$$\displaystyle\frac{d}{d x } \left( { \left( \sqrt{ x+1 } + \frac{ 2 }{ x } -1 \right) }^{ 3 } \right) \frac{ 1 }{ 2 }$$
$\frac{3\left(x^{2}-4\sqrt{x+1}\right)\left(\sqrt{x+1}x-x+2\right)^{2}}{4\sqrt{x+1}x^{4}}$
$\frac{3\left(\sqrt{x+1}x-x+2\right)\left(24\left(x+1\right)^{\frac{3}{2}}x^{3}-23\sqrt{x+1}x^{4}+x^{4}+48\left(x+1\right)^{\frac{3}{2}}x^{2}-72\sqrt{x+1}x^{3}-2x^{3}-64x\left(x+1\right)^{\frac{3}{2}}+16x^{2}+48\sqrt{x+1}x+64\left(x+1\right)^{\frac{3}{2}}+16x\right)}{8\left(x+1\right)^{\frac{3}{2}}x^{5}}$
