# How do you find derivative of f(x)=(x^2+4x+3)/(sqrtx)?

$y ' = \frac{\left(2 x + 4\right) \cdot \sqrt{x} - \left({x}^{2} + 4 x + 3\right) \cdot \frac{1}{2 \sqrt{x}}}{\sqrt{x}} ^ 2 =$
$= \frac{\frac{\left(2 x + 4\right) \sqrt{x} \cdot 2 \sqrt{x} - {x}^{2} - 4 x - 3}{2 \sqrt{x}}}{x} =$
$= \frac{\left(2 x + 4\right) \cdot 2 x - {x}^{2} - 4 x - 3}{2 x \sqrt{x}} =$
$= \frac{4 {x}^{2} + 8 x - {x}^{2} - 4 x - 3}{2 x \sqrt{x}} = \frac{3 {x}^{2} + 4 x - 3}{2 x \sqrt{x}}$.