A variable base and exponent

Take natural logarithms: $\ln y=\sqrt{x}\ln x$. Differentiate using the product rule: $\dfrac1y\dfrac{dy}{dx}=\sqrt{x}\cdot\dfrac1x+\ln x\cdot\dfrac12x^{-1/2}=\dfrac1{\sqrt{x}}+\dfrac{\ln x}{2\sqrt{x}}$. Multiply by $y=x^{\sqrt{x}}$ to obtain $\dfrac{dy}{dx}=\dfrac{x^{\sqrt{x}}(2+\ln x)}{2\sqrt{x}}$.