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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/u using proofs. To show this, we will use<a>algebraic identities</a>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 1/u using proofs. To show this, we will use<a>algebraic identities</a>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 Power Rule</li>
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<li>Using Power 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 1/u results in -1/u² using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of 1/u results in -1/u² 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/u 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/u 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/u using the first principle, we will consider f(u) = 1/u. Its derivative can be expressed as the following limit. f'(u) = limₕ→₀ [f(u + h) - f(u)] / h … (1)</p>
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<p>To find the derivative of 1/u using the first principle, we will consider f(u) = 1/u. Its derivative can be expressed as the following limit. f'(u) = limₕ→₀ [f(u + h) - f(u)] / h … (1)</p>
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<p>Given that f(u) = 1/u, we write f(u + h) = 1/(u + h).</p>
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<p>Given that f(u) = 1/u, we write f(u + h) = 1/(u + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(u) = limₕ→₀ [1/(u + h) - 1/u] / h = limₕ→₀ [u - (u + h)] / [h(u + h)u] = limₕ→₀ [-h] / [h(u + h)u] = limₕ→₀ -1 / [(u + h)u]</p>
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<p>Substituting these into<a>equation</a>(1), f'(u) = limₕ→₀ [1/(u + h) - 1/u] / h = limₕ→₀ [u - (u + h)] / [h(u + h)u] = limₕ→₀ [-h] / [h(u + h)u] = limₕ→₀ -1 / [(u + h)u]</p>
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<p>As h approaches 0, f'(u) = -1/u²</p>
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<p>As h approaches 0, f'(u) = -1/u²</p>
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<p>Hence, proved.</p>
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<p>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>To prove the differentiation of 1/u using the power rule, We rewrite the function: 1/u = u⁻¹</p>
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<p>To prove the differentiation of 1/u using the power rule, We rewrite the function: 1/u = u⁻¹</p>
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<p>Using the power rule: d/du (uⁿ) = n·uⁿ⁻¹ d/du (u⁻¹) = -1·u⁻² Thus, d/du (1/u) = -1/u²</p>
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<p>Using the power rule: d/du (uⁿ) = n·uⁿ⁻¹ d/du (u⁻¹) = -1·u⁻² Thus, d/du (1/u) = -1/u²</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 1/u using the quotient rule. The step-by-step process is demonstrated below:</p>
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<p>We will now prove the derivative of 1/u using the quotient rule. The step-by-step process is demonstrated below:</p>
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<p>Here, we use the identity, 1/u = 1 / u</p>
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<p>Here, we use the identity, 1/u = 1 / u</p>
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<p>Given that, f(u) = 1 and g(u) = u</p>
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<p>Given that, f(u) = 1 and g(u) = u</p>
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<p>Using the quotient rule formula: d/du [f(u) / g(u)] = [f '(u) g(u) - f(u) g'(u)] / [g(u)]²… (1)</p>
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<p>Using the quotient rule formula: d/du [f(u) / g(u)] = [f '(u) g(u) - f(u) g'(u)] / [g(u)]²… (1)</p>
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<p>Let’s substitute f(u) = 1 and g(u) = u in equation (1), d/du (1/u) = [0·u - 1·1] / u² = -1/u²</p>
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<p>Let’s substitute f(u) = 1 and g(u) = u in equation (1), d/du (1/u) = [0·u - 1·1] / u² = -1/u²</p>
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<p>Thus, d/du (1/u) = -1/u².</p>
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<p>Thus, d/du (1/u) = -1/u².</p>
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