Question

Let $F\left(x\right)=e^x,G\left(x\right)=e^{-x}$ and $H\left(x\right)=G\left(F\right(x\left)\right),$ where $x$ is a real variable. Then $\frac{dH}{dx}$ at $x=0$ is

• A. $1$
• B. $-1$
• C. $-\frac{1}{e}$
• D. $-e$

Answer 1

The correct option is C $-\frac{1}{e}$

$F^{\prime }\left(x\right)=e^x$ and $G^{\prime }\left(x\right)=-e^{-x}$
$H\left(x\right)=G\left(F\right(x\left)\right)$
$\Rightarrow \frac{dH}{dx}=G^{\prime }\left(F\right(x\left)\right)\cdot F^{\prime }\left(x\right)$

$\therefore \frac{dH}{dx}{|}_{x=0}=G^{\prime }\left(F\right(0\left)\right)\cdot F^{\prime }\left(0\right)$

$=-e^{-1}\cdot 1=-\frac{1}{e}$