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Question:

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)=

-8

-6

0

4

Solution:

g'(x) = 2(f(2f(x)+2))(d/dx(f(2f(x)+2))) = 2f(2f(x)+2)f'(2f(x)+2) * (2f'(x))
=> g'(0) = 2f(2f(0)+2)f'(2f(0)+2)*2(f'(0)) = 4f(0)f'(2) = 4(-1)(f'(2))