Use the product rule and factor the smaller power

With $u=3x^2+1$ and $v=(2x+7)^{1/2}$, $u^{\prime}=6x$ and $v^{\prime}=\frac12(2x+7)^{-1/2}(2)=(2x+7)^{-1/2}$. Therefore $y^{\prime}=(3x^2+1)(2x+7)^{-1/2}+6x(2x+7)^{1/2}$.

The source advises taking out the smaller power $(2x+7)^{-1/2}$. The remaining bracket is $(3x^2+1)+6x(2x+7)=15x^2+42x+1$. Thus $y^{\prime}=(2x+7)^{-1/2}(15x^2+42x+1)$. The original function is real for $x\ge-7/2$; this finite derivative formula is valid for $x>-7/2$.