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anonymous

  • 5 years ago

h(u)=square (9-u^2) I want to get the derivative. Can't I take the 9-u^2 out of the square and make it 3-u?

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  1. anonymous
    • 5 years ago
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    o yes it can be written as (3-u)(3+u)

  2. anonymous
    • 5 years ago
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    I thought so. Thanks

  3. amistre64
    • 5 years ago
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    no... it isnt a square

  4. amistre64
    • 5 years ago
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    (3-u) (3-u) does NOT equal (9-u^2) it equals (9 -6u +u^2)

  5. amistre64
    • 5 years ago
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    try using trig substitutions... sqrt(9 - u^2); sqrt(9 - (3sin)(3sin)) ; u=3sin sqrt(9 -9sin^2) = sqrt(9(1-sin^2)) = sqrt(9(cos^2)) = 3cos

  6. amistre64
    • 5 years ago
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    the derivative of 3 cos is -3sin right? plug in your sin value now..

  7. amistre64
    • 5 years ago
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    sin = u/3 ..thats what I meant :)

  8. amistre64
    • 5 years ago
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    -3(u/3) = -u if I did it right....

  9. amistre64
    • 5 years ago
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    or am i confusing integrals and derivatives.... one of these days Ill know when Im right :)

  10. amistre64
    • 5 years ago
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    I get... -1/sqrt(9-u^2) is that better?

  11. amistre64
    • 5 years ago
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    -u/sqrt(9-u^2) ??

  12. anonymous
    • 5 years ago
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    (9-u^2) is not equals (9 -6u +u^2) (9-u^2) is equal to (3-u )(3+u)

  13. amistre64
    • 5 years ago
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    and the sqrt((3-u)(3+u)) is not (3-u)...

  14. anonymous
    • 5 years ago
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    formula is a^2 - b^2 = (a+b)( a-b)

  15. amistre64
    • 5 years ago
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    Is the sqrt(9-u^2) = (3-u)? No.

  16. amistre64
    • 5 years ago
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    I might be wrong on everything else I typed, but I know im right on this one :)

  17. anonymous
    • 5 years ago
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    the sqrt(9-u^2) is not equal to (3-u) but sqrt(9-u^2) is equal to (3-u)(3+u)

  18. amistre64
    • 5 years ago
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    How so? 9-u^2 is equal to (3-u)(3+u) sqrt(9-u^2) is NOT equal to (3-u) square (3-u) and get (9-u^2) for me, just try it.... try it.... cant do it...

  19. amistre64
    • 5 years ago
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    is [(3-u)(3+u)]^2 equal to..... where was I?

  20. radar
    • 5 years ago
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    Couldn't you just square \[(9-u ^{2})^{2}\]

  21. amistre64
    • 5 years ago
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    maybe..... just maybe...... are we allowed to? :)

  22. anonymous
    • 5 years ago
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    [(3-u)(3+u)]^2 equal to (3-u)^2 (3+u)^2

  23. radar
    • 5 years ago
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    Wouldn't that give you \[u ^{4}-18u ^{2}+81\]

  24. radar
    • 5 years ago
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    Then do the derivative\[4u ^{3}-36u\]

  25. amistre64
    • 5 years ago
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    .... im sure wali is right about something in this....

  26. amistre64
    • 5 years ago
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    do we have to unsquare the derivative?

  27. anonymous
    • 5 years ago
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    ok u win the game but i m not wrong

  28. amistre64
    • 5 years ago
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    lol ... but I was gonna concede :)

  29. amistre64
    • 5 years ago
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    maybe

  30. amistre64
    • 5 years ago
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    "Can I take the 9-u^2 out of the square and make it 3-u?" What was the answer?

  31. anonymous
    • 5 years ago
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    no no ...

  32. amistre64
    • 5 years ago
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    I chose "no" as well, but im not a reliable witness :)

  33. radar
    • 5 years ago
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    Then simplifying getting\[1/4u[(u-3)(u+3)]\]

  34. anonymous
    • 5 years ago
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    ok by we will meet again because i have to prepare for tomorrow calculus paper by

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