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anonymous

  • 5 years ago

use the implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=10at point (5,1)

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  1. amistre64
    • 5 years ago
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    what do you know about implicit differentation?

  2. anonymous
    • 5 years ago
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    differentiate it and subs the point (5,1) . am i right? @amistre

  3. amistre64
    • 5 years ago
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    yep, thats right think

  4. amistre64
    • 5 years ago
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    differentiate like normal, just keep your x' and y' in the problem...

  5. anonymous
    • 5 years ago
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    y-1=-(x-5)/10

  6. amistre64
    • 5 years ago
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    D(xy^3) + D(xy) = D(10) product rule ... product rule....constant rule

  7. anonymous
    • 5 years ago
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    simplify and differentiate, 10y-10=-x-5 => x+10y=5now try to differentiate...will that do @amistre?

  8. amistre64
    • 5 years ago
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    you CAN try to seperate it all, but its really just easier to work the problam as it...

  9. amistre64
    • 5 years ago
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    (x'y + xy') + (x'y + xy') = D(10) = 0

  10. amistre64
    • 5 years ago
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    x'(y^3) + x(3y^2)y' + x'y + xy' = 0

  11. anonymous
    • 5 years ago
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    yeah this is fine even..whatever u find easy diddy :)

  12. amistre64
    • 5 years ago
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    x' = dx/dx = 1 like any good fraction... now solve for y'

  13. anonymous
    • 5 years ago
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    hey catch u later ..gotta go :) bye all

  14. amistre64
    • 5 years ago
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    Ciao think :)

  15. anonymous
    • 5 years ago
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    thanks @ thinker

  16. anonymous
    • 5 years ago
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    ur welcome ..@diddy, all the best and ciao amistre...ttyl maybe after 2 days :)

  17. amistre64
    • 5 years ago
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    Its important to realize that there is nothing that special when doing implicits.. for example: D(x^4) = x' 4x^3 D(3y^2) = y' 6y D(3xy) = 3xy' + x'3y D(y) = y' D(x) = x' its all normal rules

  18. anonymous
    • 5 years ago
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    oki so i used this form amistre x'(y^3) + x(3y^2)y' + x'y + xy' = 0 and i came up with 2y^3/3xy^2+1, but i'm not sure that's right

  19. amistre64
    • 5 years ago
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    Let me double check the work... x'(y^3) + x(3y^2)y' + x'y + xy' = 0 y^3 + y'(3xy^2) + y + y'x = 0 y'(3xy^2) + y'x = -y^3 - y y'(3xy^2 + x) = -y^3 -y -y^3 -y y' = ----------- 3xy^2 + x did I make any mistakes in this?

  20. anonymous
    • 5 years ago
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    You didn't :)

  21. amistre64
    • 5 years ago
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    yay!!... lately, thats an accomplishment :)

  22. anonymous
    • 5 years ago
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    yea i think that's right, make sense now thanks :)

  23. amistre64
    • 5 years ago
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    now plug in your x and y values to determine the slope of the tangent line at that point :)

  24. anonymous
    • 5 years ago
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    ook got it

  25. amistre64
    • 5 years ago
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    good job :)

  26. anonymous
    • 5 years ago
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    YAY i got it!!! thanks u're the best :)

  27. amistre64
    • 5 years ago
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    now, now... youre only saying that cause its true lol :)

  28. anonymous
    • 5 years ago
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    lol

  29. anonymous
    • 5 years ago
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    i'm having some trouble with this one... Find the slope of the tangent line to the curve -1x^2-2xy-3y^3=-185 at the point (-1,4).

  30. anonymous
    • 5 years ago
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    never mind, got it!!!

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