If y = frac{e^x log x}{x^2},then frac{dy}{dx} | vedclass

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(D) Given $y = \frac{e^x \log x}{x^2}$.Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^x \log x$ and $v = x

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$\frac{e^x[1 + (x - 2)\log x]}{x^3}$

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(D) Given $y = \frac{e^x \log x}{x^2}$.Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^x \log x$ and $v = x^2$:$\frac{dy}{dx} = \frac{x^2 \frac{d}{dx}(e^x \log x) - e^x \log x \frac{d}{dx}(x^2)}{(x^2)^2}$$\frac{dy}{dx} = \frac{x^2 (e^x \log x + e^x \cdot \frac{1}{x}) - e^x \log x (2x)}{x^4}$$\frac{dy}{dx} = \frac{x^2 e^x \log x + x e^x - 2x e^x \log x}{x^4}$$\frac{dy}{dx} = \frac{x e^x (x \log x + 1 - 2 \log x)}{x^4}$$\frac{dy}{dx} = \frac{e^x [1 + (x - 2) \log x]}{x^3}$.

If $y = \frac{e^x \log x}{x^2}$,then $\frac{dy}{dx} = $

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