) = x v′ = d/dx (x) = 1</p>
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<p>Let’s differentiate each term, u′ = d/dx (1/2x²) = x v′ = d/dx (x) = 1</p>
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<p>Substituting into the given equation, f'(x) = (x)(x) + (1/2x²)(1)</p>
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<p>Substituting into the given equation, f'(x) = (x)(x) + (1/2x²)(1)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = x² + 1/2x²</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = x² + 1/2x²</p>
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<p>Thus, the derivative of the specified function is x² + 1/2x².</p>
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<p>Thus, the derivative of the specified function is x² + 1/2x².</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.</p>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>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>The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>A company calculates the area of a square plot using the formula A = 1/2x², where A represents the area in square meters, and x is the side length of the plot in meters. If x = 4 meters, find the rate of change of the area with respect to the side length.</p>
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<p>A company calculates the area of a square plot using the formula A = 1/2x², where A represents the area in square meters, and x is the side length of the plot in meters. If x = 4 meters, find the rate of change of the area with respect to the side length.</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 A = 1/2x² (area of the plot)...(1) Now, we will differentiate the equation (1) Take the derivative 1/2x²: dA/dx = x Given x = 4 (substitute this into the derivative), dA/dx = 4 Hence, we get the rate of change of the area with respect to the side length as 4 square meters per meter.</p>
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<p>We have A = 1/2x² (area of the plot)...(1) Now, we will differentiate the equation (1) Take the derivative 1/2x²: dA/dx = x Given x = 4 (substitute this into the derivative), dA/dx = 4 Hence, we get the rate of change of the area with respect to the side length as 4 square meters per meter.</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 the area with respect to the side length x= 4 as 4, which means that at a given point, the area of the plot would increase at a rate of 4 square meters per meter of increase in side length.</p>
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<p>We find the rate of change of the area with respect to the side length x= 4 as 4, which means that at a given point, the area of the plot would increase at a rate of 4 square meters per meter of increase in side length.</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 A = 1/2x².</p>
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<p>Derive the second derivative of the function A = 1/2x².</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, dA/dx = x...(1) Now we will differentiate equation (1) to get the second derivative: d²A/dx² = d/dx [x] = 1</p>
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<p>The first step is to find the first derivative, dA/dx = x...(1) Now we will differentiate equation (1) to get the second derivative: d²A/dx² = d/dx [x] = 1</p>
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<p>Therefore, the second derivative of the function A = 1/2x² is 1.</p>
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<p>Therefore, the second derivative of the function A = 1/2x² is 1.</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>We then differentiate x to find the second derivative, which is a constant 1, representing the constant rate of change in the rate of change.</p>
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<p>We then differentiate x to find the second derivative, which is a constant 1, representing the constant rate of change in the rate of change.</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 ((1/2x²)²) = 2x(x²).</p>
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<p>Prove: d/dx ((1/2x²)²) = 2x(x²).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the chain rule: Consider A = (1/2x²)² = (1/2)²(x²)²</p>
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<p>Let’s start using the chain rule: Consider A = (1/2x²)² = (1/2)²(x²)²</p>
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<p>To differentiate, we use the chain rule: dA/dx = 2(1/2)²(x²)(d/dx [x²]) Since d/dx (x²) = 2x, dA/dx = (1/2)(2x)(x²) = x(x²)</p>
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<p>To differentiate, we use the chain rule: dA/dx = 2(1/2)²(x²)(d/dx [x²]) Since d/dx (x²) = 2x, dA/dx = (1/2)(2x)(x²) = x(x²)</p>
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<p>Substituting A = (1/2x²)², d/dx ((1/2x²)²) = 2x(x²) Hence proved.</p>
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<p>Substituting A = (1/2x²)², d/dx ((1/2x²)²) = 2x(x²) 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 x² with its derivative.</p>
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<p>Then, we replace x² with its derivative.</p>
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<p>As a final step, we substitute A = (1/2x²)² to derive the equation.</p>
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<p>As a final step, we substitute A = (1/2x²)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 (1/2x²/x)</p>
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<p>Solve: d/dx (1/2x²/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 (1/2x²/x) = (d/dx (1/2x²)·x - 1/2x²·d/dx(x))/ x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (1/2x²/x) = (d/dx (1/2x²)·x - 1/2x²·d/dx(x))/ x²</p>
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<p>We will substitute d/dx (1/2x²) = x and d/dx (x) = 1 = (x·x - 1/2x²·1) / x² = (x² - 1/2x²) / x² = (1/2x²) / x² = 1/2.</p>
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<p>We will substitute d/dx (1/2x²) = x and d/dx (x) = 1 = (x·x - 1/2x²·1) / x² = (x² - 1/2x²) / x² = (1/2x²) / x² = 1/2.</p>
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<p>Therefore, d/dx (1/2x²/x) = 1/2</p>
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<p>Therefore, d/dx (1/2x²/x) = 1/2</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 1/2x²</h2>
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<h2>FAQs on the Derivative of 1/2x²</h2>
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<h3>1.Find the derivative of 1/2x².</h3>
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<h3>1.Find the derivative of 1/2x².</h3>
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<p>Using the power rule on 1/2x² gives d/dx (1/2x²) = x.</p>
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<p>Using the power rule on 1/2x² gives d/dx (1/2x²) = x.</p>
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<h3>2.Can we use the derivative of 1/2x² in real life?</h3>
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<h3>2.Can we use the derivative of 1/2x² in real life?</h3>
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<p>Yes, we can use the derivative of 1/2x² in real life in calculating the rate of change of area or other quadratic relationships in fields such as physics and engineering.</p>
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<p>Yes, we can use the derivative of 1/2x² in real life in calculating the rate of change of area or other quadratic relationships in fields such as physics and engineering.</p>
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<h3>3.Is it possible to take the derivative of 1/2x² at any point?</h3>
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<h3>3.Is it possible to take the derivative of 1/2x² at any point?</h3>
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<p>Yes, the derivative of 1/2x² is defined for all x.</p>
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<p>Yes, the derivative of 1/2x² is defined for all x.</p>
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<h3>4.What rule is used to differentiate 1/2x²/x?</h3>
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<h3>4.What rule is used to differentiate 1/2x²/x?</h3>
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<p>We use the quotient rule to differentiate 1/2x²/x, d/dx (1/2x²/x) = (x·x - 1/2x²·1) / x² = 1/2.</p>
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<p>We use the quotient rule to differentiate 1/2x²/x, d/dx (1/2x²/x) = (x·x - 1/2x²·1) / x² = 1/2.</p>
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<h3>5.Are the derivatives of 1/2x² and (x²)/2 the same?</h3>
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<h3>5.Are the derivatives of 1/2x² and (x²)/2 the same?</h3>
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<p>Yes, they are the same. Both represent the same function, and their derivative is x.</p>
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<p>Yes, they are the same. Both represent the same function, and their derivative is x.</p>
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<h3>6.How do you find the derivative of 1/2x² using the first principle?</h3>
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<h3>6.How do you find the derivative of 1/2x² using the first principle?</h3>
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<p>To find it, consider f(x) = 1/2x².</p>
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<p>To find it, consider f(x) = 1/2x².</p>
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<p>Using the first principle: f'(x) = limₕ→₀ [1/2(x + h)² - 1/2x²] / h = limₕ→₀ [xh + 1/2h²] / h = x</p>
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<p>Using the first principle: f'(x) = limₕ→₀ [1/2(x + h)² - 1/2x²] / h = limₕ→₀ [xh + 1/2h²] / h = x</p>
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<h2>Important Glossaries for the Derivative of 1/2x²</h2>
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<h2>Important Glossaries for the Derivative of 1/2x²</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>Quadratic Function</strong>: A function of the form ax² + bx + c, where a, b, and c are constants.</li>
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</ul><ul><li><strong>Quadratic Function</strong>: A function of the form ax² + bx + c, where a, b, and c are constants.</li>
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</ul><ul><li><strong>Power Rule</strong>: A rule for finding the derivative of a power function: d/dx [xⁿ] = n*xⁿ⁻¹.</li>
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</ul><ul><li><strong>Power Rule</strong>: A rule for finding the derivative of a power function: d/dx [xⁿ] = n*xⁿ⁻¹.</li>
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</ul><ul><li><strong>First Principle</strong>: The foundational definition of a derivative as a limit of the difference quotient.</li>
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</ul><ul><li><strong>First Principle</strong>: The foundational definition of a derivative as a limit of the difference quotient.</li>
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</ul><ul><li><strong>Higher-Order Derivatives</strong>: Derivatives obtained by differentiating a function multiple times.</li>
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</ul><ul><li><strong>Higher-Order Derivatives</strong>: Derivatives obtained by differentiating a function multiple times.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<p>▶</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>