If h(x) = x^{x^x},then at x = 1,frac{h'(x)}{h | 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 $x$ u

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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 chain rule:$\frac{h'(x)}{h(x)} = \frac{d}{dx}(x^x) \cdot \log x + x^x \cdot \frac{d}{dx}(\log x)$.We know that $\frac{d}{dx}(x^x) = x^x(1 + \log x)$.Substituting this into the equation:$\frac{h'(x)}{h(x)} = x^x(1 + \log x) \log x + x^x \cdot \frac{1}{x} = x^x(1 + \log x) \log x + x^{x-1}$.At $x = 1$,we have $h(1) = 1^{1^1} = 1$,so $\log h(1) = \log 1 = 0$.Substituting $x = 1$ into the expression for $\frac{h'(x)}{h(x)}$:$\frac{h'(1)}{h(1)} = 1^1(1 + \log 1) \log 1 + 1^{1-1} = 1(1 + 0)(0) + 1^0 = 0 + 1 = 1$.Since $\log h(1) = 0$,the expression $1 + \log h(1) = 1 + 0 = 1$.Thus,at $x = 1$,$\frac{h'(x)}{h(x)} = 1 + \log h(x)$.

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

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