Derivation help!

Ok  Namodesto

I have made some changes to the original post, let me know how you get on.

Why did I divide by x:

I used the quotient rule.  The denominator of the quotient rule is v^2

v=sqrt{x}

v^2 = x

Anyway let me know if youdo or if you don't understand :))

The question is not written properly Namodesto

It should be like this

$f(x)=\frac{x+2}{\sqrt{x}}\\ \mbox{Use quotient rule}\\ \boxed{If\;\;y=\frac{u(x)}{v(x)}\;\;then\;\;y'=\frac{vu'-uv'}{v^2} }\\ u=x+2 \quad v=x^{0.5}\\ u'=1 \qquad v'=0.5v^{-0.5}=\frac{1}{2\sqrt2}\\ $

$(1)\qquad f'(x)=\frac{\sqrt{x}-\frac{1}{2\sqrt{x}}(x-2)}{x}\\ (2)\qquad f'(x)=\left[\sqrt{x}-\frac{1}{2\sqrt{x}}(x-2)\right] \div x\\ (3) \qquad f'(x)=\left[\frac{\sqrt{x}}{1}-\frac{(x-2)}{2\sqrt{x}}\right] \div x\\ (4) \qquad f'(x)=\left[\frac{2\sqrt{x}}{2\sqrt{x}}*\frac{\sqrt{x}}{1}-\frac{(x-2)}{2\sqrt{x}}\right] \div x\\ (5) \qquad f'(x)=\left[\frac{2\sqrt{x}}{2\sqrt{x}}*\frac{\sqrt{x}}{1}-\frac{(x-2)}{2\sqrt{x}}\right] \div x\\ (6) \qquad f'(x)=\left[\frac{2\sqrt{x}\sqrt{x}}{2\sqrt{x}*1}-\frac{(x-2)}{2\sqrt{x}}\right] \times \frac{1}{x}\\(7) \qquad f'(x)=\left[\frac{2x}{2\sqrt{x}}-\frac{(x-2)}{2\sqrt{x}}\right] \times \frac{1}{x}\\ (8) \qquad f'(x)=\frac{2x-(x-2)}{2\sqrt{x}}\times \frac{1}{x}\\ (9) \qquad f'(x)=\frac{2x-(x-2)}{2x\sqrt{x}}\\ (10) \qquad f'(x)=\frac{x+2}{2x\sqrt{x}}\\$

I have added lines to help you understand and I have put line numbers so you can tell me which bit you do not understand.

So let me know how you get on :)

Hi Melody, 

thank you so much for your help, however I still don't understand how you got 2x. And also why do you divide by x? 

Thank you Melody! I finally understand haha :) 

That is great.  I am glad I could help :)