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Math Help - implicit darivations.

  1. #1
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    implicit darivations.

    yep, yall are handy... love ya's much.

    So, it asks for us to find the y'' of x2 + xy + y3 = 1 implicitly.

    Pointers?
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  2. #2
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    I'm assuming you meant "Find y''(x) where

    x^{3}+xy+y^{2}=1."

    Is that correct? If so, what would be the first step?
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  3. #3
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    yes thats correct, so i just take it like i would a regular derv?
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  4. #4
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    Yes. Rather than do that problem for you, let me find the second derivative of y when y^2- e^x+ xy- 2y= 4.

    Differentiating once, 2yy'- e^x+ y+ xy'- 2y'= 0. Differentiating again, 2y'y'+ 2yy''- e^x+ y'+ y'+ xy''- 2y''= 2(y')^2+ 2yy''- e^x+ 2y'+ xy''- 2y''= 0. Now, you can solve for y'':
    y''= \frac{e^x- 2(y')^2- 2y'}{2y+ x- 2}.

    You could, now, replace y' from the first derivative but typically, that is not necessary.
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  5. #5
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    if im lookin for y'' i do nothing to the x's?
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  6. #6
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    What do you mean by "do nothing to the x's"?
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  7. #7
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    I must be missing something...
    You never took the derivative of the x's.
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  8. #8
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    HallsofIvy did do the differentiation correctly, taking into account that y = y(x). Let's say I have the term xy. Its derivative w.r.t. x is y + xy'. Do you see that?
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  9. #9
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    6x+2=y''(x)... is that right?
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  10. #10
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    Quote Originally Posted by FreaKariDunk View Post
    6x+2=y''(x)... is that right?
    From where did that come?
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  11. #11
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    With a reaction like that, it must be wrong. I don't know what I'm missing. I'm gonna re-read the section and re-explore that problem.
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  12. #12
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    Take a good look at HallsofIvy's Post # 4. Start with the same equation, and see if you can't reproduce those results.
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  13. #13
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    as i retry this... Im workin on finding the formula for dy/dx
    and i got to here... 3x^2+[x(dy/dx) + y(dx/dx)]+2y=0

    now, the stuff in the brackets.. how do I simplify that... in the first term can I 'cancel' the x leaving y in the numerator. but then that leaves the second term...
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  14. #14
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    No, you can't cancel inside the brackets. The symbol dy/dx is one symbol. It is not a fraction. However, the dx/dx = 1, so that simplifies a bit. Incidentally, I would agree with your differentiation except for the last term. You should use the chain rule on it thus:

    3x^{2}+[xy'+y]+2yy'=0.

    Can you continue?
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  15. #15
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    So now I have to solve for y'...
    so it would be (-3x^2-y)/(x2y) ?
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