Let f:(-1,1) rightarrow mathbb{R} be a differ | vedclass

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(D) Given $g(x) = (f(2f(x)+2))^2$. Applying the chain rule,we differentiate $g(x)$ with respect to $x$: $g^{\prime}(x) = 2(f(2f(x)+2)) \cdot \frac{d}{

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-$4$

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

Let $f:(-1,1) \rightarrow \mathbb{R}$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. If $g(x)=(f(2f(x)+2))^2$,then $g^{\prime}(0)=$

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