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  4. If f:(-1,1) arrow R be a differentiable function with f(0)=-1 and f prime(0)=1. Let g(x)=[f(2 f(x)+2)]2. Then, g prime(0) is equal to

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Q. If $f:(-1,1) \rightarrow R$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. Let $g(x)=[f(2 f(x)+2)]^2$. Then, $g^{\prime}(0)$ is equal to

Continuity and Differentiability

Solution:

Given, $g(x)=\left[f(2 f(x)+2]^2\right.$
$ \therefore g^{\prime}(x)=2 f(2 f(x)+2) \cdot f^{\prime}(2 f(x)+2) \cdot 2 f^{\prime}(x) $
$ =4 f(2 f(x)+2) f^{\prime}(2 f(x)+2) f^{\prime}(x) $
$g^{\prime}(0)=4 f(2 f(0)+2) f^{\prime}(2 f(0)+2) f^{\prime}(0) $
$=4 f(0) f^{\prime}(0) f^{\prime}(0) $
$ =-4 $
$\therefore g^{\prime}(0)=4 f(0) f^{\prime}(0) f^{\prime}(0)=-4$