#1**+15 **

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 :)

Melody
Dec 2, 2015

#2**0 **

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?

Namodesto
Dec 2, 2015