Let f(x) = (x^x)^x and g(x) = x^{(x^x)},then: | vedclass

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(D) Given $f(x) = (x^x)^x = x^{x^2}$.Taking the natural logarithm on both sides,$\ln(f(x)) = x^2 \ln(x)$.Differentiating with respect to $x$,$\frac{f'

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$f'(1) = 1$ and $g'(1) = 1$

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(D) Given $f(x) = (x^x)^x = x^{x^2}$.Taking the natural logarithm on both sides,$\ln(f(x)) = x^2 \ln(x)$.Differentiating with respect to $x$,$\frac{f'(x)}{f(x)} = 2x \ln(x) + x^2 \cdot \frac{1}{x} = 2x \ln(x) + x$.So,$f'(x) = f(x) (2x \ln(x) + x) = x^{x^2} (2x \ln(x) + x)$.At $x = 1$,$f'(1) = 1^1 (2(1) \ln(1) + 1) = 1(0 + 1) = 1$.Now,$g(x) = x^{(x^x)}$.Taking the natural logarithm,$\ln(g(x)) = x^x \ln(x)$.Differentiating with respect to $x$,$\frac{g'(x)}{g(x)} = \frac{d}{dx}(x^x) \ln(x) + x^x \cdot \frac{1}{x}$.We know $\frac{d}{dx}(x^x) = x^x(1 + \ln(x))$.So,$\frac{g'(x)}{g(x)} = x^x(1 + \ln(x)) \ln(x) + x^{x-1}$.At $x = 1$,$g(1) = 1^{(1^1)} = 1$.$g'(1) = g(1) [1^1(1 + \ln(1)) \ln(1) + 1^{1-1}] = 1 [1(1 + 0)(0) + 1] = 1(0 + 1) = 1$.Thus,$f'(1) = 1$ and $g'(1) = 1$.

Let $f(x) = (x^x)^x$ and $g(x) = x^{(x^x)}$,then:

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