Let y = f≤ft(x+e^x/e^x) satisfy f'(1) = 2 . Then the value of dy/dx at x = 0 equals: - Mathematics MCQ Mathematics Mock Test - 11 2026 | ExamTest.live

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1 mark

Let y = f (\frac{x + e ^{x}}{e ^{x}}) satisfy f^{'} (1) = 2 . Then the value of \frac{d y}{d x} at x = 0 equals:

Solution & Step-by-step Explanation

Let u = \frac{x + e ^{x}}{e ^{x}} = x e^{- x} + 1 .Then y = f (u) . By Chain Rule: \frac{d y}{d x} = f^{'} (u) \cdot \frac{d u}{d x} Differentiating u : \frac{d u}{d x} = \frac{d}{d x} (x e^{- x} + 1) = 1 \cdot e^{- x} + x \cdot (- e^{- x}) = e^{- x} (1 - x) At x = 0 : u = \frac{0 + e ^{0}}{e ^{0}} = \frac{1}{1} = 1 \frac{d u}{d x}_{x = 0} = e^{0} (1 - 0) = 1 Given f^{'} (1) = 2 : \frac{d y}{d x}_{x = 0} = f^{'} (1) \cdot 1 = 2 \cdot 1 = 2

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Let y = f (\frac{x + e ^{x}}{e ^{x}}) satisfy f^{'} (1) = 2 . Then the value of \frac{d y}{d x} at x = 0 equals:

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