What is h^{\prime}(0) equal to?
Direction: Consider the following for the two (02) items that follow : Let f:(-1,1)\rightarrow R be a differentiable function with f(0)=-1 and f^{\prime}(0)=1. Let h(x)=f(2f(x)+2) and g(x)=(h(x))^{2}.
- A. -2
- B. -1
- C. 0
- D. 2 ✓
Correct Answer: D. 2
Explanation
Using the chain rule, h'(x) = f'(2f(x)+2) \cdot (2f'(x)). At x=0, h'(0) = f'(2f(0)+2) \cdot 2f'(0). Substituting f(0)=-1 gives h'(0) = f'(2(-1)+2) \cdot 2(1) = f'(0) \cdot 2. Since f'(0)=1, h'(0) = 1 \cdot 2 = 2.
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