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anonymous

  • 5 years ago

compute d^2/dx^2 at the point (4,2) x^2+y^2=20

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  1. amistre64
    • 5 years ago
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    is that our f(x) sitting there?

  2. amistre64
    • 5 years ago
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    2x x' + 2y y' = 0 2x + 2y y' = 0 y' = -x/y

  3. amistre64
    • 5 years ago
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    y'' = -1/y + y'x/y^2

  4. amistre64
    • 5 years ago
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    \[y'' = -\frac{1}{y} + \frac{{-x^2}}{y^3}\]

  5. amistre64
    • 5 years ago
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    (x=4, y=2) y'' = -1/2 - 16/8

  6. amistre64
    • 5 years ago
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    y'' = -32/8 = -4 perhaps?

  7. amistre64
    • 5 years ago
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    thats wrong

  8. anonymous
    • 5 years ago
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    why not 5/2?

  9. amistre64
    • 5 years ago
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    -4/8 - 16/8 = -20/8 y'' = -5/2

  10. amistre64
    • 5 years ago
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    i was gettin there, my head on holds so much at once lol

  11. anonymous
    • 5 years ago
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    cool, thanks!

  12. amistre64
    • 5 years ago
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    i was wondering if i could have done that from this step like this: 2x + 2y y' = 0 2 + 2y' + 2y y'' = 0 y'' = -2 -2(-x/y) -1 -(-x/y) -1 -(-2) ---------- = -------- = ------- = doesnt look like it 2y y 2

  13. anonymous
    • 5 years ago
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    that looks more complicated, do you happen to also know about definite integrals?

  14. amistre64
    • 5 years ago
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    i can do them with somewhat of skill and mostly luck ;)

  15. amistre64
    • 5 years ago
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    definite integrals are integrating within an interval

  16. anonymous
    • 5 years ago
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    lol okay, let's see how lucky you are, I'm about to post one =D

  17. amistre64
    • 5 years ago
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    yay!!

  18. anonymous
    • 5 years ago
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    okay, for some reason it won't let me ask a question like normal

  19. amistre64
    • 5 years ago
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    refresh your browser

  20. amistre64
    • 5 years ago
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    f5

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