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Original
2026-01-01
Modified
2026-02-28
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<p>We can derive the derivative of ln(x+2) using proofs.</p>
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<p>We can derive the derivative of ln(x+2) using proofs.</p>
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<p>To show this, we will use the rules of differentiation.</p>
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<p>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:</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>Using Chain Rule</p>
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<p>Using Chain Rule</p>
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<p>By First Principle</p>
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<p>By First Principle</p>
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<p>We will now demonstrate that the differentiation of ln(x+2) results in 1/(x+2) using the above-mentioned methods:</p>
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<p>We will now demonstrate that the differentiation of ln(x+2) results in 1/(x+2) using the above-mentioned methods:</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 prove the differentiation of ln(x+2) using the chain rule, Let u = x+2, so ln(x+2) = ln(u).</p>
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<p>To prove the differentiation of ln(x+2) using the chain rule, Let u = x+2, so ln(x+2) = ln(u).</p>
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<p>By chain rule: d/dx [ln(u)] = 1/u * du/dx.</p>
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<p>By chain rule: d/dx [ln(u)] = 1/u * du/dx.</p>
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<p>Since u = x+2, du/dx = 1.</p>
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<p>Since u = x+2, du/dx = 1.</p>
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<p>Thus, d/dx [ln(x+2)] = 1/(x+2).</p>
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<p>Thus, d/dx [ln(x+2)] = 1/(x+2).</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(x+2) 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(x+2) 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(x+2) using the first principle, we will consider f(x) = ln(x+2).</p>
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<p>To find the derivative of ln(x+2) using the first principle, we will consider f(x) = ln(x+2).</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(x+2), we write f(x + h) = ln(x + 2 + h).</p>
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<p>Given that f(x) = ln(x+2), we write f(x + h) = ln(x + 2 + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(x + 2 + h) - ln(x+2)] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [ln(x + 2 + h) - ln(x+2)] / h</p>
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<p>Using the logarithm property ln(a) - ln(b) = ln(a/b), f'(x) = limₕ→₀ ln[(x + 2 + h)/(x + 2)] / h</p>
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<p>Using the logarithm property ln(a) - ln(b) = ln(a/b), f'(x) = limₕ→₀ ln[(x + 2 + h)/(x + 2)] / h</p>
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<p>Using the well-known limit property, we find: f'(x) = 1/(x+2)</p>
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<p>Using the well-known limit property, we find: f'(x) = 1/(x+2)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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