Let f: (-1, 1) → R be a differentiable function with f(0) = -1 and f'(0) = 1 . Let g(x) = [f(2f(x) + 2)]^2 . Then g'(0) = - Mathematics MCQ AIEEE 2026 | ExamTest.live

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Let f : (- 1, 1) \to R be a differentiable function with f (0) = - 1 and f^{'} (0) = 1 . Let g (x) = [f (2 f (x) + 2)]^{2} . Then g^{'} (0) =

Solution & Step-by-step Explanation

Using chain rule: g^{'} (x) = 2 [f (2 f (x) + 2)] \cdot f^{'} (2 f (x) + 2) \cdot (2 f^{'} (x)) .At x = 0 : g^{'} (0) = 2 [f (2 f (0) + 2)] \cdot f^{'} (2 f (0) + 2) \cdot 2 f^{'} (0) .Substitute f (0) = - 1 : g^{'} (0) = 2 [f (2 (- 1) + 2)] \cdot f^{'} (2 (- 1) + 2) \cdot 2 (1) g^{'} (0) = 2 [f (0)] \cdot f^{'} (0) \cdot 2 g^{'} (0) = 4 \cdot (- 1) \cdot 1 = - 4 .

Try it yourself before checking the explanation above.

Let f : (- 1, 1) \to R be a differentiable function with f (0) = - 1 and f^{'} (0) = 1 . Let g (x) = [f (2 f (x) + 2)]^{2} . Then g^{'} (0) =

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