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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 sin x using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of sin x using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. 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 sin x results in cos x using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of sin x results in cos x 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 sin 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>The derivative of sin 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 sin x using the first principle, we will consider f(x) = sin 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 sin x using the first principle, we will consider f(x) = sin 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) = sin x, we write f(x + h) = sin (x + h).</p>
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<p>Given that f(x) = sin x, we write f(x + h) = sin (x + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sin(x + h) - sin x] / h = limₕ→₀ [ [sin x cos h + cos x sin h] - sin x ] / h = limₕ→₀ [ sin x (cos h - 1) + cos x sin h ] / h = limₕ→₀ [ sin x (cos h - 1)/h + cos x sin h/h ]</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sin(x + h) - sin x] / h = limₕ→₀ [ [sin x cos h + cos x sin h] - sin x ] / h = limₕ→₀ [ sin x (cos h - 1) + cos x sin h ] / h = limₕ→₀ [ sin x (cos h - 1)/h + cos x sin h/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) = sin x(0) + cos x(1) = cos x</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) = sin x(0) + cos x(1) = cos x</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Chain Rule</h3>
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<h3>Using Chain Rule</h3>
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<p>To prove the differentiation of sin x using the chain rule, We consider a composite function, but for simple sin x, it directly gives: d/dx (sin x) = cos x</p>
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<p>To prove the differentiation of sin x using the chain rule, We consider a composite function, but for simple sin x, it directly gives: d/dx (sin x) = cos x</p>
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<h3>Using Product Rule</h3>
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<h3>Using Product Rule</h3>
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<p>We will now prove the derivative of sin x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, sin x = (1).(sin x) Given that, u = 1 and v = sin x</p>
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<p>We will now prove the derivative of sin x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, sin x = (1).(sin x) Given that, u = 1 and v = sin x</p>
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<p>Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (1) = 0 v' = d/dx (sin x) = cos x</p>
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<p>Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (1) = 0 v' = d/dx (sin x) = cos x</p>
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<p>Again, use the product rule formula: d/dx (sin x) = u'. v + u. v'</p>
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<p>Again, use the product rule formula: d/dx (sin x) = u'. v + u. v'</p>
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<p>Let’s substitute u = 1, u' = 0, v = sin x, and v' = cos x</p>
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<p>Let’s substitute u = 1, u' = 0, v = sin x, and v' = cos x</p>
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<p>When we simplify each<a>term</a>: We get, d/dx (sin x) = 0 + 1(cos x)</p>
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<p>When we simplify each<a>term</a>: We get, d/dx (sin x) = 0 + 1(cos x)</p>
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<p>Thus: d/dx (sin x) = cos x.</p>
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<p>Thus: d/dx (sin x) = cos x.</p>
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