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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/t using proofs. To show this, we will use the rules of differentiation.</p>
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<p>We can derive the derivative of 1/t using proofs. 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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<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/t results in -1/t² using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of 1/t results in -1/t² 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/t can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 1/t using the first principle, we will consider f(t) = 1/t. Its derivative can be expressed as the following limit.</p>
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<p>The derivative of 1/t can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 1/t using the first principle, we will consider f(t) = 1/t. Its derivative can be expressed as the following limit.</p>
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<p>f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1)</p>
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<p>f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1)</p>
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<p>Given that f(t) = 1/t, we write f(t + h) = 1/(t + h).</p>
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<p>Given that f(t) = 1/t, we write f(t + h) = 1/(t + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(t) = limₕ→₀ [1/(t + h) - 1/t] / h = limₕ→₀ [(t - (t + h)) / (t(t + h))] / h = limₕ→₀ [-h / (t² + th)] / h = limₕ→₀ [-1 / (t² + th)]</p>
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<p>Substituting these into<a>equation</a>(1), f'(t) = limₕ→₀ [1/(t + h) - 1/t] / h = limₕ→₀ [(t - (t + h)) / (t(t + h))] / h = limₕ→₀ [-h / (t² + th)] / h = limₕ→₀ [-1 / (t² + th)]</p>
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<p>As h approaches 0, we get: f'(t) = -1/t². Hence, proved.</p>
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<p>As h approaches 0, we get: f'(t) = -1/t². 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/t using the<a>power</a>rule, We rewrite 1/t as t⁻¹. d/dt (t⁻¹) = -1 * t⁻² Since t⁻² = 1/t², d/dt (1/t) = -1/t².</p>
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<p>To prove the differentiation of 1/t using the<a>power</a>rule, We rewrite 1/t as t⁻¹. d/dt (t⁻¹) = -1 * t⁻² Since t⁻² = 1/t², d/dt (1/t) = -1/t².</p>
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<p>Using Quotient Rule We will now prove the derivative of 1/t</p>
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<p>Using Quotient Rule We will now prove the derivative of 1/t</p>
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<h3>using the quotient rule.</h3>
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<h3>using the quotient rule.</h3>
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<p>The step-by-step process is demonstrated below: Here, we use the formula, 1/t = 1 * (t)⁻¹ Given that, u = 1 and v = t Using the quotient rule formula: d/dt [u/v] = (v * u' - u * v') / v² u' = d/dt (1) = 0 (substitute u = 1) v' = d/dt (t) = 1 (substitute v = t)</p>
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<p>The step-by-step process is demonstrated below: Here, we use the formula, 1/t = 1 * (t)⁻¹ Given that, u = 1 and v = t Using the quotient rule formula: d/dt [u/v] = (v * u' - u * v') / v² u' = d/dt (1) = 0 (substitute u = 1) v' = d/dt (t) = 1 (substitute v = t)</p>
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<p>Again, use the quotient rule formula: d/dt (1/t) = (t * 0 - 1 * 1) / t² = -1/t²</p>
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<p>Again, use the quotient rule formula: d/dt (1/t) = (t * 0 - 1 * 1) / t² = -1/t²</p>
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