Here's the question you clicked on:
abz_tech
f(x)' √x(2x+4)
\[\sqrt{x}(2x+4) \ \ or \ \ \sqrt{x(2x+4)}\]
First can you try it out by yourself?
hmm i got : \[2√x + 1/2√x(2x+4)\]
dunno how to go from there =/
Yes thats Right!!! :) Hope you understood :)
You can simplify it , 1second ..
\[2\sqrt{x}+ \frac{(2x+4)}{\sqrt{2x}}=\frac{2\sqrt{x}}{1}+ \frac{(2x+4)}{2\sqrt{x}}\]\[\frac{2\sqrt{x}\times 2\sqrt{x}}{1\times 2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}=\frac{4x}{2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}\]\[ \frac{(4x+2x+4)}{2\sqrt{x}}= \frac{(6x+4)}{2\sqrt{x}}\]Now factor out 2 from 6x+4 6x+4 =2(3x+2)\[ \frac{2(3x+2)}{2\sqrt{x}}= \frac{(3x+2)}{\sqrt{x}}\]
If you want you can rationalise \[ \frac{(3x+2) \times \sqrt{x}}{\sqrt{x} \times \sqrt{x}} = \frac{\sqrt{x}(3x+2)}{x}\]\[\frac{\sqrt{x}3x}{x}+\frac{2\sqrt{x}}{x}= 3\sqrt{x}+ \frac{ 2}{\sqrt{x}}\]
thx diya.. omg that was kinda easy -.-.. can't belive didn't get that -.-
lol Take your time :) Is there any step which you didn't follow ?
No problem ,You're Welcome :)
poke u later when im stuck.. kinda ok for now :p