If g(x) = [f(2f(x) + 2)]^2 and f(0) = -1, f'( | vedclass

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(A) Given $g(x) = [f(2f(x) + 2)]^2$. Applying the chain rule to differentiate $g(x)$ with respect to $x$: $g'(x) = 2[f(2f(x) + 2)] \cdot f'(2f(x) + 2)

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-$4$

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

If $g(x) = [f(2f(x) + 2)]^2$ and $f(0) = -1, f'(0) = 1$,then $g'(0)$ is

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