Let f: (-1, 1) to R be a differentiable funct | vedclass

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(A) Given $g(x) = [f(2f(x) + 2)]^2$. Using the chain rule,we differentiate $g(x)$ with respect to $x$: $g'(x) = 2[f(2f(x) + 2)] \cdot \frac{d}{dx}[f(2

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$-4$

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(A) Given $g(x) = [f(2f(x) + 2)]^2$. Using the chain rule,we differentiate $g(x)$ with respect to $x$: $g'(x) = 2[f(2f(x) + 2)] \cdot \frac{d}{dx}[f(2f(x) + 2)]$ $g'(x) = 2[f(2f(x) + 2)] \cdot f'(2f(x) + 2) \cdot \frac{d}{dx}(2f(x) + 2)$ $g'(x) = 2[f(2f(x) + 2)] \cdot f'(2f(x) + 2) \cdot 2f'(x)$ $g'(x) = 4[f(2f(x) + 2)] \cdot f'(2f(x) + 2) \cdot f'(x)$ Now,substitute $x = 0$: $g'(0) = 4[f(2f(0) + 2)] \cdot f'(2f(0) + 2) \cdot f'(0)$ Given $f(0) = -1$ and $f'(0) = 1$: $g'(0) = 4[f(2(-1) + 2)] \cdot f'(2(-1) + 2) \cdot f'(0)$ $g'(0) = 4[f(0)] \cdot f'(0) \cdot f'(0)$ $g'(0) = 4(-1) \cdot (1) \cdot (1) = -4$.

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

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