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Derivatives, chain rule 5

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the second derivative of $$9x^2+y^2=9$$

    2. Relevant equations

    Chain rule

    3. The attempt at a solution

    I find the first derivative first.

    $$18x+2y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=-9\frac{x}{y}$$

    I then find the second derivative.

    $$-9(\frac{y-x\frac{d'y}{dx''}}{y^2})$$ $$y^2=-9y+9x\frac{d'y}{dx''}$$ $$y^2+9y=9x\frac{d'y}{dx''}$$ $$\frac{d'y}{dx''}=\frac{y^2+9y}{9x}=\frac{y(y+9)}{9x}$$

    My textbook says that the answer should be $$\frac{-81}{y^3}$$ What did I do wrong?
     
  2. jcsd
  3. Nov 10, 2013 #2
    Nevermind I got it! I should be plugging in y' into my equation for y''.
     
  4. Nov 10, 2013 #3
    I actually still need help.

    When I plug in the y' into y'' I still don't get the right answer.

    $$\frac{-9y-9x\frac{dy}{dx}}{y^2}$$ $$=\frac{-9y-9x(-\frac{9x}{y})}{y^2}$$ $$=\frac{-9y+\frac{81x^2}{y}}{y^2}$$

    This is still not the answer. What did I do wrong now?
     
  5. Nov 10, 2013 #4

    Dick

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    First correct a sign error in your numerator. Then multiply numerator and denominator by y. Then use your original function to simplify the numerator.
     
  6. Nov 10, 2013 #5
    I'm sorry, but I don't get what you mean. When I multiply the numerator and denominator by y I get

    $$\frac{-9y^2+81x^2}{y^3}$$
     
  7. Nov 10, 2013 #6

    Dick

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    I edited my post when I realized you had a sign error in the numerator. Find it and fix it.
     
  8. Nov 10, 2013 #7
    Okay, so I've fixed the sign error and multiplied the top and bottom by y. Now I have this

    $$\frac{-9y^2-81x^2}{y^3}$$

    The most I can do is factor out the -9, but I still don't get the correct answer.
     
  9. Nov 10, 2013 #8

    Dick

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    Your original function says y^2+9x^2=9. Use that to simplify the numerator.
     
  10. Nov 10, 2013 #9
    Okay I got it! Thanks so much!

    $$\frac{-9(y^2+9x^2)}{y^3}=\frac{-9(9)}{y^3}=\frac{-81}{y^3}$$
     
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