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

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(C) Given $y = ({x^x})^x$. Using the property of exponents $(a^m)^n = a^{mn}$,we have $y = x^{x^2}$. Taking the natural logarithm on both sides: $\ln

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

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(C) Given $y = ({x^x})^x$. Using the property of exponents $(a^m)^n = a^{mn}$,we have $y = x^{x^2}$. Taking the natural logarithm on both sides: $\ln y = \ln(x^{x^2}) = x^2 \ln x$. Differentiating both sides with respect to $x$ using the product rule: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x)$. $\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x$. $\frac{1}{y} \frac{dy}{dx} = x(1 + 2 \ln x)$. Therefore,$\frac{dy}{dx} = y \cdot x(1 + 2 \ln x) = x({x^x})^x(1 + 2 \ln x)$.

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

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