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Original
2026-01-01
Modified
2026-02-28
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<p>We can derive the derivative of sin(x/2) using proofs.</p>
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<p>We can derive the derivative of sin(x/2) using proofs.</p>
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<p>To show this, we will use the trigonometric identities along with the rules of differentiation.</p>
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<p>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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<p>By First Principle</p>
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<p>By First Principle</p>
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<p>Using Chain Rule</p>
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<p>Using Chain Rule</p>
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<p>We will now demonstrate that the differentiation of sin(x/2) results in (1/2)cos(x/2) using the above-mentioned methods:</p>
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<p>We will now demonstrate that the differentiation of sin(x/2) results in (1/2)cos(x/2) using the above-mentioned methods:</p>
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<p>By First Principle</p>
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<p>By First Principle</p>
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<p>The derivative of sin(x/2) 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/2) 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/2) using the first principle, we will consider f(x) = sin(x/2).</p>
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<p>To find the derivative of sin(x/2) using the first principle, we will consider f(x) = sin(x/2).</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)</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)</p>
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<p>Given that f(x) = sin(x/2), we write f(x + h) = sin((x + h)/2).</p>
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<p>Given that f(x) = sin(x/2), we write f(x + h) = sin((x + h)/2).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sin((x + h)/2) - sin(x/2)] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sin((x + h)/2) - sin(x/2)] / h</p>
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<p>Using the trigonometric identity for the difference of sines: sin(A) - sin(B) = 2cos((A + B)/2)sin((A - B)/2), f'(x) = limₕ→₀ [2cos((x + h)/4 + x/4)sin(h/4)] / h = limₕ→₀ [cos((x + h)/4 + x/4)·(1/2)sin(h/4)/(h/4)] · (1/2)</p>
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<p>Using the trigonometric identity for the difference of sines: sin(A) - sin(B) = 2cos((A + B)/2)sin((A - B)/2), f'(x) = limₕ→₀ [2cos((x + h)/4 + x/4)sin(h/4)] / h = limₕ→₀ [cos((x + h)/4 + x/4)·(1/2)sin(h/4)/(h/4)] · (1/2)</p>
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<p>Using the limit formula limₕ→₀ (sin(h/4)/(h/4)) = 1, f'(x) = (1/2)cos(x/2)</p>
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<p>Using the limit formula limₕ→₀ (sin(h/4)/(h/4)) = 1, f'(x) = (1/2)cos(x/2)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using Chain Rule</p>
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<p>Using Chain Rule</p>
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<p>To prove the differentiation of sin(x/2) using the chain rule,</p>
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<p>To prove the differentiation of sin(x/2) using the chain rule,</p>
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<p>We consider the function sin(u) where u = x/2.</p>
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<p>We consider the function sin(u) where u = x/2.</p>
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<p>The chain rule states: d/dx(sin(u)) = cos(u)·du/dx</p>
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<p>The chain rule states: d/dx(sin(u)) = cos(u)·du/dx</p>
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<p>Here, u = x/2, so du/dx = 1/2. Therefore, d/dx(sin(x/2)) = cos(x/2)·(1/2)</p>
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<p>Here, u = x/2, so du/dx = 1/2. Therefore, d/dx(sin(x/2)) = cos(x/2)·(1/2)</p>
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<p>Thus, the result is (1/2)cos(x/2).</p>
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<p>Thus, the result is (1/2)cos(x/2).</p>
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