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Original
2026-01-01
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2026-02-28
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<p>We can derive the derivative of csc²(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 csc²(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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<ul><li>By First Principle </li>
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<ul><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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</ul><p>We will now demonstrate that the differentiation of csc²(x) results in -2csc²(x)cot(x) using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of csc²(x) results in -2csc²(x)cot(x) using the above-mentioned methods:</p>
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<h2><strong>By First Principle</strong></h2>
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<h2><strong>By First Principle</strong></h2>
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<p>The derivative of csc²(x) 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 csc²(x) using the first principle, we will consider f(x) = csc²(x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = csc²(x), we write f(x + h) = csc²(x + h).</p>
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<p>The derivative of csc²(x) 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 csc²(x) using the first principle, we will consider f(x) = csc²(x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = csc²(x), we write f(x + h) = csc²(x + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [csc²(x + h) - csc²(x)] / h = limₕ→₀ [1/sin²(x + h) - 1/sin²(x)] / h = limₕ→₀ [[sin²(x) - sin²(x + h)] / [sin²(x)sin²(x + h)]] / h We now use the formula sin²(A) - sin²(B) = (A - B)(A + B). f'(x) = limₕ→₀ [(sin(x + h) - sin(x))(sin(x + h) + sin(x))] / [h sin²(x)sin²(x + h)] = limₕ→₀ [(sin h) cos((x + x + h)/2) 2cos((x - x - h)/2)] / [h sin²(x)sin²(x + h)] = limₕ→₀ (sin h)/ h · limₕ→₀ [cos((x + x + h)/2) · 2cos((x - x - h)/2)] / [sin²(x)sin²(x + h)] Using limit formulas, limₕ→₀ (sin h)/ h = 1. f'(x) = -2csc²(x)cot(x) Hence, proved.</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [csc²(x + h) - csc²(x)] / h = limₕ→₀ [1/sin²(x + h) - 1/sin²(x)] / h = limₕ→₀ [[sin²(x) - sin²(x + h)] / [sin²(x)sin²(x + h)]] / h We now use the formula sin²(A) - sin²(B) = (A - B)(A + B). f'(x) = limₕ→₀ [(sin(x + h) - sin(x))(sin(x + h) + sin(x))] / [h sin²(x)sin²(x + h)] = limₕ→₀ [(sin h) cos((x + x + h)/2) 2cos((x - x - h)/2)] / [h sin²(x)sin²(x + h)] = limₕ→₀ (sin h)/ h · limₕ→₀ [cos((x + x + h)/2) · 2cos((x - x - h)/2)] / [sin²(x)sin²(x + h)] Using limit formulas, limₕ→₀ (sin h)/ h = 1. f'(x) = -2csc²(x)cot(x) Hence, proved.</p>
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<h2><strong>Using Chain Rule</strong></h2>
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<h2><strong>Using Chain Rule</strong></h2>
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<p>To prove the differentiation of csc²(x) using the chain rule, We use the formula: csc²(x) = (csc(x))² Let u = csc(x), so we have u² By chain rule: d/dx (u²) = 2u u' Let’s substitute u = csc(x) and u' = -csc(x)cot(x), d/dx (csc²(x)) = 2(csc(x))(-csc(x)cot(x)) = -2csc²(x)cot(x)</p>
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<p>To prove the differentiation of csc²(x) using the chain rule, We use the formula: csc²(x) = (csc(x))² Let u = csc(x), so we have u² By chain rule: d/dx (u²) = 2u u' Let’s substitute u = csc(x) and u' = -csc(x)cot(x), d/dx (csc²(x)) = 2(csc(x))(-csc(x)cot(x)) = -2csc²(x)cot(x)</p>
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<h2><strong>Using Product Rule</strong></h2>
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<h2><strong>Using Product Rule</strong></h2>
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<p>We will now prove the derivative of csc²(x) using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, csc²(x) = (csc(x))(csc(x)) Given that, u = csc(x) and v = csc(x) Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (csc(x)) = -csc(x)cot(x) (substitute u = csc(x)) v' = d/dx (csc(x)) = -csc(x)cot(x) (substitute v = csc(x))</p>
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<p>We will now prove the derivative of csc²(x) using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, csc²(x) = (csc(x))(csc(x)) Given that, u = csc(x) and v = csc(x) Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (csc(x)) = -csc(x)cot(x) (substitute u = csc(x)) v' = d/dx (csc(x)) = -csc(x)cot(x) (substitute v = csc(x))</p>
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<p>Again, use the product rule formula: d/dx (csc²(x)) = u'.v + u.v' Let’s substitute u = csc(x), u' = -csc(x)cot(x), v = csc(x), and v' = -csc(x)cot(x) When we simplify each<a>term</a>: We get, d/dx (csc²(x)) = -csc(x)cot(x)csc(x) + csc(x)(-csc(x)cot(x)) = -2csc²(x)cot(x) Thus: d/dx (csc²(x)) = -2csc²(x)cot(x).</p>
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<p>Again, use the product rule formula: d/dx (csc²(x)) = u'.v + u.v' Let’s substitute u = csc(x), u' = -csc(x)cot(x), v = csc(x), and v' = -csc(x)cot(x) When we simplify each<a>term</a>: We get, d/dx (csc²(x)) = -csc(x)cot(x)csc(x) + csc(x)(-csc(x)cot(x)) = -2csc²(x)cot(x) Thus: d/dx (csc²(x)) = -2csc²(x)cot(x).</p>
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