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2026-01-01
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2026-02-28
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<p>We can derive the derivative of x|x| using proofs. To show this, we will use the piecewise definition of |x| along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of x|x| using proofs. To show this, we will use the piecewise definition of |x| along with the rules of differentiation. 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 Piecewise Differentiation</li>
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<li>Using Piecewise Differentiation</li>
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</ol><p>We will now demonstrate that the differentiation of x|x| results in 2x for x > 0 and 0 for x < 0 using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of x|x| results in 2x for x > 0 and 0 for x < 0 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 x|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>The derivative of x|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 x|x| using the first principle, we will consider f(x) = x|x|. Its derivative can be expressed as the following limit. f'(x) = lim(h→0) [f(x + h) - f(x)] / h</p>
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<p>To find the derivative of x|x| using the first principle, we will consider f(x) = x|x|. Its derivative can be expressed as the following limit. f'(x) = lim(h→0) [f(x + h) - f(x)] / h</p>
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<p>Given that f(x) = x|x|, we write f(x + h) = (x + h)|x + h|.</p>
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<p>Given that f(x) = x|x|, we write f(x + h) = (x + h)|x + h|.</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = lim(h→0) [(x + h)|x + h| - x|x|] / h For x > 0, |x| = x, hence |x + h| = x + h</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = lim(h→0) [(x + h)|x + h| - x|x|] / h For x > 0, |x| = x, hence |x + h| = x + h</p>
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<p>when h is small and positive. f'(x) = lim(h→0) [(x + h)(x + h) - x²] / h = lim(h→0) [x² + 2xh + h² - x²] / h = lim(h→0) [2xh + h²] / h = lim(h→0) [2x + h] = 2x For x < 0, |x| = -x,</p>
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<p>when h is small and positive. f'(x) = lim(h→0) [(x + h)(x + h) - x²] / h = lim(h→0) [x² + 2xh + h² - x²] / h = lim(h→0) [2xh + h²] / h = lim(h→0) [2x + h] = 2x For x < 0, |x| = -x,</p>
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<p>hence |x + h| = -(x + h) when h is small. f'(x) = lim(h→0) [-(x + h)(x + h) - (-x)x] / h = lim(h→0) [-x² - 2xh - h² + x²] / h = lim(h→0) [-2xh - h²] / h = lim(h→0) [-2x - h] = 0 Hence, proved.</p>
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<p>hence |x + h| = -(x + h) when h is small. f'(x) = lim(h→0) [-(x + h)(x + h) - (-x)x] / h = lim(h→0) [-x² - 2xh - h² + x²] / h = lim(h→0) [-2xh - h²] / h = lim(h→0) [-2x - h] = 0 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 x|x| using the chain rule, We use the formula: x|x| = x * (x for x ≥ 0 and -x for x < 0) Consider f(x) = x and g(x) = |x|</p>
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<p>To prove the differentiation of x|x| using the chain rule, We use the formula: x|x| = x * (x for x ≥ 0 and -x for x < 0) Consider f(x) = x and g(x) = |x|</p>
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<p>By<a>product</a>rule: d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) Let’s substitute f(x) = x and g(x) = |x| d/dx (x|x|) = 1 · |x| + x · (d/dx |x|) For x > 0, |x| = x and d/dx |x| = 1,</p>
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<p>By<a>product</a>rule: d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) Let’s substitute f(x) = x and g(x) = |x| d/dx (x|x|) = 1 · |x| + x · (d/dx |x|) For x > 0, |x| = x and d/dx |x| = 1,</p>
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<p>hence d/dx (x|x|) = x + x · 1 = 2x For x < 0, |x| = -x and d/dx |x| = -1, hence d/dx (x|x|) = -x + x · (-1) = 0</p>
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<p>hence d/dx (x|x|) = x + x · 1 = 2x For x < 0, |x| = -x and d/dx |x| = -1, hence d/dx (x|x|) = -x + x · (-1) = 0</p>
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<h3>Using Piecewise Differentiation</h3>
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<h3>Using Piecewise Differentiation</h3>
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<p>We will now prove the derivative of x|x| using piecewise differentiation: Here, we use the formula, x|x| = { x², for x ≥ 0 -x², for x < 0 }</p>
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<p>We will now prove the derivative of x|x| using piecewise differentiation: Here, we use the formula, x|x| = { x², for x ≥ 0 -x², for x < 0 }</p>
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<p>Differentiate each piece: For x ≥ 0: d/dx (x²) = 2x For x < 0: d/dx (-x²) = 0 Thus, d/dx (x|x|) = { 2x, for x > 0 0, for x < 0 }</p>
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<p>Differentiate each piece: For x ≥ 0: d/dx (x²) = 2x For x < 0: d/dx (-x²) = 0 Thus, d/dx (x|x|) = { 2x, for x > 0 0, for x < 0 }</p>
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