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If $h(x)=x^{x^x}$, then at $x=1, \frac{h^{\prime}(x)}{h(x)}$ is equal to
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Verified Answer
The correct answer is:
$1+\log h(x)$
We have,
$\begin{aligned} h(x) & =x^{x^{x^*}} \\ \Rightarrow \quad \log h(x) & =x^x \log x\end{aligned}$
On differentiating w. r. t. $x$, we get
as $\begin{aligned} \frac{h^{\prime}(x)}{h(x)} & =x^x \cdot \frac{1}{x}+x^x(1+\log x) \log x \\ & =x^{x-1}+(1+\log x) \log h(x)[\text { from Eq. (i) }] \\ & =1+(1+\log 1) \log h(x) \\ & =1+\log h(x)\end{aligned}$
$\begin{aligned} h(x) & =x^{x^{x^*}} \\ \Rightarrow \quad \log h(x) & =x^x \log x\end{aligned}$
On differentiating w. r. t. $x$, we get
as $\begin{aligned} \frac{h^{\prime}(x)}{h(x)} & =x^x \cdot \frac{1}{x}+x^x(1+\log x) \log x \\ & =x^{x-1}+(1+\log x) \log h(x)[\text { from Eq. (i) }] \\ & =1+(1+\log 1) \log h(x) \\ & =1+\log h(x)\end{aligned}$
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