Question

F(x) = (x^2 - 4) / √x F'(x) = ...?

Answer 1

$f(x)=\dfrac{x^2-4}{\sqrt{x}}\\\\f'(x)=\dfrac{(x^2-4)'(\sqrt{x})-(x^2-4)(\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{2x\sqrt{x}-\dfrac{1}{2\sqrt{x}}(x^2-4)}{x}\\\\=\dfrac{\dfrac{2x\sqrt{x}\cdot2\sqrt{x}}{2\sqrt{x}}-\dfrac{x^2-4}{2\sqrt{x}}}{x}=\dfrac{4x^2-x^2+4}{2x\sqrt{x}}=\dfrac{3x^2+4}{2x\sqrt{x}}$