\(\frac{d}{d\mathrm x}\frac{\mathrm x^2}{\sqrt{\mathrm x+1}}\) \(=\frac{\sqrt{\mathrm x+1}\frac{d}{d\mathrm x}\mathrm x^2-\mathrm x^2\frac{d}{d\mathrm x}\sqrt{\mathrm x+1}}{\mathrm x+1}\) (By \(\frac{U}{V}\) formula)

\(\mathrm{=\frac{2x\sqrt{x+1}-\frac{x^2}{2\sqrt{x+1}}}{x+1}}\)   \(\Big(\because\) \(\frac{d}{d\mathrm x}\mathrm x^2=2\mathrm x\) and \(\frac{d}{d\mathrm x}\sqrt{\mathrm x}=\frac{1}{2\sqrt{\mathrm x}}\Big)\)

\(\mathrm{=\frac{4x(x+1)-x^2}{x+1}}\)

\(\mathrm{\frac{3x^2+4x}{2(x+1)^{^{3}/_{2}}}}\)

Now, \(\frac{d^2}{d\mathrm x^2}\mathrm{\frac{x^2}{\sqrt{x+1}}}\) \(=\frac{d}{d\mathrm x}\mathrm{\frac{3x^2+4x}{2(x+1)^{^3/_2}}}\)

\(=\frac{1}{2}\) \(\times\) \(\mathrm{\frac{(x+1)^{\frac{3}{2}}\frac{d}{d\mathrm x}(3x^2+4x)-(3x^2+4x)\frac{d}{d\mathrm x}(x+1)^{\frac{3}{2}}}{(x+1)^3}}\)   

\(=\frac{1}{2}\) \(\times\) \(\mathrm{\frac{(6x+4)(x+1)^{\frac{3}{2}}-\frac{3}{2}\times (3x^2+4x)\times (x+1)^{\frac{1}{2}}}{(x+1)^3}}\) \((\because\) \(\frac{d}{d\mathrm x}\mathrm x^4 = n\mathrm x^{n-1})\)

\(\mathrm{=\frac{\frac{1}{2}(\mathrm x+1)^{\frac{1}{2}}((6x+4)(x+1)-\frac{3}{2}(3x^2+4x))}{(x+1)^3}}\)

\(=\frac{1}{4(\mathrm x+1)^{\frac{5}{2}}}\)\(\mathrm{(12x^2+20x+8}\) \(\mathrm{-9x^2-12x})\)

\(\mathrm{=\frac{3x^2+8x+8}{4(x+1)^{\frac{5}{2}}}}\)

Hence, second derivative of \(\frac{\mathrm x^2}{\sqrt{\mathrm x+1}}\) is \(\mathrm{\frac{3x^2+8x+8}{4(x+1)^{\frac{5}{2}}}}\).