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

Answer: (C)

$-\frac{1}{e}$

Given : $F\left(x\right)=e^x$ and $G\left(x\right)=e^{-x}$

$H\left(x\right)=G\left(F\right(x\left)\right)=G\left(e^x\right)=e^{-e^x}$
$\therefore \frac{dH}{dx}=H^{\prime }\left(x\right)=-e^{-e^x}.e^x$
$⟹H^{\prime }\left(0\right)=-e^{-e^0}\cdot e^0=-\frac{1}{e}$

Hence, $\frac{dH}{dx}$ at $x=0$ is $-\frac{1}{e}$.