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

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(A) Given $y = x^{\sqrt{x}}$.Taking the natural logarithm on both sides,we get $\ln y = \ln(x^{\sqrt{x}})$.Using the property $\ln(a^b) = b \ln a$,we

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$x^{\sqrt{x}} \left( \frac{2 + \log x}{2\sqrt{x}} \right)$

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(A) Given $y = x^{\sqrt{x}}$.Taking the natural logarithm on both sides,we get $\ln y = \ln(x^{\sqrt{x}})$.Using the property $\ln(a^b) = b \ln a$,we have $\ln y = \sqrt{x} \ln x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x}) \cdot \ln x + \sqrt{x} \cdot \frac{d}{dx}(\ln x)$.$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \ln x + \sqrt{x} \cdot \frac{1}{x}$.Since $\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$,we have $\frac{1}{y} \frac{dy}{dx} = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}$.To combine the terms,write $\frac{1}{\sqrt{x}}$ as $\frac{2}{2\sqrt{x}}$:$\frac{1}{y} \frac{dy}{dx} = \frac{\ln x + 2}{2\sqrt{x}}$.Therefore,$\frac{dy}{dx} = y \left( \frac{2 + \ln x}{2\sqrt{x}} \right) = x^{\sqrt{x}} \left( \frac{2 + \ln x}{2\sqrt{x}} \right)$.

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

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