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Implicit Derivative

  1. Nov 14, 2004 #1
    Hi all:

    How would you find the second derivative of

    2x^3 - 3y^2 = 8?

    I know the first derivative is x^2 / y. Would I use the quotient rule, or would I use some type of substitution and then use the product rule?

    Any help is greatly appreciated!


    Thanks
     
  2. jcsd
  3. Nov 14, 2004 #2

    nrqed

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    Maybe I am wrong because it sounds too easy (so I might be forgetting something), but isn't it symply given by differentiating dy/dx a second time, which will give

    [tex] {d^2 y\over dx^2} = {d \over dx} {x^2 \over y} = {2 x \over y} - {x^2 \over y^2} {dy \over dx} = {2 x \over y} -{ x^4 \over y^3} [/tex]

    I might be missing something because it looked too simple!

    Pat
     
  4. Nov 14, 2004 #3

    Hurkyl

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    You're missing a bit of specification here... I'm guessing from your partial result you're looking for the second derivative of y with respect to x.


    When I'm unsure about a question, I often do two things. Here, I would:

    (a) Try to come up with reasons why I couldn't use the quotient rule.
    (b) Look for justification for using the quotient rule.


    Have you gotten anywhere on either of these?


    (edit: *sigh* there goes my attempt at a confidence building exercise)
     
  5. Nov 14, 2004 #4
    ok Thanks a lot
     
  6. Nov 14, 2004 #5

    nrqed

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    I am sorry, really.

    Pat :frown:
     
  7. Nov 15, 2004 #6

    Hurkyl

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    (don't be sorry)
     
  8. Nov 15, 2004 #7

    HallsofIvy

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    2x3- 3y2= 8 so
    6x2- 6y y'= 0.

    Differentiate again:

    12x- (6y')y'- 6yy"= 0 or

    12x- 6y'2- 6yy"=0.

    Solve for y".
     
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