Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 16

Find $y'$ of $\displaystyle \frac{3x-2}{\sqrt{2x+1}}$

$\begin{equation} \begin{aligned} y' &= \frac{d}{dx} \left( \frac{3x-2}{\sqrt{2x+1}} \right)\\ \\ y' &= \frac{(\sqrt{2x+1}) \frac{d}{dx}(3x-2)-(3x-2)\frac{d}{dx}(\sqrt{2x+1}) }{(\sqrt{2x+1})^2}\\ \\ y' &= \frac{(2x+1)^{\frac{1}{2}}\left[ 3\frac{d}{dx} (x) - \frac{d}{dx}(2) \right] - (3x-2) \frac{d}{dx}(2x+1)^{\frac{1}{2}} }{\left[ (2x + 1) ^{\frac{1}{2}}\right]^2}\\ \\ y' &= \frac{(2x+1)^{\frac{1}{2}}(3-0) - (3x-2)\left(\frac{1}{2} \right)(2x+1)^{\frac{-1}{2}}\frac{d}{dx}(2x+1) }{2x+1}\\ \\ y' &= \frac{(2x+1)^{\frac{1}{2}}(3) - (3x-2)\left(\frac{1}{\cancel{2}}\right) (2x+1)^{\frac{-1}{2}} (\cancel{2}) }{2x+1}\\ \\ y' &= \frac{(2x+1)^{\frac{1}{2}}(3) - \frac{(3x-2)}{(2x+1)^{\frac{1}{2}}} }{2x+1}\\ \\ y' &= \frac{\frac{(2x+1)(3)-(3x-2)}{(2x+1)^{\frac{1}{2}}} }{2x+1}\\ \\ y' &= \frac{6x+3-3x+2}{(2x+1)(2x+1)^{\frac{1}{2}}}\\ \\ y' &= \frac{3x+5}{(2x+1)^{3/2}} \end{aligned} \end{equation}$