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anonymous

  • 4 years ago

derivative of y=x(9-x)^1/2?

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  1. anonymous
    • 4 years ago
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    \[y=x\sqrt{(9-x)}\]?

  2. anonymous
    • 4 years ago
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    yes

  3. anonymous
    • 4 years ago
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    product rule for this one

  4. anonymous
    • 4 years ago
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    this is what I have: x[-1/x(9-x)^1/2]+(9-x)^1/2(1)

  5. anonymous
    • 4 years ago
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    is that right?

  6. anonymous
    • 4 years ago
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    if so how to simplify

  7. anonymous
    • 4 years ago
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    \[y'=1\times \sqrt{9-x}+x\times \frac{1}{\sqrt{9-x}}\times -1\] and then clean up with some algebra

  8. anonymous
    • 4 years ago
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    really not much to clean up, just \[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]

  9. anonymous
    • 4 years ago
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    simplifiedmore?

  10. anonymous
    • 4 years ago
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    or if you want to put it over one denominator, you can

  11. anonymous
    • 4 years ago
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    \[\frac{9-x-x}{\sqrt{9-x}}\] \[\frac{9-2x}{\sqrt{9-x}}\]

  12. anonymous
    • 4 years ago
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    if you want to "simplify" things usually best not to use exponents, but rather radicals

  13. anonymous
    • 4 years ago
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    so√(9-x) *x equals 9-2x right

  14. anonymous
    • 4 years ago
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    on no

  15. anonymous
    • 4 years ago
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    so how did u getthe numerator

  16. anonymous
    • 4 years ago
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    \[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]

  17. anonymous
    • 4 years ago
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    is that part ok?

  18. anonymous
    • 4 years ago
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    oh so u subtract

  19. anonymous
    • 4 years ago
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    you want to put i to over the same denominator, so you multiply top and bottom of the first term by \[\sqrt{9-x}\]

  20. anonymous
    • 4 years ago
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    \[y'=\sqrt{9-x}-\frac{x}{\sqrt{9-x}}\] \[y'=\frac{\sqrt{9-x}}{\sqrt{9-x}}\times \sqrt{9-x}-\frac{x}{\sqrt{9-x}}\]

  21. anonymous
    • 4 years ago
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    how come wolfram has: -3(x-6)/2(9-x)^1/2 ?

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