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Original
2026-01-01
Modified
2026-02-28
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<p>We can derive the derivative of 1/sin x using proofs.</p>
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<p>We can derive the derivative of 1/sin x 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>Using Product Rule</p>
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<p>Using Product Rule</p>
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<p>We will now demonstrate that the differentiation of 1/sin x results in -cot(x)csc(x) using the above-mentioned methods:</p>
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<p>We will now demonstrate that the differentiation of 1/sin x results in -cot(x)csc(x) 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 1/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 1/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 1/sin x using the first principle, we will consider f(x) = 1/sin x.</p>
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<p>To find the derivative of 1/sin x using the first principle, we will consider f(x) = 1/sin x.</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) = 1/sin x, we write f(x + h) = 1/sin (x + h).</p>
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<p>Given that f(x) = 1/sin x, we write f(x + h) = 1/sin (x + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/sin(x + h) - 1/sin x] / h = limₕ→₀ [sin x - sin(x + h)] / [h sin x sin(x + h)]</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/sin(x + h) - 1/sin x] / h = limₕ→₀ [sin x - sin(x + h)] / [h sin x sin(x + h)]</p>
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<p>We now use the formula sin A - sin B = 2 cos((A+B)/2) sin((A-B)/2). f'(x) = limₕ→₀ [-2 cos((2x + h)/2) sin(h/2)] / [h sin x sin(x + h)] = limₕ→₀ [-2 cos(x + h/2) sin(h/2)] / [h sin x sin(x + h)]</p>
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<p>We now use the formula sin A - sin B = 2 cos((A+B)/2) sin((A-B)/2). f'(x) = limₕ→₀ [-2 cos((2x + h)/2) sin(h/2)] / [h sin x sin(x + h)] = limₕ→₀ [-2 cos(x + h/2) sin(h/2)] / [h sin x sin(x + h)]</p>
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<p>Using limit formulas, limₕ→₀ (sin(h/2))/(h/2) = 1. f'(x) = [-2 cos(x)] / [sin²x]</p>
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<p>Using limit formulas, limₕ→₀ (sin(h/2))/(h/2) = 1. f'(x) = [-2 cos(x)] / [sin²x]</p>
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<p>As cotangent is cos(x)/sin(x) and cosecant is 1/sin(x), we have, f'(x) = -cot(x)csc(x).</p>
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<p>As cotangent is cos(x)/sin(x) and cosecant is 1/sin(x), we have, f'(x) = -cot(x)csc(x).</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 1/sin x using the chain rule, We use the formula: 1/sin x = (sin x)⁻¹</p>
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<p>To prove the differentiation of 1/sin x using the chain rule, We use the formula: 1/sin x = (sin x)⁻¹</p>
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<p>Consider f(x) = sin x Using the chain rule: d/dx [f(x)⁻¹] = -[f(x)]⁻² f'(x)</p>
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<p>Consider f(x) = sin x Using the chain rule: d/dx [f(x)⁻¹] = -[f(x)]⁻² f'(x)</p>
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<p>Let’s substitute f(x) = sin x, d/dx (sin x)⁻¹ = -[sin x]⁻² cos x = -cos x/sin²x</p>
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<p>Let’s substitute f(x) = sin x, d/dx (sin x)⁻¹ = -[sin x]⁻² cos x = -cos x/sin²x</p>
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<p>Since cot x = cos x/sin x and csc x = 1/sin x, d/dx (1/sin x) = -cot(x)csc(x)</p>
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<p>Since cot x = cos x/sin x and csc x = 1/sin x, d/dx (1/sin x) = -cot(x)csc(x)</p>
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<p>Using Product Rule</p>
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<p>Using Product Rule</p>
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<p>We will now prove the derivative of 1/sin x using the<a>product</a>rule.</p>
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<p>We will now prove the derivative of 1/sin x using the<a>product</a>rule.</p>
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<p>The step-by-step process is demonstrated below:</p>
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<p>The step-by-step process is demonstrated below:</p>
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<p>Here, we use the formula, 1/sin x = (sin x)⁻¹</p>
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<p>Here, we use the formula, 1/sin x = (sin x)⁻¹</p>
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<p>Given that, u = 1 and v = (sin x)⁻¹</p>
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<p>Given that, u = 1 and v = (sin x)⁻¹</p>
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<p>Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (1) = 0</p>
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<p>Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (1) = 0</p>
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<p>Here we use the chain rule: v = (sin x)⁻¹ = (sin x)⁻¹ v' = -1. (sin x)⁻² (cos x) v' = -cos x/sin²x Again, use the product rule formula: d/dx (1/sin x) = u'. v + u. v'</p>
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<p>Here we use the chain rule: v = (sin x)⁻¹ = (sin x)⁻¹ v' = -1. (sin x)⁻² (cos x) v' = -cos x/sin²x Again, use the product rule formula: d/dx (1/sin x) = u'. v + u. v'</p>
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<p>Let’s substitute u = 1, u' = 0, v = (sin x)⁻¹, and v' = -cos x/sin²x</p>
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<p>Let’s substitute u = 1, u' = 0, v = (sin x)⁻¹, and v' = -cos x/sin²x</p>
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<p>When we simplify each<a>term</a>: We get, d/dx (1/sin x) = -cos x/sin²x = -cot(x)csc(x)</p>
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<p>When we simplify each<a>term</a>: We get, d/dx (1/sin x) = -cos x/sin²x = -cot(x)csc(x)</p>
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