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

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(B) Given $y = \log(x^x)$.Using the property of logarithms,$\log(a^b) = b \log a$,we have:$y = x \log x$.Differentiating with respect to $x$ using the

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$\log(ex)$

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(B) Given $y = \log(x^x)$.Using the property of logarithms,$\log(a^b) = b \log a$,we have:$y = x \log x$.Differentiating with respect to $x$ using the product rule:$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x)$$\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x}$$\frac{dy}{dx} = \log x + 1$.Since $\log e = 1$,we can write:$\frac{dy}{dx} = \log x + \log e$.Using the property $\log a + \log b = \log(ab)$,we get:$\frac{dy}{dx} = \log(ex)$.Thus,the correct option is $B$.

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

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