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Let $F(x)=e^{x}, G(x)=e^{-x}$ and $H(x)=G(F(x)),$ where $x$ is a real variable. Then, $\frac{d H}{d x}$ at $x=0$ is
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Verified Answer
The correct answer is:
$-\frac{1}{e}$
We have, $F(x)=e^{x}, G(x)=e^{-x}$ $\therefore \quad
H(x)=G(F(x))$ $\begin{aligned} &=G\left(e^{x}\right) \\ &=e^{-e^{x}} \\ \frac{d H}{d x} &=e^{-e^{x}} \cdot\left(-e^{x}\right) \\ &=-e^{x} e^{-e^{x}} \end{aligned}$
$\left.\therefore \quad \frac{d H}{d x}\right|_{x=0}=-e^{0} \cdot e^{-e^{0}}$
$=-1 . e^{-1}$
$=-e^{-1}$
$=\frac{-1}{e}$
H(x)=G(F(x))$ $\begin{aligned} &=G\left(e^{x}\right) \\ &=e^{-e^{x}} \\ \frac{d H}{d x} &=e^{-e^{x}} \cdot\left(-e^{x}\right) \\ &=-e^{x} e^{-e^{x}} \end{aligned}$
$\left.\therefore \quad \frac{d H}{d x}\right|_{x=0}=-e^{0} \cdot e^{-e^{0}}$
$=-1 . e^{-1}$
$=-e^{-1}$
$=\frac{-1}{e}$
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