Question

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

• A. $-4$
• B. $0$
• C. $-2$
• D. $4$

Answer 1

The correct option is A $-4$

$g^{\prime }\left(x\right)=2\left(f\right(2f\left(x\right)+2\left)\right)\left(\frac{d}{dx}\right(f\left(2f\right(x)+2)\left)\right)=2f\left(2f\right(x)+2)f^{\prime }\left(2f\right(x)+2).\left(2f^{\prime }\right(x\left)\right)$

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

$=4(-1)\left(1\right)=-4$

Hence, option 'A' is correct.