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If $f(x)=e^x, h(x)=($ fof $)(x)$, then $\frac{h^{\prime}(x)}{h(x)}=$
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The correct answer is:
$\log h(x)$
$f(x)=e^x$
$\begin{aligned} & h(x)=f(f(x))=e^{e^x} \\ & h^{\prime}(x)=e^{e^x} \cdot e^x=h(x) \cdot e^x \Rightarrow \frac{h^{\prime}(x)}{h(x)}=e^x \\ & \log h(x)=\log e^{e^x}=e^x \cdot \log e=e^x \\ & \therefore \frac{h^{\prime}(x)}{h(x)}=\log h(x)\end{aligned}$
$\begin{aligned} & h(x)=f(f(x))=e^{e^x} \\ & h^{\prime}(x)=e^{e^x} \cdot e^x=h(x) \cdot e^x \Rightarrow \frac{h^{\prime}(x)}{h(x)}=e^x \\ & \log h(x)=\log e^{e^x}=e^x \cdot \log e=e^x \\ & \therefore \frac{h^{\prime}(x)}{h(x)}=\log h(x)\end{aligned}$
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