If h(x) = x^{x^x}, then at x = 1, frac{h^{pri | vedclass

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(C) Given $h(x) = x^{x^x}$.Taking the natural logarithm on both sides, we get $\log h(x) = x^x \log x$.Differentiating both sides with respect to $x$

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$1 + \log h(x)$

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(C) Given $h(x) = x^{x^x}$.Taking the natural logarithm on both sides, we get $\log h(x) = x^x \log x$.Differentiating both sides with respect to $x$ using the product rule and the chain rule:$\frac{d}{dx}(\log h(x)) = \frac{d}{dx}(x^x \log x)$$\frac{h^{\prime}(x)}{h(x)} = \frac{d}{dx}(x^x) \cdot \log x + x^x \cdot \frac{d}{dx}(\log x)$Since $\frac{d}{dx}(x^x) = x^x(1 + \log x)$, we have:$\frac{h^{\prime}(x)}{h(x)} = x^x(1 + \log x) \log x + x^x \cdot \frac{1}{x}$$\frac{h^{\prime}(x)}{h(x)} = x^x(1 + \log x) \log x + x^{x-1}$At $x = 1$:$\frac{h^{\prime}(1)}{h(1)} = 1^1(1 + \log 1) \log 1 + 1^{1-1}$$\frac{h^{\prime}(1)}{h(1)} = 1(1 + 0)(0) + 1^0 = 0 + 1 = 1$.Now, check the options at $x = 1$:$h(1) = 1^{1^1} = 1$.Option $C$ is $1 + \log h(x) = 1 + \log(1) = 1 + 0 = 1$.Thus, the value is $1 + \log h(x)$.

If $h(x) = x^{x^x}$, then at $x = 1$, $\frac{h^{\prime}(x)}{h(x)}$ is equal to

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