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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(1+x) using proofs.</p>
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<p>We can derive the derivative of ln(1+x) using proofs.</p>
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<p>To demonstrate this, we utilize properties of<a>logarithms</a>and rules of differentiation. Some methods to prove this include:</p>
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<p>To demonstrate this, we utilize properties of<a>logarithms</a>and rules of differentiation. Some methods to prove this include:</p>
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<p>By First Principle</p>
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<p>By First Principle</p>
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<p>Using Chain Rule</p>
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<p>Using Chain Rule</p>
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<p>We will demonstrate how to derive the derivative of ln(1+x) using these methods:</p>
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<p>We will demonstrate how to derive the derivative of ln(1+x) using these methods:</p>
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<p>By First Principle</p>
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<p>By First Principle</p>
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<p>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>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, consider f(x) = ln(1+x).</p>
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<p>To find the derivative of ln(1+x) using the first principle, consider f(x) = ln(1+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 f(x) = ln(1+x), we write f(x+h) = ln(1+(x+h)).</p>
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<p>Given 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 Using the property ln(a/b) = ln(a) - ln(b), = limₕ→₀ ln[1 + 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 Using the property ln(a/b) = ln(a) - ln(b), = limₕ→₀ ln[1 + h/(1+x)] / h</p>
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<p>Using the limit property, limₕ→₀ ln(1+y)/y = 1 as y→0, f'(x) = 1/(1+x)</p>
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<p>Using the limit property, limₕ→₀ ln(1+y)/y = 1 as y→0, f'(x) = 1/(1+x)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using Chain Rule</p>
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<p>Using Chain Rule</p>
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<p>To demonstrate the differentiation of ln(1+x) using the chain rule, Consider u = 1+x.</p>
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<p>To demonstrate the differentiation of ln(1+x) using the chain rule, Consider u = 1+x.</p>
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<p>Then, ln(1+x) = ln(u). Using the chain rule formula: d/dx [ln(u)] = 1/u · du/dx</p>
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<p>Then, ln(1+x) = ln(u). Using the chain rule formula: d/dx [ln(u)] = 1/u · du/dx</p>
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<p>Since u = 1+x, du/dx = 1.</p>
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<p>Since u = 1+x, du/dx = 1.</p>
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<p>Substitute into the chain rule formula: d/dx (ln(1+x)) = 1/(1+x) · 1 = 1/(1+x)</p>
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<p>Substitute into the chain rule formula: d/dx (ln(1+x)) = 1/(1+x) · 1 = 1/(1+x)</p>
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<p>Thus, the derivative is 1/(1+x).</p>
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<p>Thus, the derivative is 1/(1+x).</p>
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