Question

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

• A. 4
• B. -4
• C. 0.0
• D. -2

Answer 1

The correct option is B -4

$g\left(x\right)=\left(f\right(2f\left(x\right)+2){)}^2\Rightarrow g^{\prime }(x)=2(f\left(2f\right(x)+2))\times f^{\prime }(2f\left(x\right)+2)\times 2f^{\prime }(x)\Rightarrow g^{\prime }(0)=2(f\left(2f\right(0)+2))\times f^{\prime }(2f\left(0\right)+2)\times 2f^{\prime }(0\left)g^{\prime }\right(0)=-4$