For starters, you can write
[tex]\dfrac{x\sqrt{x+1}}{x+1}=\dfrac{x\sqrt{x+1}}{(\sqrt{x+1})^2}=\dfrac x{\sqrt{x+1}}=\dfrac x{(x+1)^{1/2}}[/tex]
since the original function is only defined and continuous for [tex]x>-1[/tex], so you never divide by 0.
Now to differentiate, you can use the quotient rule:
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac x{(x+1)^{1/2}}\right]=\dfrac{(x+1)^{1/2}\frac{\mathrm d}{\mathrm dx}x-x\frac{\mathrm d}{\mathrm dx}\left[(x+1)^{1/2}\right]}{\left((x+1)^{1/2}\right)^2}=\dfrac{(x+1)^{1/2}-\frac x{2(x+1)^{1/2}}}{x+1}[/tex]
Next, factor out [tex]\dfrac1{2(x+1)^{1/2}}[/tex] to get
[tex]\dfrac{2(x+1)-x}{2(x+1)^{1/2}(x+1)}=\dfrac{x+2}{2(x+1)^{3/2}}[/tex]
