If y = x^2 log x + frac{2}{sqrt{x}}, then fra | vedclass

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(B) Given $y = x^2 \log x + 2x^{-1/2}$.Applying the product rule to $x^2 \log x$ and the power rule to $2x^{-1/2}$:$\frac{dy}{dx} = \frac{d}{dx}(x^2 \

Vedclass · Accepted Answer

$x + 2x \log x - \frac{1}{x^{3/2}}$

Vedclass · Answer

(B) Given $y = x^2 \log x + 2x^{-1/2}$.Applying the product rule to $x^2 \log x$ and the power rule to $2x^{-1/2}$:$\frac{dy}{dx} = \frac{d}{dx}(x^2 \log x) + \frac{d}{dx}(2x^{-1/2})$$\frac{dy}{dx} = (x^2 \cdot \frac{1}{x} + \log x \cdot 2x) + 2(-\frac{1}{2}x^{-3/2})$$\frac{dy}{dx} = x + 2x \log x - x^{-3/2}$$\frac{dy}{dx} = x + 2x \log x - \frac{1}{x^{3/2}}$Thus,the correct option is $B$.

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

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