If y = (x^x)^x,then frac{dy}{dx} = | vedclass

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(D) Given $y = (x^x)^x$. Using the exponent rule $(a^m)^n = a^{mn}$,we have $y = x^{x \cdot x} = x^{x^2}$. Taking the natural logarithm on both sides:

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$x \cdot x^{x^2}(2 \log x+1)$

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(D) Given $y = (x^x)^x$. Using the exponent rule $(a^m)^n = a^{mn}$,we have $y = x^{x \cdot x} = x^{x^2}$. Taking the natural logarithm on both sides: $\log y = \log(x^{x^2}) = x^2 \log x$. Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^2 \log x)$. Using the product rule: $\frac{1}{y} \frac{dy}{dx} = x^2 \cdot \frac{1}{x} + (\log x) \cdot (2x) = x + 2x \log x$. $\frac{dy}{dx} = y(x + 2x \log x) = x^{x^2} \cdot x(1 + 2 \log x)$. Thus,$\frac{dy}{dx} = x \cdot x^{x^2}(1 + 2 \log x)$.

If $y = (x^x)^x$,then $\frac{dy}{dx} = $

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