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

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(A) Given $y = x^x$. Taking $\log$ on both sides,we get $\log y = x \log x$. Differentiating both sides with respect to $x$,we use the product rule on

Vedclass · Accepted Answer

$x^x \log(ex)$

Vedclass · Answer

(A) Given $y = x^x$. Taking $\log$ on both sides,we get $\log y = x \log x$. Differentiating both sides with respect to $x$,we use the product rule on the right side: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x)$ $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x}$ $\frac{1}{y} \frac{dy}{dx} = \log x + 1$ Therefore,$\frac{dy}{dx} = y(1 + \log x) = x^x(1 + \log x)$. Since $1 = \log e$,we can write $1 + \log x = \log e + \log x = \log(ex)$. Thus,$\frac{dy}{dx} = x^x \log(ex)$.

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

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