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abz_tech

  • 2 years ago

f(x)' √x(2x+4)

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  1. Diyadiya
    • 2 years ago
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    \[\sqrt{x}(2x+4) \ \ or \ \ \sqrt{x(2x+4)}\]

  2. abz_tech
    • 2 years ago
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    first one

  3. Diyadiya
    • 2 years ago
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    First can you try it out by yourself?

  4. Diyadiya
    • 2 years ago
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    Use the product rule

  5. abz_tech
    • 2 years ago
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    hmm i got : \[2√x + 1/2√x(2x+4)\]

  6. abz_tech
    • 2 years ago
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    dunno how to go from there =/

  7. Diyadiya
    • 2 years ago
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    Yes thats Right!!! :) Hope you understood :)

  8. Diyadiya
    • 2 years ago
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    You can simplify it , 1second ..

  9. Diyadiya
    • 2 years ago
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    \[2\sqrt{x}+ \frac{(2x+4)}{\sqrt{2x}}=\frac{2\sqrt{x}}{1}+ \frac{(2x+4)}{2\sqrt{x}}\]\[\frac{2\sqrt{x}\times 2\sqrt{x}}{1\times 2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}=\frac{4x}{2\sqrt{x}}+ \frac{(2x+4)}{2\sqrt{x}}\]\[ \frac{(4x+2x+4)}{2\sqrt{x}}= \frac{(6x+4)}{2\sqrt{x}}\]Now factor out 2 from 6x+4 6x+4 =2(3x+2)\[ \frac{2(3x+2)}{2\sqrt{x}}= \frac{(3x+2)}{\sqrt{x}}\]

  10. abz_tech
    • 2 years ago
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    0_o

  11. Diyadiya
    • 2 years ago
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    If you want you can rationalise \[ \frac{(3x+2) \times \sqrt{x}}{\sqrt{x} \times \sqrt{x}} = \frac{\sqrt{x}(3x+2)}{x}\]\[\frac{\sqrt{x}3x}{x}+\frac{2\sqrt{x}}{x}= 3\sqrt{x}+ \frac{ 2}{\sqrt{x}}\]

  12. abz_tech
    • 2 years ago
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    thx diya.. omg that was kinda easy -.-.. can't belive didn't get that -.-

  13. Diyadiya
    • 2 years ago
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    lol Take your time :) Is there any step which you didn't follow ?

  14. Diyadiya
    • 2 years ago
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    No problem ,You're Welcome :)

  15. abz_tech
    • 2 years ago
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    poke u later when im stuck.. kinda ok for now :p

  16. Diyadiya
    • 2 years ago
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    Haha Alrighht!~

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