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Original
2026-01-01
Modified
2026-02-28
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<p>We can derive the derivative of -ex using proofs.</p>
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<p>We can derive the derivative of -ex using proofs.</p>
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<p>To show this, we will use the properties of exponential functions along with the basic rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>To show this, we will use the properties of exponential functions along with the basic rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>By First Principle Using Constant Multiplier Rule</p>
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<p>By First Principle Using Constant Multiplier Rule</p>
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<p>We will now demonstrate that the differentiation of -ex results in -e^x using the above-mentioned methods:</p>
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<p>We will now demonstrate that the differentiation of -ex results in -e^x using the above-mentioned methods:</p>
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<p>By First Principle The derivative of -ex 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 -ex 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^x using the first principle, we will consider f(x) = -ex.</p>
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<p>To find the derivative of -e^x using the first principle, we will consider f(x) = -ex.</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</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</p>
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<p>Given that f(x) = -ex, we write f(x + h) = -e(x + h).</p>
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<p>Given that f(x) = -ex, we write f(x + h) = -e(x + h).</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [-e^(x + h) + e^x] / h = limₕ→₀ [-ex * (eh - 1)] / h</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [-e^(x + h) + e^x] / h = limₕ→₀ [-ex * (eh - 1)] / h</p>
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<p>Using the limit properties, limₕ→₀ (eh - 1)/h = 1, f'(x) = -ex * 1 = -ex.</p>
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<p>Using the limit properties, limₕ→₀ (eh - 1)/h = 1, f'(x) = -ex * 1 = -ex.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using Constant Multiplier Rule</p>
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<p>Using Constant Multiplier Rule</p>
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<p>To prove the differentiation of -ex using the constant<a>multiplier</a>rule, We use the formula:</p>
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<p>To prove the differentiation of -ex using the constant<a>multiplier</a>rule, We use the formula:</p>
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<p>The derivative of a constant times a function is the constant times the derivative of the function.</p>
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<p>The derivative of a constant times a function is the constant times the derivative of the function.</p>
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<p>Let u(x) = ex, so -ex = -1 * u(x).</p>
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<p>Let u(x) = ex, so -ex = -1 * u(x).</p>
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<p>By the constant multiplier rule: d/dx(-1 * u(x)) = -1 * d/dx(u(x)) Since d/dx(ex) = ex, we have: d/dx(-ex) = -1 * ex = -ex. Thus, the derivative is -ex.</p>
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<p>By the constant multiplier rule: d/dx(-1 * u(x)) = -1 * d/dx(u(x)) Since d/dx(ex) = ex, we have: d/dx(-ex) = -1 * ex = -ex. Thus, the derivative is -ex.</p>
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