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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 tan(2x) 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 tan(2x) 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 tan(2x) results in 2sec²(2x) using the above-mentioned methods:</p>
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</ul><p>We will now demonstrate that the differentiation of tan(2x) results in 2sec²(2x) 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 tan(2x) 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 tan(2x) using the first principle, we will consider f(x) = tan(2x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = tan(2x), we write f(x + h) = tan(2(x + h)). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [tan(2(x + h)) - tan(2x)] / h = limₕ→₀ [ [sin(2(x + h)) / cos(2(x + h))] - [sin(2x) / cos(2x)] ] / h = limₕ→₀ [ [sin(2x + 2h) cos(2x) - cos(2x + 2h) sin(2x)] / [cos(2x) · cos(2(x + h))] ] / h We now use the formula sin A cos B - cos A sin B = sin(A - B). f'(x) = limₕ→₀ [ sin(2h) ] / [ h cos(2x) · cos(2(x + h))] = limₕ→₀ [ sin(2h) ] / [ h cos(2x) · cos(2(x + h))] = limₕ→₀ (sin(2h)) / h · limₕ→₀ 1 / [cos(2x) · cos(2(x + h))] Using limit formulas, limₕ→₀ (sin(2h)) / h = 2. f'(x) = 2 [ 1 / (cos(2x) · cos(2x))] = 2/cos²(2x) As the reciprocal of cosine is secant, we have, f'(x) = 2sec²(2x). Hence, proved.</p>
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<p>The derivative of tan(2x) 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 tan(2x) using the first principle, we will consider f(x) = tan(2x). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = tan(2x), we write f(x + h) = tan(2(x + h)). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [tan(2(x + h)) - tan(2x)] / h = limₕ→₀ [ [sin(2(x + h)) / cos(2(x + h))] - [sin(2x) / cos(2x)] ] / h = limₕ→₀ [ [sin(2x + 2h) cos(2x) - cos(2x + 2h) sin(2x)] / [cos(2x) · cos(2(x + h))] ] / h We now use the formula sin A cos B - cos A sin B = sin(A - B). f'(x) = limₕ→₀ [ sin(2h) ] / [ h cos(2x) · cos(2(x + h))] = limₕ→₀ [ sin(2h) ] / [ h cos(2x) · cos(2(x + h))] = limₕ→₀ (sin(2h)) / h · limₕ→₀ 1 / [cos(2x) · cos(2(x + h))] Using limit formulas, limₕ→₀ (sin(2h)) / h = 2. f'(x) = 2 [ 1 / (cos(2x) · cos(2x))] = 2/cos²(2x) As the reciprocal of cosine is secant, we have, f'(x) = 2sec²(2x). 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 tan(2x) using the chain rule, We use the formula: Tan(2x) = sin(2x)/cos(2x) Consider f(x) = sin(2x) and g(x) = cos(2x) So we get, tan(2x) = 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) = sin(2x) and g(x) = cos(2x) in equation (1), d/dx (tan(2x)) = [(cos(2x) (2cos(2x)) - (sin(2x) (-2sin(2x)))] / (cos(2x))² = (2cos²(2x) + 2sin²(2x)) / cos²(2x) …(2) Here, we use the formula: (cos²(2x)) + (sin²(2x)) = 1 (Pythagorean identity) Substituting this into (2), d/dx (tan(2x)) = 2/ (cos²(2x)) Since sec(2x) = 1/cos(2x), we write: d/dx(tan(2x)) = 2sec²(2x)</p>
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<p>To prove the differentiation of tan(2x) using the chain rule, We use the formula: Tan(2x) = sin(2x)/cos(2x) Consider f(x) = sin(2x) and g(x) = cos(2x) So we get, tan(2x) = 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) = sin(2x) and g(x) = cos(2x) in equation (1), d/dx (tan(2x)) = [(cos(2x) (2cos(2x)) - (sin(2x) (-2sin(2x)))] / (cos(2x))² = (2cos²(2x) + 2sin²(2x)) / cos²(2x) …(2) Here, we use the formula: (cos²(2x)) + (sin²(2x)) = 1 (Pythagorean identity) Substituting this into (2), d/dx (tan(2x)) = 2/ (cos²(2x)) Since sec(2x) = 1/cos(2x), we write: d/dx(tan(2x)) = 2sec²(2x)</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 tan(2x) using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, Tan(2x) = sin(2x)/cos(2x) tan(2x) = (sin(2x)). (cos(2x))⁻¹ Given that, u = sin(2x) and v = (cos(2x))⁻¹ Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (sin(2x)) = 2cos(2x). (substitute u = sin(2x)) Here we use the chain rule: v = (cos(2x))⁻¹ = (cos(2x))⁻¹ (substitute v = (cos(2x))⁻¹) v' = -1. (cos(2x))⁻². d/dx (cos(2x)) v' = 2sin(2x)/ (cos(2x))² Again, use the product rule formula: d/dx (tan(2x)) = u'. v + u. v' Let’s substitute u = sin(2x), u' = 2cos(2x), v = (cos(2x))⁻¹, and v' = 2sin(2x)/ (cos(2x))² When we simplify each<a>term</a>: We get, d/dx (tan(2x)) = 2 + 2sin²x / (cos(2x))² Sin²x / (cos(2x))² = tan²(2x) (we use the identity sin²x + cos²x =1) Thus: d/dx (tan(2x)) = 2 + 2tan²(2x) Since, 2 + 2tan²(2x) = 2sec²(2x) d/dx (tan(2x)) = 2sec²(2x).</p>
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<p>We will now prove the derivative of tan(2x) using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, Tan(2x) = sin(2x)/cos(2x) tan(2x) = (sin(2x)). (cos(2x))⁻¹ Given that, u = sin(2x) and v = (cos(2x))⁻¹ Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (sin(2x)) = 2cos(2x). (substitute u = sin(2x)) Here we use the chain rule: v = (cos(2x))⁻¹ = (cos(2x))⁻¹ (substitute v = (cos(2x))⁻¹) v' = -1. (cos(2x))⁻². d/dx (cos(2x)) v' = 2sin(2x)/ (cos(2x))² Again, use the product rule formula: d/dx (tan(2x)) = u'. v + u. v' Let’s substitute u = sin(2x), u' = 2cos(2x), v = (cos(2x))⁻¹, and v' = 2sin(2x)/ (cos(2x))² When we simplify each<a>term</a>: We get, d/dx (tan(2x)) = 2 + 2sin²x / (cos(2x))² Sin²x / (cos(2x))² = tan²(2x) (we use the identity sin²x + cos²x =1) Thus: d/dx (tan(2x)) = 2 + 2tan²(2x) Since, 2 + 2tan²(2x) = 2sec²(2x) d/dx (tan(2x)) = 2sec²(2x).</p>
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