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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Aicharya · Accepted Answer

We have, $F(x)=e^{x}, G(x)=e^{-x}$ $	herefore \quad H(x)=G(F(x))$ $\begin{aligned} &=G\left(e^{x}ight) \ &=e^{-e^{x}} \ \frac{d H}{d x} &=e^{-e^{x}} \cdot\left(-e^{x}ight) \ &=-e^{x} e^{-e^{x}} \end{aligned}$ $\left.	herefore \quad \frac{d H}{d x}ight|_{x=0}=-e^{0} \cdot e^{-e^{0}}$ $=-1 . e^{-1}$ $=-e^{-1}$ $=\frac{-1}{e}$

Let $F(x)=e^{x}, G(x)=e^{-x}$ and $H(x)=G(F(x)),$ where $x$ is a real variable. Then, $\frac{d H}{d x}$ at…

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