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anonymous
 3 years ago
f(x)' √x(2x+4)
anonymous
 3 years ago
f(x)' √x(2x+4)

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Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2\[\sqrt{x}(2x+4) \ \ or \ \ \sqrt{x(2x+4)}\]

Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2First can you try it out by yourself?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm i got : \[2√x + 1/2√x(2x+4)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dunno how to go from there =/

Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2Yes thats Right!!! :) Hope you understood :)

Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2You can simplify it , 1second ..

Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2\[2\sqrt{x}+ \frac{(2x+4)}{\sqrt{2x}}=\frac{2\sqrt{x}}{1}+ \frac{(2x+4)}{2\sqrt{x}}\]\[\frac{2\sqrt{x}\times 2\sqrt{x}}{1\times 2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}=\frac{4x}{2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}\]\[ \frac{(4x+2x+4)}{2\sqrt{x}}= \frac{(6x+4)}{2\sqrt{x}}\]Now factor out 2 from 6x+4 6x+4 =2(3x+2)\[ \frac{2(3x+2)}{2\sqrt{x}}= \frac{(3x+2)}{\sqrt{x}}\]

Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2If you want you can rationalise \[ \frac{(3x+2) \times \sqrt{x}}{\sqrt{x} \times \sqrt{x}} = \frac{\sqrt{x}(3x+2)}{x}\]\[\frac{\sqrt{x}3x}{x}+\frac{2\sqrt{x}}{x}= 3\sqrt{x}+ \frac{ 2}{\sqrt{x}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thx diya.. omg that was kinda easy ... can't belive didn't get that .

Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2lol Take your time :) Is there any step which you didn't follow ?

Diyadiya
 3 years ago
Best ResponseYou've already chosen the best response.2No problem ,You're Welcome :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0poke u later when im stuck.. kinda ok for now :p
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