Let F(x)=e^{x},G(x)=e^{-x} and H(x)=G(F(x)),w | vedclass

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(C) Given functions are $F(x)=e^{x}$ and $G(x)=e^{-x}$.We define $H(x) = G(F(x))$.Substituting $F(x)$ into $G(x)$,we get $H(x) = G(e^{x}) = e^{-(e^{x}

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$-\frac{1}{e}$

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(C) Given functions are $F(x)=e^{x}$ and $G(x)=e^{-x}$.We define $H(x) = G(F(x))$.Substituting $F(x)$ into $G(x)$,we get $H(x) = G(e^{x}) = e^{-(e^{x})}$.Now,we differentiate $H(x)$ with respect to $x$ using the chain rule:$\frac{dH}{dx} = \frac{d}{dx}(e^{-e^{x}}) = e^{-e^{x}} \cdot \frac{d}{dx}(-e^{x}) = e^{-e^{x}} \cdot (-e^{x}) = -e^{x} \cdot e^{-e^{x}}$.To find the value at $x=0$,we substitute $x=0$ into the derivative:$\left. \frac{dH}{dx} \right|_{x=0} = -e^{0} \cdot e^{-e^{0}} = -1 \cdot e^{-1} = -\frac{1}{e}$.

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

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