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IsTim

  • 4 years ago

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

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  1. CoCoTsoi
    • 4 years ago
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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)

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

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

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

  5. IsTim
    • 4 years ago
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    The Power Rule. Ok.

  6. Ishaan94
    • 4 years ago
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    Sorry.

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

  8. IsTim
    • 4 years ago
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    If you can, please describe each step thoroughly. Thank you.

  9. IsTim
    • 4 years ago
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    OH well.

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

  11. IsTim
    • 4 years ago
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    Or, how did it come to that?

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

  13. IsTim
    • 4 years ago
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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.

  14. IsTim
    • 4 years ago
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    @CoCoTsoi When you have time, of course.

  15. CoCoTsoi
    • 4 years ago
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    |dw:1335423732321:dw|

  16. IsTim
    • 4 years ago
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    Thanks.

  17. Maekin
    • 4 years ago
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    |dw:1335423870842:dw||dw:1335423900291:dw|

  18. Maekin
    • 4 years ago
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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

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

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

  21. Maekin
    • 4 years ago
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    |dw:1335424089162:dw| was the original question

  22. IsTim
    • 4 years ago
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    Yes.

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