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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 1/(x+2) using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation.</p>
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<p>We can derive the derivative of 1/(x+2) using proofs. To show this, we will use algebraic manipulation along with 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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<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 Power Rule</li>
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<li>Using Power Rule</li>
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</ol><p>We will now demonstrate that the differentiation of 1/(x+2) results in -1/(x+2)² using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of 1/(x+2) results in -1/(x+2)² 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 1/(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 1/(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 1/(x+2) using the first principle, we will consider f(x) = 1/(x+2). 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>To find the derivative of 1/(x+2) using the first principle, we will consider f(x) = 1/(x+2). 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) = 1/(x+2), we write f(x + h) = 1/(x+h+2).</p>
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<p>Given that f(x) = 1/(x+2), we write f(x + h) = 1/(x+h+2).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/(x+h+2) - 1/(x+2)] / h = limₕ→₀ [(x+2) - (x+h+2)] / [(x+2)(x+h+2)h] = limₕ→₀ [-h] / [(x+2)(x+h+2)h] = limₕ→₀ -1 / [(x+2)(x+h+2)]</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/(x+h+2) - 1/(x+2)] / h = limₕ→₀ [(x+2) - (x+h+2)] / [(x+2)(x+h+2)h] = limₕ→₀ [-h] / [(x+2)(x+h+2)h] = limₕ→₀ -1 / [(x+2)(x+h+2)]</p>
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<p>As h approaches 0, we simplify the<a>expression</a>: f'(x) = -1 / (x+2)²Hence, proved.</p>
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<p>As h approaches 0, we simplify the<a>expression</a>: f'(x) = -1 / (x+2)²Hence, 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 1/(x+2) using the chain rule, We use the formula: 1/(x+2) = (x+2)⁻¹</p>
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<p>To prove the differentiation of 1/(x+2) using the chain rule, We use the formula: 1/(x+2) = (x+2)⁻¹</p>
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<p>Let u = x+2</p>
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<p>Let u = x+2</p>
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<p>Thus, y = u⁻¹</p>
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<p>Thus, y = u⁻¹</p>
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<p>Using the chain rule formula: d/dx [uⁿ] = nuⁿ⁻¹ du/dx dy/dx = -1 * (x+2)⁻² * d/dx (x+2) dy/dx = -1/(x+2)² Hence, proved.</p>
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<p>Using the chain rule formula: d/dx [uⁿ] = nuⁿ⁻¹ du/dx dy/dx = -1 * (x+2)⁻² * d/dx (x+2) dy/dx = -1/(x+2)² Hence, proved.</p>
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<h3>Using Power Rule</h3>
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<h3>Using Power Rule</h3>
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<p>We will now prove the derivative of 1/(x+2) using the<a>power</a>rule. The step-by-step process is demonstrated below:</p>
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<p>We will now prove the derivative of 1/(x+2) using the<a>power</a>rule. The step-by-step process is demonstrated below:</p>
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<p>Rewrite the function: y = (x+2)⁻¹</p>
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<p>Rewrite the function: y = (x+2)⁻¹</p>
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<p>Using the power rule to differentiate: d/dx (y) = -1 * (x+2)⁻² * d/dx (x+2) d/dx (y) = -1/(x+2)²</p>
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<p>Using the power rule to differentiate: d/dx (y) = -1 * (x+2)⁻² * d/dx (x+2) d/dx (y) = -1/(x+2)²</p>
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<p>Thus, the derivative of 1/(x+2) is -1/(x+2)².</p>
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<p>Thus, the derivative of 1/(x+2) is -1/(x+2)².</p>
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