Question

Let $F(x)=e^x,G(x)=e^{-x}$ and $H(x)=G(F(x))$, 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)

We have, $F(x)=e^x,G(x)=e^{-x}$
$\therefore H(x)=G(F(x))$
$=G\left(e^x\right)$
$=e^{-e^x}$
$\therefore \frac{dH}{dx}=e^{-e^x}\cdot \left(-e^x\right)$
$=-e^x{e}^{-e^x}$
$\therefore {\frac{dH}{dx}|}_{x=0}=-e^0\cdot e^{-e^{\circ }}$
$=-1\cdot e^{-1}$
$=-e^{-1}$
$=\frac{-1}{e}$