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Second Derivative (Implicit Differentiation)

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Find y''

    2. Relevant equations
    9x^2 +y^2 = 9

    3. The attempt at a solution


    For the second derivative, I get the correct answer (same as the book) up until the very last step.
    Here's where I'm left at:

    -9( (-9x^2 - y^2) / y^3 )

    The book then takes this answer to:

    y''= -9 (9/y^3) [since x and y must satisfy the original equation, 9x^2+y^2=9] Thus y''= - 81/y^3

    I'm not sure how they got here as my final answer ends up something like -81x^2/y^3 -9y (probably with some sign mistakes in there but I was too distracted with the book's justification for their answer to worry about that).

    Last edited by a moderator: Oct 15, 2016
  2. jcsd
  3. Oct 15, 2016 #2


    Staff: Mentor

    I think you have a sign error somewhere. As you haven't shown your intermediate steps, I cannot say where. Only that your last equation for ##y'## is wrong. Now, why are you stuck? What is ##(-9x^2-y^2)##?
  4. Oct 15, 2016 #3
    Alright I'll show my intermediate steps for the first to second derivative.


    Y''=-9y^2/y^3 -9(-81x^2)/y^3
    Y''=-9/y + 9(81x^2)/y^3

    Slightly different from my end answer but still no closer to the book's.
    Apologies for the plaintext as I am on mobile.
  5. Oct 15, 2016 #4


    Staff: Mentor

    I don't understand your third row. Did you pull ##-9## out of the parentheses or not? And there is still a sign error.
  6. Oct 15, 2016 #5
    [QUOTE="quicksilver123, post: 5594654, member

    Y''=-9y^2/y^3 -9(-9x^2)/y^3
    Y''=-9/y + 81x^2/y^3

    Slightly different from my end answer but still no closer to the book's.

    Made some corrections.
  7. Oct 15, 2016 #6


    Staff: Mentor

    This sign error is pretty resistant. May have to do with your linear notation.
    You have ##y''=(-9y-(-9x)y')/y^2## which is
    $$y''=\frac{-9y-(-9x)y'}{y^2}=\frac{-9y+9xy'}{y^2}=(-9) \cdot \frac{y-xy'}{y^2}= (-9) \cdot \frac{y-(-9)\frac{x^2}{y}}{y^2}$$
    $$ \quad =(-9) \cdot \frac{y+9\frac{x^2}{y}}{y^2} = (-9) \cdot \frac{y^2+9x^2}{y^3}$$
  8. Oct 16, 2016 #7
    All I see in your post is "math processing error"
  9. Oct 16, 2016 #8


    Staff: Mentor

    Hmmm, not in mine.

    Sorry, I tried to upload it as an image, but it did not work.

    Perhaps a repetition works.
    If not, here is the source code:

    You have y''=(-9y-(-9x)y')/y^2 which is
    y''=\frac{-9y-(-9x)y'}{y^2}=\frac{-9y+9xy'}{y^2}=(-9) \cdot \frac{y-xy'}{y^2}= (-9) \cdot \frac{y-(-9)\frac{x^2}{y}}{y^2}
    \quad =(-9) \cdot \frac{y+9\frac{x^2}{y}}{y^2} = (-9) \cdot \frac{y^2+9x^2}{y^3}
  10. Oct 16, 2016 #9
    Thanks, but it still looks nothing like the answer given in the book.. why is that?
  11. Oct 16, 2016 #10


    Staff: Mentor

    Because I left out the final step where the knowledge about the value of ##9x^2 + y^2##
    (9x^2 + y^2)
    has to be taken into account.
  12. Oct 16, 2016 #11
  13. Oct 16, 2016 #12


    Staff: Mentor

    By definition ##9x^2 + y^2 = 9##
    So ##y'' = \dots = (-9) \cdot \frac{y^2+9x^2}{y^3}= (-9) \cdot \frac{9}{y^3}= -81 \frac{1}{y^3}##.

    By definition 9x^2 + y^2 = 9.
    So y'' = \dots = (-9) \cdot \frac{y^2+9x^2}{y^3}= (-9) \cdot \frac{9}{y^3}= -81 \frac{1}{y^3}.
  14. Oct 16, 2016 #13
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