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

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(D) Given,$y = \frac{e^x \log x}{x^2}$.Applying the quotient rule for differentiation,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given,$y = \frac{e^x \log x}{x^2}$.Applying the quotient rule for differentiation,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.Here,$u = e^x \log x$ and $v = x^2$.$\frac{du}{dx} = e^x \log x + e^x \cdot \frac{1}{x} = e^x \left( \log x + \frac{1}{x} \right)$.$\frac{dv}{dx} = 2x$.Therefore,$\frac{dy}{dx} = \frac{x^2 \left[ e^x \left( \log x + \frac{1}{x} \right) \right] - (e^x \log x)(2x)}{(x^2)^2}$.$\frac{dy}{dx} = \frac{x^2 e^x \log x + x e^x - 2x e^x \log x}{x^4}$.$\frac{dy}{dx} = \frac{x e^x \log x + e^x - 2 e^x \log x}{x^3}$.$\frac{dy}{dx} = \frac{e^x [x \log x + 1 - 2 \log x]}{x^3}$.$\frac{dy}{dx} = \frac{e^x [1 + (x - 2) \log x]}{x^3}$.Thus,option $D$ is correct.

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

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