[tex]y=\dfrac{3x-2}{\sqrt{2x+1}}=\dfrac{3x-2}{(2x+1)^{1/2}}[/tex]
The quotient rule gives
[tex]y'=\dfrac{(2x+1)^{1/2}(3x-2)'-(3x-2)\left((2x+1)^{1/2}\right)'}{\left((2x+1)^{1/2}\right)^2}=\dfrac{3(2x+1)^{1/2}-\dfrac{2(3x-2)}{2(2x+1)^{1/2}}}{2x+1}[/tex]
Factor out [tex](2x+1)^{-1/2}[/tex] to get
[tex]\dfrac{(2x+1)^{-1/2}\left(3(2x+1)^{2/2}-(3x-2)\right)}{2x+1}=\dfrac{6x+3-6x+2}{(2x+1)^{3/2}}=\dfrac{3x+5}{(2x+1)^{3/2}}[/tex]
