$y=\sqrt[5]{2x+1}+\dfrac{\sqrt{x+7}+2x}{5x+2}\\ =(2x+1)^{\tfrac{1}{5}}+\dfrac{\sqrt{x+7}+2x}{5x+2}\\ y'=2.\dfrac{1}{5}.(2x+1)^{\tfrac{-4}{5}}+\dfrac{\left(\sqrt{x+7}+2x\right)'(5x+2)-\left(\sqrt{x+7}+2x\right)(5x+2)'}{(5x+2)^2}\\ =\dfrac{2}{5}\dfrac{1}{\sqrt[5]{(2x+1)^4}}+\dfrac{\left(\dfrac{1}{2\sqrt{x+7}}+2\right)(5x+2)-5\left(\sqrt{x+7}+2x\right)}{(5x+2)^2}\\ =\dfrac{2}{5}\dfrac{1}{\sqrt[5]{(2x+1)^4}}+\dfrac{\dfrac{5x+2}{2\sqrt{x+7}}+10x+4-5\sqrt{x+7}-10x}{(5x+2)^2}\\ =\dfrac{2}{5}\dfrac{1}{\sqrt[5]{(2x+1)^4}}+\dfrac{\dfrac{5x+2}{2\sqrt{x+7}}+4-5\sqrt{x+7} }{(5x+2)^2}\\ =\dfrac{2}{5}\dfrac{1}{\sqrt[5]{(2x+1)^4}}+\dfrac{-5x+8\sqrt{x+7}-68}{2\sqrt{x+7}(5x+2)^2}\\ $
