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{d H}{d x} =e^{-e^{x}} \cdot\left(-e^{x}\right) $
$=-e^{x} e^{-e^{x}} $
$\therefore \left.\frac{d H}{d x}\right|_{x=0} =-e^{0} \cdot e^{-e^{\circ}}$
$=-1 \cdot e^{-1}$
$=-e^{-1}$
$=\frac{-1}{e}$