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Homework Help: Hard? implicit differentiation

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data

    Find d^2y/dx^2 in terms of x and y.

    2. Relevant equations

    property rule, chain rule, quotient rule,

    3. The attempt at a solution

    I can do this the long way, but there must be a shorter solution. Can I simplify it? I've found [tex]dy/dx=(-xy^2 +2x)/(2x^2y)[/tex]. but the derivative of that takes a ridiculous amount of steps. If there isn't a shorter way then I'll just take my time.
  2. jcsd
  3. Oct 19, 2007 #2
    this is how i would simplify


    there aren't that many steps, it only took me 3 steps, maybe you're doing something wrong, if you want show your steps and i'll be happy to help further.
  4. Oct 19, 2007 #3
    OK thanks, that simplification made the first derivative easier and pointed out to me that my original calculation had a mistake. My math skills have quite a few holes. So now I have [tex]dy/dx= (-xy^2+1)/(x^2y)[/tex] so is there another simplification I'm overlooking? If I differentiate that it's still a bit messy and I'm learning there is always an easier way... well sometimes.
  5. Oct 19, 2007 #4


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    Your answer for dy/dx in post #3 is correct. I'm not sure how you want to simplify this. You could write [tex]\frac{1-xy^2}{x^2y}=\frac{1}{x^2y}-\frac{y}{x}[/tex] but I'm not sure that this is simpler to differentiate.
  6. Oct 19, 2007 #5


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    Science Advisor

    I'm wondering what "long way" you used. Straight forward implicit differentiation does seem that long to me!

    [tex]x^2y^2- 2x= 3[/tex]
    Differentiating once:
    [tex]2xy^2+ 2x^2yy'- 2= 0[/tex]
    Differenitiating again:
    [tex]2y^2+ 4xyy'+ 4xyy'+ 2x^2(y')^2+ 2x^2yy"= 0[/tex]
    [tex]-2x^2yy"= 2y^2+ 8xyy'+ 2x^2(y')^2[/tex]
    Now divide by [itex]-2x^2y[/itex] to get
    [tex]-\frac{2y^2+ 8xyy'+ 2x^2(y')^2}{2x^2y}[/tex]

    Since the derivative will necessarily have "y" in it, I would see no reason to write y' as a function of x which I think is the "hard" part of what you did.
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