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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 4sin x using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation.</p>
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<p>We can derive the derivative of 4sin x using proofs. To show this, we will use the 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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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Chain Rule</li>
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<li>Using Chain Rule</li>
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<li>Using Product Rule</li>
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<li>Using Product Rule</li>
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</ol><p>We will now demonstrate that the differentiation of 4sin x results in 4cos x using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of 4sin x results in 4cos x using the above-mentioned methods:</p>
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<p><strong>By First Principle</strong>The derivative of 4sin x 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><strong>By First Principle</strong>The derivative of 4sin x 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 4sin x using the first principle, we will consider f(x) = 4sin x. 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 4sin x using the first principle, we will consider f(x) = 4sin x. 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) = 4sin x, we write f(x + h) = 4sin (x + h).</p>
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<p>Given that f(x) = 4sin x, we write f(x + h) = 4sin (x + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [4sin(x + h) - 4sin x] / h = 4 · limₕ→₀ [sin(x + h) - sin x] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [4sin(x + h) - 4sin x] / h = 4 · limₕ→₀ [sin(x + h) - sin x] / h</p>
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<p>Using the trigonometric identity sin(A + B) = sinAcosB + cosAsinB, f'(x) = 4 · limₕ→₀ [(sin x cos h + cos x sin h) - sin x] / h = 4 · limₕ→₀ [cos x sin h + (cos h - 1) sin x] / h = 4 · [cos x · limₕ→₀ (sin h)/ h + sin x · limₕ→₀ (cos h - 1)/ h]</p>
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<p>Using the trigonometric identity sin(A + B) = sinAcosB + cosAsinB, f'(x) = 4 · limₕ→₀ [(sin x cos h + cos x sin h) - sin x] / h = 4 · limₕ→₀ [cos x sin h + (cos h - 1) sin x] / h = 4 · [cos x · limₕ→₀ (sin h)/ h + sin x · limₕ→₀ (cos h - 1)/ h]</p>
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<p>Using limit formulas, limₕ→₀ (sin h)/ h = 1 and limₕ→₀ (cos h - 1)/ h = 0. f'(x) = 4 · [cos x · 1 + sin x · 0] f'(x) = 4cos x. Hence, proved.</p>
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<p>Using limit formulas, limₕ→₀ (sin h)/ h = 1 and limₕ→₀ (cos h - 1)/ h = 0. f'(x) = 4 · [cos x · 1 + sin x · 0] f'(x) = 4cos x. Hence, proved.</p>
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<p><strong>Using Chain Rule</strong>To prove the differentiation of 4sin x using the chain rule, We use the formula: 4sin x = 4 · sin x Consider y = sin x, So we get, 4sin x = 4y By the chain rule: d/dx (4y) = 4 · d/dx (y) … (1)</p>
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<p><strong>Using Chain Rule</strong>To prove the differentiation of 4sin x using the chain rule, We use the formula: 4sin x = 4 · sin x Consider y = sin x, So we get, 4sin x = 4y By the chain rule: d/dx (4y) = 4 · d/dx (y) … (1)</p>
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<p>Let’s substitute y = sin x in equation (1), d/dx (4sin x) = 4 · d/dx (sin x) = 4cos x</p>
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<p>Let’s substitute y = sin x in equation (1), d/dx (4sin x) = 4 · d/dx (sin x) = 4cos x</p>
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<p><strong>Using Product Rule</strong>We will now prove the derivative of 4sin x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, 4sin x = 4 · sin x Given that, u = 4 and v = sin x</p>
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<p><strong>Using Product Rule</strong>We will now prove the derivative of 4sin x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, 4sin x = 4 · sin x Given that, u = 4 and v = sin x</p>
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<p><strong>Using the product rule formula:</strong>d/dx [u·v] = u'·v + u·v' u' = d/dx (4) = 0. (substitute u = 4) v' = d/dx (sin x) = cos x (substitute v = sin x)</p>
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<p><strong>Using the product rule formula:</strong>d/dx [u·v] = u'·v + u·v' u' = d/dx (4) = 0. (substitute u = 4) v' = d/dx (sin x) = cos x (substitute v = sin x)</p>
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<p>Again, use the product rule formula: d/dx (4sin x) = u'·v + u·v'</p>
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<p>Again, use the product rule formula: d/dx (4sin x) = u'·v + u·v'</p>
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<p>Let’s substitute u = 4, u' = 0, v = sin x, and v' = cos x When we simplify each<a>term</a>: We get, d/dx (4sin x) = 0 · sin x + 4 · cos x Thus: d/dx (4sin x) = 4cos x</p>
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<p>Let’s substitute u = 4, u' = 0, v = sin x, and v' = cos x When we simplify each<a>term</a>: We get, d/dx (4sin x) = 0 · sin x + 4 · cos x Thus: d/dx (4sin x) = 4cos x</p>
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