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IsTim

Determine dy/dx for y=5sqrt(x^3)

  • one year ago
  • one year ago

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  1. CoCoTsoi
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    y=5sqrt(x^3) = 5(x^(3/2)) dy/dx = 5(3/2)(x^(3/2 -1)) = 15/2 (x^1/2)

    • one year ago
  2. Ishaan94
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    IsTim now you're here for math too? giggle, shiny and math?

    • one year ago
  3. CoCoTsoi
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    y=x^n dy/dx = nx^(n-1) this is the property I use

    • one year ago
  4. IsTim
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    Truly, I don't really understand what derivatives is. @Ishaan94 Pardon?

    • one year ago
  5. IsTim
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    The Power Rule. Ok.

    • one year ago
  6. Ishaan94
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    Sorry.

    • one year ago
  7. IsTim
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    It's ok. I don't understand (your speech and derivatives; either/or) though.

    • one year ago
  8. IsTim
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    If you can, please describe each step thoroughly. Thank you.

    • one year ago
  9. IsTim
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    OH well.

    • one year ago
  10. IsTim
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    What's happening at dy/dx = 5(3/2)(x^(3/2 -1)) ?

    • one year ago
  11. IsTim
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    Or, how did it come to that?

    • one year ago
  12. CoCoTsoi
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    y=ax^n dy/dx =a( nx^(n-1)) put a=1 and n=3/2

    • one year ago
  13. IsTim
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    Uhm, sorry to ask, but could you type dy/dx = 5(3/2)(x^(3/2 -1)) in Latex (The Equation Editor)? I don't think I'm understanding it correctly.

    • one year ago
  14. IsTim
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    @CoCoTsoi When you have time, of course.

    • one year ago
  15. CoCoTsoi
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    |dw:1335423732321:dw|

    • one year ago
  16. IsTim
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    Thanks.

    • one year ago
  17. Maekin
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    |dw:1335423870842:dw||dw:1335423900291:dw|

    • one year ago
  18. Maekin
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    That was just showing you an of how to work the power rule. I rewrote the sqrt as x^1/2

    • one year ago
  19. IsTim
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    |dw:1335434722081:dw| That was the answer in the textbook. How do I convert to that?

    • one year ago
  20. Maekin
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    A-ha, that wasn't 5 times sqrt X (etc...) It was the fifth root

    • one year ago
  21. Maekin
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    |dw:1335424089162:dw| was the original question

    • one year ago
  22. IsTim
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    Yes.

    • one year ago
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