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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 -csc 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 -csc 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 -csc x results in csc(x) cot(x) using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of -csc x results in csc(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 -csc 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 -csc x using the first principle, we will consider f(x) = -csc 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) = -csc x, we write f(x + h) = -csc (x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [-csc(x + h) + csc x] / h = limₕ→₀ [-1/sin(x + h) + 1/sin x] / h Using the identity 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 + h) sin x] = limₕ→₀ [-2 cos((2x + h)/2) (sin(h/2)/(h/2))] / [sin(x + h) sin x] Using limit formulas, limₕ→₀ (sin(h/2)/(h/2)) = 1. f'(x) = -2 cos(x) / [sin^2 x] = -2 cot x / sin x = csc(x) cot(x). Hence, proved.</p>
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<p>The derivative of -csc 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 -csc x using the first principle, we will consider f(x) = -csc 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) = -csc x, we write f(x + h) = -csc (x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [-csc(x + h) + csc x] / h = limₕ→₀ [-1/sin(x + h) + 1/sin x] / h Using the identity 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 + h) sin x] = limₕ→₀ [-2 cos((2x + h)/2) (sin(h/2)/(h/2))] / [sin(x + h) sin x] Using limit formulas, limₕ→₀ (sin(h/2)/(h/2)) = 1. f'(x) = -2 cos(x) / [sin^2 x] = -2 cot x / sin x = csc(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 -csc x using the chain rule, We use the formula: -csc x = -1/sin x Consider f(x) = -1 and g(x) = sin x So, we get -csc x = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]²… (1) Let’s substitute f(x) = -1 and g(x) = sin x in equation (1), d/dx (-csc x) = [(0)(sin x) - (-1)(cos x)] / (sin x)² = cos x / sin² x = cot x / sin x Since csc x = 1/sin x, we write: d/dx(-csc x) = csc(x) cot(x)</p>
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<p>To prove the differentiation of -csc x using the chain rule, We use the formula: -csc x = -1/sin x Consider f(x) = -1 and g(x) = sin x So, we get -csc x = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]²… (1) Let’s substitute f(x) = -1 and g(x) = sin x in equation (1), d/dx (-csc x) = [(0)(sin x) - (-1)(cos x)] / (sin x)² = cos x / sin² x = cot x / sin x Since csc x = 1/sin x, we write: d/dx(-csc x) = csc(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 -csc x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, -csc x = -1/sin x -csc x = (-1) · (sin x)⁻¹ Given that, u = -1 and v = (sin x)⁻¹ Using the product rule formula: d/dx [u·v] = u'·v + u·v' u' = d/dx (-1) = 0. Here we use the chain rule: v = (sin x)⁻¹ (substitute v = (sin x)⁻¹) v' = -1(sin x)⁻² (cos x) v' = -cos x / sin² x Again, use the product rule formula: d/dx (-csc x) = u'·v + u·v' Let’s substitute u = -1, u' = 0, v = (sin x)⁻¹, and v' = -cos x / sin² x When we simplify each<a>term</a>: We get, d/dx (-csc x) = -(-cos x / sin² x) = cos x / sin² x = cot x / sin x = csc(x) cot(x).</p>
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<p>We will now prove the derivative of -csc x using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, -csc x = -1/sin x -csc x = (-1) · (sin x)⁻¹ Given that, u = -1 and v = (sin x)⁻¹ Using the product rule formula: d/dx [u·v] = u'·v + u·v' u' = d/dx (-1) = 0. Here we use the chain rule: v = (sin x)⁻¹ (substitute v = (sin x)⁻¹) v' = -1(sin x)⁻² (cos x) v' = -cos x / sin² x Again, use the product rule formula: d/dx (-csc x) = u'·v + u·v' Let’s substitute u = -1, u' = 0, v = (sin x)⁻¹, and v' = -cos x / sin² x When we simplify each<a>term</a>: We get, d/dx (-csc x) = -(-cos x / sin² x) = cos x / sin² x = cot x / sin x = csc(x) cot(x).</p>
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