Let y be an implicit function of x defined by x^2x - 2x^x y - 1 = 0 . Then y'(1) equals: - Mathematics MCQ AIEEE 2009 2026 | ExamTest.live

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Let y be an implicit function of x defined by x^{2 x} - 2 x^{x} cot y - 1 = 0 . Then y^{'} (1) equals:

Solution & Step-by-step Explanation

At x = 1 : 1^{2} - 2 (1) cot y - 1 = 0 ⟹ - 2 cot y = 0 ⟹ cot y = 0 ⟹ y = π /2 .Differentiating x^{2 x} - 2 x^{x} cot y - 1 = 0 : \frac{d}{d x} (x^{2 x}) - 2 [\frac{d}{d x} (x^{x}) cot y + x^{x} (- cosec^{2} y) y^{'}] = 0 .We know \frac{d}{d x} (x^{a x}) = x^{a x} \cdot a (1 + lo g x) .At (1, π /2) : 1^{2} \cdot 2 (1 + 0) - 2 [1^{1} (1 + 0) \cdot 0 + 1^{1} (- 1) y^{'}] = 0 2 - 2 [- y^{'}] = 0 ⟹ 2 + 2 y^{'} = 0 ⟹ y^{'} = - 1 .

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Let y be an implicit function of x defined by x^{2 x} - 2 x^{x} cot y - 1 = 0 . Then y^{'} (1) equals:

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lo g 2

- lo g 2

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