Question

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

• A. 4
• B. $-4$
• C. 0
• D. $-2$

Answer 1

Answer: (B)

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