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Mathematics
OpenStudy (anonymous):

f(x)' √x(2x+4)

OpenStudy (diyadiya):

\[\sqrt{x}(2x+4) \ \ or \ \ \sqrt{x(2x+4)}\]

OpenStudy (anonymous):

first one

OpenStudy (diyadiya):

First can you try it out by yourself?

OpenStudy (diyadiya):

Use the product rule

OpenStudy (anonymous):

hmm i got : \[2√x + 1/2√x(2x+4)\]

OpenStudy (anonymous):

dunno how to go from there =/

OpenStudy (diyadiya):

Yes thats Right!!! :) Hope you understood :)

OpenStudy (diyadiya):

You can simplify it , 1second ..

OpenStudy (diyadiya):

\[2\sqrt{x}+ \frac{(2x+4)}{\sqrt{2x}}=\frac{2\sqrt{x}}{1}+ \frac{(2x+4)}{2\sqrt{x}}\]\[\frac{2\sqrt{x}\times 2\sqrt{x}}{1\times 2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}=\frac{4x}{2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}\]\[ \frac{(4x+2x+4)}{2\sqrt{x}}= \frac{(6x+4)}{2\sqrt{x}}\]Now factor out 2 from 6x+4 6x+4 =2(3x+2)\[ \frac{2(3x+2)}{2\sqrt{x}}= \frac{(3x+2)}{\sqrt{x}}\]

OpenStudy (anonymous):

0_o

OpenStudy (diyadiya):

If you want you can rationalise \[ \frac{(3x+2) \times \sqrt{x}}{\sqrt{x} \times \sqrt{x}} = \frac{\sqrt{x}(3x+2)}{x}\]\[\frac{\sqrt{x}3x}{x}+\frac{2\sqrt{x}}{x}= 3\sqrt{x}+ \frac{ 2}{\sqrt{x}}\]

OpenStudy (anonymous):

thx diya.. omg that was kinda easy -.-.. can't belive didn't get that -.-

OpenStudy (diyadiya):

lol Take your time :) Is there any step which you didn't follow ?

OpenStudy (diyadiya):

No problem ,You're Welcome :)

OpenStudy (anonymous):

poke u later when im stuck.. kinda ok for now :p

OpenStudy (diyadiya):

Haha Alrighht!~

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