Question

If $f:(–1,1)\to R$ be a differentiable function with $f\left(0\right)=–1andf\text{'}\left(0\right)=1.$ Let $g\left(x\right)={\left[f\right\{2f\left(x\right)+2\}]}^2$. Then,$g\text{'}\left(0\right)$is equal to

Answer 1

Finding the value of $g\text{'}\left(0\right)$:

Given that $f:(–1,1)\to R$ is a differentiable function and $f\left(0\right)=–1andf\text{'}\left(0\right)=1.$ $$\begin{gathered} g\left(x\right)={[f\{2f\left(x\right)+2\}]}^2 \\ {g}^{\text{'}}\left(x\right)=2(f\{2f\left(x\right)+2\})f^{\text{'}}\{2f(x)+2\})2f^{\text{'}}\left(x\right) \\ {g}^{\text{'}}\left(0\right)=2(f\{2f\left(0\right)+2\})f^{\text{'}}\{2f\left(0\right)+2\left\}\right)2f^{\text{'}}\left(0\right) \\ {g}^{\text{'}}\left(0\right)=2(f\{2\times -1+2\left\}\right)f^{\text{'}}\{2\times -1+2\})2\times 1 \\ {g}^{\text{'}}(0)=2f(0)f^{\text{'}}(0)2 \\ {g}^{\text{'}}(0)=2\times -1\times 1\times 2 \\ \therefore g^{\text{'}}(0)=-4 \end{gathered}$$

Hence, The value of $g\text{'}\left(0\right)$is equal to $-4$