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abz_tech

f(x)' x√x

  • one year ago
  • one year ago

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  1. abz_tech
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    hmm?

    • one year ago
  2. Diyadiya
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    \[\Large \sqrt{x}=x^ \frac{1}{2}\]Do you know this?

    • one year ago
  3. abz_tech
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    ya .. i know derivite of that is 1/2√x

    • one year ago
  4. Diyadiya
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    The easier way is to do by this metho\[\Large x \sqrt{x}= x.x^ \frac{1}{2}\]

    • one year ago
  5. abz_tech
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    ok.. then?

    • one year ago
  6. Diyadiya
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    \[\Large x^m \times x^n = x^{m+n}\]

    • one year ago
  7. Diyadiya
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    \[\Large x \times x^ \frac{1}{2}= ?\]

    • one year ago
  8. annas
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    x^3/2

    • one year ago
  9. abz_tech
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    0_o.. didn;t think of that..but aren't u supposed to use roduct rule?

    • one year ago
  10. abz_tech
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    product rule

    • one year ago
  11. annas
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    no product rule is not applicable here 3/2 x^1/2 answer

    • one year ago
  12. abz_tech
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    now that makes sense 0_o

    • one year ago
  13. Diyadiya
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    Yes you can use product rule ,You'll get the same answer

    • one year ago
  14. abz_tech
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    why not tough?

    • one year ago
  15. abz_tech
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    :o ok

    • one year ago
  16. abz_tech
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    thx

    • one year ago
  17. annas
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    \[3/2x^{1/2}\]

    • one year ago
  18. Diyadiya
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    \[\frac{d}{dx}x \sqrt{x}=x \frac{d}{dx}\sqrt{x}+\sqrt{x}\frac{d}{dx}x\]\[=x .\frac{1}{2\sqrt{x}}+\sqrt{x}=\frac{x}{2\sqrt{x}}+\sqrt{x} =\frac{\sqrt{x}}{2}+\sqrt{x}\]\[=\frac{\sqrt{x}}{2}+ \frac{\sqrt{x}}{1}=\frac{\sqrt{x}}{2}+\frac{\sqrt{x} \times 2 }{1 \times 2} =\frac{\sqrt{x}}{2}+\frac{2\sqrt{x}}{2}=\frac{3\sqrt{x}}{2}\]

    • one year ago
  19. Diyadiya
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    Using the Product Rule^^

    • one year ago
  20. annas
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    wow diya you are really good at maths :)

    • one year ago
  21. Diyadiya
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    \[\Large \frac{3}{2}x^ \frac{1}{2}= \frac{3}{2}\sqrt{x}= \frac{3\sqrt{x}}{2}\]Same answer! :)

    • one year ago
  22. Diyadiya
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    Lol Thanks Annas!

    • one year ago
  23. annas
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    welcome !

    • one year ago
  24. abz_tech
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    lol ye.. ur really good

    • one year ago
  25. abz_tech
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    hmm hey, if u dont mind can u help me with another question =/?

    • one year ago
  26. Diyadiya
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    Post another question :)

    • one year ago
  27. abz_tech
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    ok

    • one year ago
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