Answer
$\dfrac{1}{2}x^{-1/2}(3x+4)^{1/2}-\dfrac{3}{2}x^{1/2}(3x+4)^{-1/2}=\dfrac{2}{\sqrt{3x^{2}+4x}}$
Work Step by Step
$\dfrac{1}{2}x^{-1/2}(3x+4)^{1/2}-\dfrac{3}{2}x^{1/2}(3x+4)^{-1/2}$
Take out common factor $\dfrac{1}{2}x^{-1/2}(3x+4)^{-1/2}$:
$\dfrac{1}{2}x^{-1/2}(3x+4)^{1/2}-\dfrac{3}{2}x^{1/2}(3x+4)^{-1/2}=...$
$...=\dfrac{1}{2}x^{-1/2}(3x+4)^{-1/2}\Big[(3x+4)-3x\Big]=...$
Simplify the expression inside brackets:
$...=\dfrac{1}{2}x^{-1/2}(3x+4)^{-1/2}(4)=2x^{-1/2}(3x+4)^{-1/2}=...$
Rearrange:
$...=\dfrac{2}{x^{1/2}(3x+4)^{1/2}}=\dfrac{2}{(\sqrt{x})(\sqrt{3x+4})}=\dfrac{2}{\sqrt{3x^{2}+4x}}$
