Answer:
[tex]y'=\frac{3x+2}{2\sqrt{x}}\,\,,\,\,\,x>0[/tex] (for real x). See more details below.
Step-by-step explanation:
Assuming x is real, we start with the observation that x>=0 (otherwise the square root will become complex). If x>=0, then we can simplify:
[tex]y=\sqrt{x(x+2)^2}=\sqrt{x}|x+2|=\sqrt{x}(x+2)\,\,,\,\,x\geq0[/tex]
and the derivative will be:
[tex]y'=\frac{1}{2\sqrt{x}}(x+2)+\sqrt{x}=\frac{3x+2}{2\sqrt{x}}\,\,,\,\,\,x>0[/tex]
If we do not make any assumptions on x, the derivative will be:
[tex]y=\sqrt{x(x+2)^2}\\y'=\frac{(x+2)^2+2x(x+2)}{2\sqrt{x(x+2)^2}}=\frac{(x+2)(3x+2)}{2\sqrt{x(x+2)^2}}[/tex]
