Question

Suppose for a differentiable function $f,f\left(0\right)=0,f\left(1\right)=1,f^{\prime }\left(0\right)=4=f^{\prime }\left(1\right)$.
If $g\left(x\right)=f\left(e^x\right)e^{f\left(x\right)}$, then $g^{\prime }\left(0\right)$ is equal to

• A. $4$
• B. $8$
• C. $2$
• D. $0$

Answer 1

Answer: (B)

$8$

$g\left(x\right)=f\left(e^x\right)e^{f\left(x\right)}$

Differentiating both sides, we get

$g^{\prime }\left(x\right)=f^{\prime }\left(e^x\right)\cdot e^x\cdot e^{f\left(x\right)}+f\left(e^x\right)\cdot e^{f\left(x\right)}\cdot f^{\prime }\left(x\right)$

$g^{\prime }\left(0\right)=f^{\prime }\left(1\right)\cdot e^{f\left(0\right)}+f\left(1\right)\cdot e^{f\left(0\right)}\cdot f^{\prime }\left(0\right)$

$g^{\prime }\left(0\right)=4+4=8$