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