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Original
2026-01-01
Modified
2026-02-28
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<p>We can derive the derivative of 2cosx using proofs.</p>
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<p>We can derive the derivative of 2cosx using proofs.</p>
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<p>To show this, we will use trigonometric identities along with the rules of differentiation.</p>
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<p>To show this, we will use trigonometric identities 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</p>
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<p>By First Principle</p>
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<p>Using the Chain Rule</p>
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<p>Using the Chain Rule</p>
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<p>Using the Product Rule</p>
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<p>Using the Product Rule</p>
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<p>We will now demonstrate that the differentiation of 2cosx results in -2sinx using the above-mentioned methods:</p>
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<p>We will now demonstrate that the differentiation of 2cosx results in -2sinx using the above-mentioned methods:</p>
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<p>By First Principle The derivative of 2cosx 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 2cosx using the first principle, we will consider f(x) = 2cosx.</p>
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<p>By First Principle The derivative of 2cosx 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 2cosx using the first principle, we will consider f(x) = 2cosx.</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 … (1) Given that f(x) = 2cosx, we write f(x + h) = 2cos(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 … (1) Given that f(x) = 2cosx, we write f(x + h) = 2cos(x + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [2cos(x + h) - 2cosx] / h = 2limₕ→₀ [cos(x + h) - cosx] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [2cos(x + h) - 2cosx] / h = 2limₕ→₀ [cos(x + h) - cosx] / h</p>
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<p>Using the trigonometric identity for cos(A + B) and limit properties, f'(x) = 2 * limₕ→₀ [(-2sin(x + h/2)sin(h/2)) / h] = 2 * (-sinx)</p>
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<p>Using the trigonometric identity for cos(A + B) and limit properties, f'(x) = 2 * limₕ→₀ [(-2sin(x + h/2)sin(h/2)) / h] = 2 * (-sinx)</p>
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<p>Therefore, f'(x) = -2sinx.</p>
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<p>Therefore, f'(x) = -2sinx.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using the Chain Rule To prove the differentiation of 2cosx using the chain rule, Consider f(x) = 2 and g(x) = cosx So, f(x) * g(x) = 2cosx</p>
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<p>Using the Chain Rule To prove the differentiation of 2cosx using the chain rule, Consider f(x) = 2 and g(x) = cosx So, f(x) * g(x) = 2cosx</p>
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<p>By the chain rule, d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) Substitute f(x) = 2 and g(x) = cosx, d/dx (2cosx) = 0 * cosx + 2 * (-sinx) = -2sinx</p>
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<p>By the chain rule, d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) Substitute f(x) = 2 and g(x) = cosx, d/dx (2cosx) = 0 * cosx + 2 * (-sinx) = -2sinx</p>
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<p>Using the Product Rule We will now prove the derivative of 2cosx using the<a>product</a>rule.</p>
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<p>Using the Product Rule We will now prove the derivative of 2cosx using the<a>product</a>rule.</p>
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<p>The step-by-step process is demonstrated below: Here, we use the formula, 2cosx = 2 * cosx Let u = 2 and v = cosx</p>
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<p>The step-by-step process is demonstrated below: Here, we use the formula, 2cosx = 2 * cosx Let u = 2 and v = cosx</p>
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<p>Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (2) = 0 v' = d/dx (cosx) = -sinx Using the product rule formula: d/dx (2cosx) = u'.v + u.v' = 0 * cosx + 2 * (-sinx) = -2sinx</p>
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<p>Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (2) = 0 v' = d/dx (cosx) = -sinx Using the product rule formula: d/dx (2cosx) = u'.v + u.v' = 0 * cosx + 2 * (-sinx) = -2sinx</p>
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