determine the derivative of $\displaystyle g(x)=\sqrt{x}(2x+3)^2$
You can clean it up with some factorisation...
$\displaystyle \frac{1}{2}x^{-\frac{1}{2}}(2x + 3)^2 + 4x^{\frac{1}{2}}(2x + 3) = (2x + 3)\left(\frac{2x + 3}{2\sqrt{x}} + 4\sqrt{x}\right)$
$\displaystyle = (2x+3)\left(\frac{2x + 3 + 8x}{2\sqrt{x}}\right)$
$\displaystyle = \frac{(2x+3)(10x+3)}{2\sqrt{x}}$