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Original
2026-01-01
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2026-02-28
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<p>We can derive the derivative of e^ax using proofs. To show this, we will use the exponential properties along with the rules of differentiation.</p>
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<p>We can derive the derivative of e^ax using proofs. To show this, we will use the exponential properties 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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<p>By First Principle Using Chain Rule We will now demonstrate that the differentiation of e^ax results in ae^ax using the above-mentioned methods:</p>
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<p>By First Principle Using Chain Rule We will now demonstrate that the differentiation of e^ax results in ae^ax using the above-mentioned methods:</p>
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<p>By First Principle The derivative of e^ax 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>By First Principle The derivative of e^ax 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 e^ax using the first principle, we will consider f(x) = e^ax. 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 e^ax using the first principle, we will consider f(x) = e^ax. 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) = e^ax, we write f(x + h) = e^(a(x + h)).</p>
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<p>Given that f(x) = e^ax, we write f(x + h) = e^(a(x + h)).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [e^(a(x + h)) - e^ax] / h = limₕ→₀ [e^ax (e^(ah) - 1)] / h = e^ax limₕ→₀ [e^(ah) - 1] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [e^(a(x + h)) - e^ax] / h = limₕ→₀ [e^ax (e^(ah) - 1)] / h = e^ax limₕ→₀ [e^(ah) - 1] / h</p>
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<p>We now make a substitution: let u = ah, then as h approaches 0, u approaches 0. f'(x) = e^ax limᵤ→₀ [e^u - 1] / (u/a) = ae^ax limᵤ→₀ [e^u - 1] / u</p>
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<p>We now make a substitution: let u = ah, then as h approaches 0, u approaches 0. f'(x) = e^ax limᵤ→₀ [e^u - 1] / (u/a) = ae^ax limᵤ→₀ [e^u - 1] / u</p>
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<p>Using limit formulas, limᵤ→₀ [e^u - 1] / u = 1. f'(x) = ae^ax. Hence, proved.</p>
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<p>Using limit formulas, limᵤ→₀ [e^u - 1] / u = 1. f'(x) = ae^ax. Hence, proved.</p>
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<p>Using Chain Rule To prove the differentiation of e^ax using the chain rule, We use the formula: Let u = ax Then e^ax = e^u</p>
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<p>Using Chain Rule To prove the differentiation of e^ax using the chain rule, We use the formula: Let u = ax Then e^ax = e^u</p>
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<p>By chain rule: d/dx [e^u] = e^u · du/dx = e^ax · a Therefore, d/dx (e^ax) = ae^ax.</p>
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<p>By chain rule: d/dx [e^u] = e^u · du/dx = e^ax · a Therefore, d/dx (e^ax) = ae^ax.</p>
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