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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 ln(2/x) using proofs. To show this, we will use logarithmic identities 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 ln(2/x) using proofs. To show this, we will use logarithmic identities 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 Quotient Rule</li>
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<li>Using Quotient Rule</li>
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</ol><p>We will now demonstrate that the differentiation of ln(2/x) results in -1/x using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of ln(2/x) results in -1/x 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 ln(2/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 ln(2/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(2/x) using the first principle, we will consider f(x) = ln(2/x).</p>
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<p>To find the derivative of ln(2/x) using the first principle, we will consider f(x) = ln(2/x).</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = ln(2/x), we write f(x + h) = ln(2/(x + h)).</p>
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<p>Given that f(x) = ln(2/x), we write f(x + h) = ln(2/(x + h)).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(2/(x + h)) - ln(2/x)] / h = limₕ→₀ [ln((2/x) / (2/(x + h)))] / h = limₕ→₀ [ln((x + h)/x)] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(2/(x + h)) - ln(2/x)] / h = limₕ→₀ [ln((2/x) / (2/(x + h)))] / h = limₕ→₀ [ln((x + h)/x)] / h</p>
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<p>We now use the logarithmic property ln(a/b) = ln(a) - ln(b). f'(x) = limₕ→₀ [ln(1 + h/x)] / h</p>
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<p>We now use the logarithmic property ln(a/b) = ln(a) - ln(b). f'(x) = limₕ→₀ [ln(1 + h/x)] / h</p>
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<p>Using the limit property ln(1 + u) ≈ u when u is small, f'(x) = limₕ→₀ (h/x) / h = limₕ→₀ 1/x f'(x) = -1/x</p>
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<p>Using the limit property ln(1 + u) ≈ u when u is small, f'(x) = limₕ→₀ (h/x) / h = limₕ→₀ 1/x f'(x) = -1/x</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 ln(2/x) using the chain rule, We use the formula: ln(2/x) = ln(2) - ln(x)</p>
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<p>To prove the differentiation of ln(2/x) using the chain rule, We use the formula: ln(2/x) = ln(2) - ln(x)</p>
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<p>The derivative of ln(x) is 1/x, and the<a>constant</a>ln(2) differentiates to 0. d/dx [ln(2/x)] = d/dx [ln(2) - ln(x)] d/dx [ln(2)] = 0 and d/dx [-ln(x)] = -1/x Therefore, d/dx [ln(2/x)] = -1/x</p>
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<p>The derivative of ln(x) is 1/x, and the<a>constant</a>ln(2) differentiates to 0. d/dx [ln(2/x)] = d/dx [ln(2) - ln(x)] d/dx [ln(2)] = 0 and d/dx [-ln(x)] = -1/x Therefore, d/dx [ln(2/x)] = -1/x</p>
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<h3>Using Quotient Rule</h3>
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<h3>Using Quotient Rule</h3>
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<p>We will now prove the derivative of ln(2/x) using the quotient rule. The step-by-step process is demonstrated below: Here, we use the formula, ln(2/x) = ln(2) - ln(x)</p>
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<p>We will now prove the derivative of ln(2/x) using the quotient rule. The step-by-step process is demonstrated below: Here, we use the formula, ln(2/x) = ln(2) - ln(x)</p>
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<p>The derivative of a constant is 0, and the derivative of ln(x) is 1/x. Using these rules: d/dx [ln(2/x)] = d/dx [ln(2) - ln(x)]</p>
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<p>The derivative of a constant is 0, and the derivative of ln(x) is 1/x. Using these rules: d/dx [ln(2/x)] = d/dx [ln(2) - ln(x)]</p>
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<p>Since d/dx [ln(2)] = 0, d/dx [-ln(x)] = -1/x Thus: d/dx [ln(2/x)] = -1/x</p>
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<p>Since d/dx [ln(2)] = 0, d/dx [-ln(x)] = -1/x Thus: d/dx [ln(2/x)] = -1/x</p>
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