Answer
$$y'=\frac{2x+4}{(x+1)^{3/2}}$$
$$y''=\frac{-x-4}{(x+1)^{5/2}}$$
Work Step by Step
$$y'=\left(\frac{4x}{\sqrt{x+1}}\right)'=\frac{(4x)'\sqrt{x+1}-4x(\sqrt{x+1})'}{(\sqrt{x+1})^2}=
\frac{4\sqrt{x+1}-4x\frac{1}{2\sqrt{x+1}}(x+1)'}{x+1}=
\frac{4(x+1)-2x\cdot1}{(x+1)^{3/2}}=\frac{2x+4}{(x+1)^{3/2}}$$
$$y''=(y')'=\left(\frac{2x+4}{(x+1)^{3/2}}\right)'=\frac{(2x+4)'(x+1)^{3/2}-(2x+4)((x+1)^{3/2})'}{(x+1)^3}=\frac{2(x+1)^{3/2}-(2x+4)\frac{3}{2}(x+1)^{1/2}(x+1)'}{(x+1)^3}=
\frac{2(x+1)^{3/2}-(x+2)\cdot3(x+1)^{1/2}\cdot1}{(x+1)^3}=
\frac{2(x+1)-3(x+2)}{(x+1)^{5/2}}=\frac{2x+2-3x-6}{(x+1)^{5/2}}=\frac{-x-4}{(x+1)^{5/2}}$$
