The derivative of y = x^{ln x} is | vedclass

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(C) Given $y = x^{\ln x}$.Taking the natural logarithm on both sides,we get $\ln y = \ln(x^{\ln x})$.Using the property $\ln(a^b) = b \ln a$,we have $

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$2x^{\ln x - 1} \ln x$

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(C) Given $y = x^{\ln x}$.Taking the natural logarithm on both sides,we get $\ln y = \ln(x^{\ln x})$.Using the property $\ln(a^b) = b \ln a$,we have $\ln y = (\ln x)(\ln x) = (\ln x)^2$.Differentiating both sides with respect to $x$ using the chain rule:$\frac{1}{y} \frac{dy}{dx} = 2(\ln x) \cdot \frac{d}{dx}(\ln x)$.$\frac{1}{y} \frac{dy}{dx} = 2(\ln x) \cdot \frac{1}{x}$.$\frac{dy}{dx} = y \cdot \frac{2 \ln x}{x}$.Substituting $y = x^{\ln x}$ back into the equation:$\frac{dy}{dx} = x^{\ln x} \cdot \frac{2 \ln x}{x} = 2 x^{\ln x} \cdot x^{-1} \cdot \ln x$.$\frac{dy}{dx} = 2 x^{\ln x - 1} \ln x$.

The derivative of $y = x^{\ln x}$ is

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